Answer:
More current will be loss through the metal wire strands if the force on them was repulsive, and more stress will be induced on the wire strands due to internal and external flexing.
Explanation:
A wire bundle is made up of wire strands bunched together to increase flexibility that is not always possible in a single solid metal wire conductor. In the strands of wire carrying a high voltage power, each strand carries a certain amount of current, and the current through the strands all travel in the same direction. It is know that for two conductors or wire, separated by a certain distance, that carries current flowing through them in the same direction, an attractive force is produced on these wires, one on the other. This effect is due to the magnetic induction of a current carrying conductor. The forces between these strands of the high voltage wire bundle, pulls the wire strands closer, creating more bond between these wire strands and reducing internal flex induced stresses.
If the case was the opposite, and the wires opposed themselves, the effect would be that a lot of cost will be expended in holding these wire strands together. Also, stress within the strands due to the repulsion, will couple with external stress from the flexing of the wire, resulting in the weakening of the material.
The biggest problem will be that more current will be lost in the wire due to increased surface area caused by the repulsive forces opening spaces between the strand. This loss is a s a result of the 'skin effect' in wire transmission, in which current tends to flow close to the surface of the metal wire. The skin effect generates power loss as heat through the exposed surface area.
When using a mercury barometer , the vapor pressure of mercury is usually assumed to be zero. At room temperature mercury's vapor pressure is about 0.0015 mm-Hg. At sea level, the height hhh of mercury in a barometer is about 760 mm.Required:a. If the vapor pressure of mercury is neglected, is the true atmospheric pressure greater or less than the value read from the barometer? b. What is the percent error? c. What is the percent error if you use a water barometer and ignore water's saturated vapor pressure at STP?
Answer:
Explanation:
(a)
The true atmospheric pressure will has more value than the reading in the barometer. If Parm is the atmospheric
pressure in the tube then the resulting vapour pressure is
Patm - pgh = Prapor
The final reading ion the barometer is
pgh = Palm - Proper
Hence, the true atmospheric pressure is greater.
you can find the answer in this book
physics principles with Applications, Global Edition Problem 67P: Chapter: CH 13 Problem:67p
An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination of the flight. The plane flies with an air speed of 120 m/s. If a constant wind blows at 10.0 m/s due west during the flight, what direction must the plane fly relative to north to arrive at the destination? Consider: east to the right, west to the left, north upwards and south downwards
Answer:
θ = 4.78º
with respect to the vertical or 4.78 to the east - north
Explanation:
This is a velocity compound exercise since it is a vector quantity.
The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed
v_fly² = v_nort² + v_air²
v_nort² = v_fly² + - v_air²
Let's use trigonometry to find the direction of the plane
sin θ = v_air / v_fly
θ = sin⁻¹ (v_air / v_fly)
let's calculate
θ = sin⁻¹ (10/120)
θ = 4.78º
with respect to the vertical or 4.78 to the north-east
Imagine you are in a small boat on a small pond that has no inflow or outflow. If you take an anchor that was sitting on the floor of the boat and lower it over the side until it sits on the ground at the bottom of the pond, will the water level rise slightly, stay the same, or lower slightly?Two students, Ian and Owen, are discussing this. Ian says that the anchor will still displace just as much water when it is sitting on the bottom of the pond as it does when it is in the boat. After all, adding the anchor to the boat causes the water level in the lake to rise, and so would immersing the anchor in the pond. So Ian reasons that both displacements would be equal, and the lake level remains unchanged.
Answer;
The pond's water level will fall.
Explanation;
Archimedes principle explains that a floating body will displace the amount of water that weighs the same as it, whereas a body resting on the bottom of the water displaces the amount of water that is equal to the body's volume.
When the anchor is in the boat it is in the category of floating body and when it is on the bottom of the pond it is in the second category.
Since anchors are naturally heavy and denser than water, the amount of water displaced when the anchor is in the boat is greater than the amount of water displaced when the anchor is on the bottom of the pond since the way anchors are doesn't make for them to have considerable volume.
When the anchor is dropped to the bottom of the pond, the water level will therefore fall. If the anchor doesn't reach the bottom it is still in the floating object category and there will be no difference to the water level, but once it touches the bottom of the pond, the water level of the pond drops.
Hope this Helps!!!
Buoyancy is an upward force exerted by a fluid on a body partially or completely immersed in it
The pond water level will lower slightly
According to Archimedes principle, the up thrust on the boat by the water is given by the volume of the water displaced
When a boat floats, the weight of the boat and all its contents and passengers is equal to the displaced water, so that larger boats with more wider opening can displace more water and therefore, carry more loadWith regards to lowering the anchor from the boat into the pond, the weight of the anchor is no longer carried by the boat but by the bottom of the pond, therefore, the weight of the boat reduces, and the boat rises, while the volume initially occupied by the boat is taken up by the water available, therefore, the water level lowers slightly
Learn more here;
https://brainly.com/question/24529607
The index of refraction for a certain type of glass is 1.645 for blue light and 1.609 for red light. A beam of white light (one that contains all colors) enters a plate of glass from the air, nair≈1, at an incidence angle of 38.55∘. What is the absolute value of ????, the angle in the glass between blue and red parts of the refracted beams?
Answer:
blue θ₂ = 22.26º
red θ₂ = 22.79º
Explanation:
When a light beam passes from one material medium to another, it undergoes a deviation from the path, described by the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the incident and transmitted media refractive indices and θ are the angles in the media
let's apply this equation to each wavelength
λ = blue
in this case n₁ = 1, n₂ = 1,645
sin θ₂ = n₁/ n₂ sin₂ θ₁
let's calculate
sin θ₂ = 1 / 1,645 sint 38.55
sin θ₂ = 0.37884
θ₂ = sin⁻¹ 0.37884
θ₂ = 22.26º
λ = red
n₂ = 1,609
sin θ₂ = 1 / 1,609 sin 38.55
sin θ₂ = 0.3873
θ₂ = sim⁻¹ 0.3873
θ₂ = 22.79º
the refracted rays are between these two angles
A 50-kg block is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 150 N. Fp is parallel to the displacement of the block. The coefficient of kinetic friction is 0.25.
a) What is the total work done on the block?
b) If the box started from rest, what is the final speed of the block?
Answer:
a) WT = 137.5 J
b) v2 = 2.34 m/s
Explanation:
a) The total work done on the block is given by the following formula:
[tex]W_T=F_pd-F_fd=(F_p-F_f)d[/tex] (1)
Fp: force parallel to the displacement of the block = 150N
Ff: friction force
d: distance = 5.0 m
Then, you first calculate the friction force by using the following relation:
[tex]F_f=\mu_k N=\mu_k Mg[/tex] (2)
μk: coefficient of kinetic friction = 0.25
M: mass of the block = 50kg
g: gravitational constant = 9.8 m/s^2
Next, you replace the equation (2) into the equation (1) and solve for WT:
[tex]W_T=(F_p-\mu_kMg)d=(150N-(0.25)(50kg)(9.8m/s^2))(5.0m)\\\\W_T=137.5J[/tex]
The work done over the block is 137.5 J
b) If the block started from rest, you can use the following equation to calculate the final speed of the block:
[tex]W_T=\Delta K=\frac{1}{2}M(v_2^2-v_1^2)[/tex] (3)
WT: total work = 137.5 J
v2: final speed = ?
v1: initial speed of the block = 0m/s
You solve the equation (3) for v2:
[tex]v_2=\sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(137.5J)}{50kg}}=2.34\frac{m}{s}[/tex]
The final speed of the block is 2.34 m/s
In a contest, two tractors pull two identical blocks of stone thesame distance over identical surfaces. However, block A is moving twice as fast as block B when it crosses the finish line. Which statement is correct?a) Block A has twiceas much kinetic energy as block B.b) Block B has losttwice as much kinetic energy to friction as block A.c) Block B has losttwice as much kinetic energy as block A.d) Both blocks havehad equal losses of energy to friction.e) No energy is lostto friction because the ground has no displacement.
Answer:
d) Both blocks have had equal losses of energy to friction
Explanation:
As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces
Moreover, the block A is twice as fast than block B at the time of crossing the finish line
So based on the above information, it contains the losses of identical friction
And we also know that
Friction energy loss is
[tex]= \mu \times m \times g \times D[/tex]
It would be the same for both the blocks
hence, the option d is correct
The correct answer will be both blocks have had equal losses of energy to friction.
What is friction?Friction is defined as when any object is slides on a surface by means of any external force then the force in the opposite direction generated between the surface and the body restrict the motion of the body this force is called as the friction.
As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces.
Moreover, the block A is twice as fast as block B at the time of crossing the finish line.
So based on the above information, it contains the losses of identical friction.
And we also know that
Friction energy loss is
[tex]E_f=\mu m g D[/tex]
It would be the same for both the blocks
Hence both blocks have had equal losses of energy to friction.
To know more about friction, follow
https://brainly.com/question/24386803
The force a spring exerts on a body is a conservative force because:
a. a spring always exerts a force parallel to the displacement of the body.
b. the work a spring does on a body is equal for compressions and extensions of equal magnitude.
c. the net work a spring does on a body is zero when the body returns to its initial position.
d. the work a spring does on a body is equal and opposite for compressions and extensions of equal magnitude.
e. a spring always exerts a force opposite to the displacement of the body.
Answer:
c. the net work a spring does on a body is zero when the body returns to its initial position
Explanation:
A force is conservative when the net work done over any path that returns to the initial position is zero. Choice C matches that definition.
An ideal spring of the kind used in physics problems has the characteristic that it applies the same force at the same distance always. So any work required to extend or compress the spring is reversed when the reverse motion takes place.
A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x
Answer:
If the particle is an electron [tex]E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton, [tex]E_y = 6.08 * 10^6 N/C[/tex]
Explanation:
Initial speed at the origin, [tex]u = 3 * 10^6 m/s[/tex]
[tex]\theta = 38^0[/tex] to +ve x-axis
The particle crosses the x-axis at , x = 1.5 cm = 0.015 m
The particle can either be an electron or a proton:
Mass of an electron, [tex]m_e = 9.1 * 10^{-31} kg[/tex]
Mass of a proton, [tex]m_p = 1.67 * 10^{-27} kg[/tex]
The electric field intensity along the positive y axis [tex]E_y[/tex], can be given by the formula:
[tex]E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\[/tex]
If the particle is an electron:
[tex]E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton:
[tex]E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 6.08 * 10^6 N/C[/tex]
when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.2525 N. What were the initial charges on the spheres
Answer:
q1 = 7.6uC , -2.3 uC
q2 = 7.6uC , -2.3 uC
( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )
Explanation:
Solution:-
- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.
- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:
[tex]F = k\frac{|q_1|.|q_2|}{r^2}[/tex]
Where,
k: The coulomb's constant = 8.99*10^9
- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.
- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.
- Therefore, the force of attraction between the spheres would be:
[tex]\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1[/tex] ... Eq 1
- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).
- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,
[tex]q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}[/tex]
- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:
[tex]\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6[/tex] .. Eq2
- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:
[tex]-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123[/tex]
[tex]q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\[/tex]
A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnitude of the impulse exerted on the ball by the floor
Answer:
9.6 Ns
Explanation:
Note: From newton's second law of motion,
Impulse = change in momentum
I = m(v-u).................. Equation 1
Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.
Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)
Substitute into equation 1
I = 2.4[2.5-(-1.5)]
I = 2.4(2.5+1.5)
I = 2.4(4)
I = 9.6 Ns
The magnitude of impulse will be "9.6 Ns".
According to the question,
Mass,
m = 2.4 kgFinal velocity,
v = 2.5 m/sInitial velocity,
u = -1.5 m/sBy using Newton's 2nd law of motion, we get
→ Impulse, [tex]I = m(v-u)[/tex]
By substituting the values, we get
[tex]= 2.4[2.5-(1.5)][/tex]
[tex]= 2.4(2.5+1.5)[/tex]
[tex]= 2.4\times 4[/tex]
[tex]= 9.6 \ Ns[/tex]
Thus the above answer is right.
Learn more about Impulse here:
https://brainly.com/question/15495020
A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m before coming to rest, determine the coefficient of kinetic friction between the puck and ice.
Answer:
μ_k = 0.1773
Explanation:
We are given;
Initial velocity;u = 20 m/s
Final velocity;v = 0 m/s (since it comes to rest)
Distance before coming to rest;s = 115 m
Let's find the acceleration using Newton's second law of motion;
v² = u² + 2as
Making a the subject, we have;
a = (v² - u²)/2s
Plugging relevant values;
a = (0² - 20²)/(2 × 115)
a = -400/230
a = -1.739 m/s²
From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:
F_k = −ma - - - (1)
We also know that F_k is defined by;
F_k = μ_k•N
Where;
μ_k is coefficient of kinetic friction
N is normal force which is (mg)
Since gravity acts in the negative direction, the normal force will be positive.
Thus;
F_k = μ_k•mg - - - (2)
where g is acceleration due to gravity.
Thus,equating equation 1 and 2,we have;
−ma = μ_k•mg
m will cancel out to give;
-a = μ_k•g
μ_k = -a/g
g has a constant value of 9.81 m/s², so;
μ_k = - (-1.739/9.81)
μ_k = 0.1773
The coefficient of kinetic friction between the hockey puck and ice is equal to 0.178
Given the following data:
Initial speed = 20 m/sFinal velocity = 0 m/s (since it came to rest)Distance = 115 mScientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]To determine the coefficient of kinetic friction between the hockey puck and ice:
First of all, we would calculate the acceleration of the hockey puck by using the third equation of motion.
[tex]V^2 = U^2 + 2aS\\\\0^2 =20^2 + 2a(115)\\\\-400=230a\\\\a=\frac{-400}{230}[/tex]
Acceleration, a = -1.74 [tex]m/s^2[/tex]
Note: The negative signs indicates that the hockey puck is slowing down or decelerating.
From Newton's Second Law of Motion, we have:
[tex]\sum F_x = F_k + F_n =0\\\\F_k =- F_n\\\\\mu mg =-ma\\\\\mu = \frac{-a}{g}\\\\\mu = \frac{-(-1.74)}{9.8}\\\\\mu = \frac{1.74}{9.8}[/tex]
Coefficient of kinetic friction = 0.178
Read more: https://brainly.com/question/13821217
Which three terms are needed to describe the energy a BASE jumper has as
she falls toward the ground?
O A. Potential
B. Electromagnetic
C. Gravitational
D. Kinetic
Answer:
I’m saying kinetic gravitational and electromagnetic and I will comment on this if I got it right
Explanation:.
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the
Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}[/tex]
k = 1.4
[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K[/tex]
Work done is given as;
[tex]W = \frac{1}{2} *m*(v_i^2 - v_e^2)[/tex]
inlet velocity is negligible;
[tex]v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s[/tex]
Therefore, the exit velocity is 629.41 m/s
what is the most likely elevation of point x?
A. 150 ft
B. 200 ft
C. 125 ft
D. 250 ft
A high-jumper clears the bar and has a downward velocity of - 5.00 m/s just before landing on an air mattress and bouncing up at 1.0 m/s. The mass of the high-jumper is 60.0 kg. What is the magnitude and direction of the impulse that the air mattress exerts on her
-- As she lands on the air mattress, her momentum is (m v)
Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down
-- As she leaves it after the bounce,
Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up
-- The impulse (change in momentum) is
Change = (60 kg-m/s up) - (300 kg-m/s down)
Magnitude of the change = 360 km-m/s
The direction of the change is up /\ .
The direction of a body or object's movement is defined by its velocity.In its basic form, speed is a scalar quantity.In essence, velocity is a vector quantity.It is the speed at which distance changes.It is the displacement change rate.
Solve the problem ?
Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is travelling along a path.In other words, velocity is a vector, whereas speed is a scalar value. We discuss the conceptive impulse in this puzzle.A high jumper weighing 60.0 kg sprints on the matrix at minus 6 meters per second in the downhill direction before falling to the mattress.her admirer.Speed drops to 0 meters/second.We must determine the impulse's size and presumed direction, which is upward and positive.The change in momentum is then equal to the impulse.The impulse therefore equals m times.the end velocity less the starting velocity.60.0kg times 0 minus minus 6 meters per second is the impulse, therefore.The impulse is 360 kilogram meters per second, or 360 newtons, to put it another way.The second is upward, and the direction.To learn more about magnitude refer
https://brainly.com/question/24468862
#SPJ2
g A mass of 2 kg is attached to a spring whose constant is 7 N/m. The mass is initially released from a point 4 m above the equilibrium position with a downward velocity of 10 m/s, and the subsequent motion takes place in a medium that imparts a damping force numerically equal to 10 times the instantaneous velocity. What is the differential equation for the mass-spring system.
Answer:
mass 20 times of an amazing and all its motion
A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 430 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 5,200 rev/min.
Required:
a. Find the kinetic energy stored in the flywheel.
b. If the flywheel is to supply energy to the car as would a 15.0-hp motor, find the length of time the car could run before the flywheel would have to be brought back up to speed.
Answer:
a
[tex]KE = 7.17 *10^{7} \ J[/tex]
b
[tex]t = 6411.09 \ s[/tex]
Explanation:
From the question we are told that
The radius of the flywheel is [tex]r = 1.50 \ m[/tex]
The mass of the flywheel is [tex]m = 430 \ kg[/tex]
The rotational speed of the flywheel is [tex]w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec[/tex]
The power supplied by the motor is [tex]P = 15.0 hp = 15 * 746 = 11190 \ W[/tex]
Generally the moment of inertia of the flywheel is mathematically represented as
[tex]I = \frac{1}{2} mr^2[/tex]
substituting values
[tex]I = \frac{1}{2} ( 430)(1.50)^2[/tex]
[tex]I = 483.75 \ kgm^2[/tex]
The kinetic energy that is been stored is
[tex]KE = \frac{1}{2} * I * w^2[/tex]
substituting values
[tex]KE = \frac{1}{2} * 483.75 * (544.61)^2[/tex]
[tex]KE = 7.17 *10^{7} \ J[/tex]
Generally power is mathematically represented as
[tex]P = \frac{KE}{t}[/tex]
=> [tex]t = \frac{KE}{P}[/tex]
substituting the value
[tex]t = \frac{7.17 *10^{7}}{11190}[/tex]
[tex]t = 6411.09 \ s[/tex]
Which circuits are parallel circuits?
Answer:
The bottom two lines.
Explanation:
They need their own line of voltage quantity. A parallel circuit has the definition of 'two or more paths for current to flow through.' The voltage does stay the same in each line.
Assume you have a rocket in Earth orbit and want to go to Mars. The required change in velocity is ΔV≈9.6km/s . There are two options for the propulsion system --- chemical and electric --- each with a different specific impulse. Recall that the relationship between specific impulse and exhaust velocity is: Vex=g0Isp Using the Ideal Rocket Equation and setting g0=9.81m/s2 , calculate the propellant fraction required to achieve the necessary ΔV for each of propulsion system. Part 1: Cryogenic Chemical Propulsion First, consider a cryogenic chemical propulsion system with Isp≈450s . Enter the required propellant fraction as a proportion with at least 2 decimal places (i.e., enter 0.25 to represent 25%): incorrect Part 2: Electric Propulsion Next, consider an electric propulsion system with Isp≈2000s . Enter the required propellant fraction as a proportion with at least 2 decimal places (i.e., enter 0.25 to represent 25%):
Answer: Part 1: Propellant Fraction (MR) = 8.76
Part 2: Propellant Fraction (MR) = 1.63
Explanation: The Ideal Rocket Equation is given by:
Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]
Where:
[tex]v_{ex}[/tex] is relationship between exhaust velocity and specific impulse
[tex]\frac{m_{f}}{m_{e}}[/tex] is the porpellant fraction, also written as MR.
The relationship [tex]v_{ex}[/tex] is: [tex]v_{ex} = g_{0}.Isp[/tex]
To determine the fraction:
Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]
[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]
Knowing that change in velocity is Δv = 9.6km/s and [tex]g_{0}[/tex] = 9.81m/s²
Note: Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.
Part 1: Isp = 450s
[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]
ln(MR) = [tex]\frac{9.6.10^{3}}{9.81.450}[/tex]
ln (MR) = 2.17
MR = [tex]e^{2.17}[/tex]
MR = 8.76
Part 2: Isp = 2000s
[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]
ln (MR) = [tex]\frac{9.6.10^{3}}{9.81.2.10^{3}}[/tex]
ln (MR) = 0.49
MR = [tex]e^{0.49}[/tex]
MR = 1.63
Jack and Jill went up the hill to fetch a pail of water. Jack, who’s mass is 75 kg, 1.5 times heavier than Jill’s mass, fell down and broke his crown after climbing a 15 m high hill. Jillcame tumbling after covering the same distance as Jack in 1/3rd of the time.Required:a. Who did the most work climbing up the hill? b. Who applied the most power?
Answer:
a) Jack does more work uphill
b) Numerically, we can see that Jill applied the most power downhill
Explanation:
Jack's mass = 75 kg
Jill's mass = [tex]1.5x = 75[/tex]
Jill's mass = [tex]x = \frac{75}{1.5}[/tex] = 50 kg
distance up hill = 15 m
a) work done by Jack uphill = mgh
where g = acceleration due to gravity= 9.81 m/s^2
work = 75 x 9.81 x 15 = 11036.25 J
similarly,
Jill's work uphill = 50 x 9.81 x 15 = 7357.5 J
this shows that Jack does more work climbing up the hill
b) assuming Jack's time downhill to be t,
then Jill's time = [tex]\frac{t}{3}[/tex]
we recall that power is the rate in which work id done, i.e
P = [tex]\frac{work}{time}[/tex]
For Jack, power = [tex]\frac{11036.25}{t}[/tex]
For Jill, power = [tex]\frac{3*7357.5}{t}[/tex] = [tex]\frac{22072.5}{t}[/tex]
Numerically, we can see that Jill applied the most power downhill
Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box of mass M. There is friction between the surfaces of blocks 2M and 3M so the three blocks accelerate together to the right.
Which block has the smallest net force acting on it?
A) M
B) 2M
C) 3M
D) The net force is the same for all three blocks Submit
Answer:
A) M
Explanation:
The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:
Box with mass M
[tex]\Sigma F = F - F' = M\cdot a[/tex]
Box with mass 2M
[tex]\Sigma F = F' - F'' = 2\cdot M \cdot a[/tex]
Box with mass 3M
[tex]\Sigma F = F'' = 3\cdot M \cdot a[/tex]
On the third equation, acceleration can be modelled in terms of F'':
[tex]a = \frac{F''}{3\cdot M}[/tex]
An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.
[tex]F' = 2\cdot M \cdot a + F''[/tex]
[tex]F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''[/tex]
[tex]F' = \frac{5}{3}\cdot F''[/tex]
Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:
[tex]F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)[/tex]
[tex]F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''[/tex]
[tex]F = 2\cdot F''[/tex]
[tex]F'' = \frac{1}{2}\cdot F[/tex]
Afterwards, F' as function of the external force can be obtained by direct substitution:
[tex]F' = \frac{5}{6}\cdot F[/tex]
The net forces of each block are now calculated:
Box with mass M
[tex]M\cdot a = F - \frac{5}{6}\cdot F[/tex]
[tex]M\cdot a = \frac{1}{6}\cdot F[/tex]
Box with mass 2M
[tex]2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F[/tex]
[tex]2\cdot M \cdot a = \frac{1}{3}\cdot F[/tex]
Box with mass 3M
[tex]3\cdot M \cdot a = \frac{1}{2}\cdot F[/tex]
As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.
A skydiver stepped out of an airplane at an altitude of 1000m fell freely for 5.00s opened her parachute and slowed to 7.00m/s in a negligible time what was the total elapsed time from leaving the airplane to landing on the ground
Answer:
t = 17.68s
Explanation:
In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:
[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex] (1)
y: height for a time t
yo: initial height = 1000m
vo: initial velocity = 0m/s
g: gravitational acceleration = 9.8m/s^2
t: time = 5.00 s
You replace the values of the parameters to get the values of the new height of the skydiver:
[tex]y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m[/tex]
Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.
You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0
[tex]0=877.5-7.00t-4.9t^2[/tex] (2)
The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:
[tex]t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s[/tex]
You use the positive value of t1 because it has physical meaning.
Finally, you sum the times of both parts of the trajectory:
total time = 5.00s + 12.68s = 17.68s
The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s
You are trying to overhear a juicy conversation, but from your distance of 25.0 m, it sounds like only an average whisper of 25.0 dB. So you decide to move closer to give the conversation a sound level of 80.0 dB instead. How close should you come?
Answer:
r₂ = 1,586 m
Explanation:
For this problem we are going to solve it by parts, let's start by finding the sound intensity when we are 25 m
β = 10 log (I / I₀)
where Io is the sensitivity threshold 10⁻¹² W / m²
I₁ / I₀ = [tex]e^{\beta/10}[/tex]
I₁ = I₀ e^{\beta/10}
let's calculate
I₁ = 10⁻¹² e^{25/10}
I₁ = 1.20 10⁻¹¹ W / m²
the other intensity in exercise is
I₂ = 10⁻¹² e^{80/10}
I₂ = 2.98 10⁻⁹ W / m²
now we use the definition of sound intensity
I = P / A
where P is the emitted power that is a constant and A the area of the sphere where the sound is distributed
P = I A
the area a sphere is
A = 4π r²
we can write this equation for two points of the found intensities
I₁ A₁ = I₂ A₂
where index 1 corresponds to 25m and index 2 to the other distance
I₁ 4π r₁² = I₂ 4π r₂²
I₁ r₁² = I₂ r₂²
r₂ = √ (I₁ / I₂) r₁
let's calculate
r₂ = √ (1.20 10⁻¹¹ / 2.98 10⁻⁹) 25
r₂ = √ (0.40268 10⁻²) 25
r₂ = 1,586 m
That 85 kg paratrooper from the 50's was moving at constant speed of 56 m/s because the air was applying a frictional drag force to him that matched his weight. If he fell this way for 40 m, how much heat was generated by this frictional drag force in J
Answer:
46648 J
Explanation:
mass m= 85 Kg
velocity v = 56 m/s
distance covered s =40 m
According to Question,
frictional drag force to him that matched his weight
[tex]\Rightarrow F_d =mg\\=85\times9.81=833 N[/tex]
Therefore, work done by practometer against the drag force = heat was generated by this frictional drag force in J
W=Q= F_d×s
=833×56 = 46648 J
A depiction of a famous scientific experiment is given. Consider how the beam changes when the magnet is off compared to when the magnet is on. A bell-shaped evacuated glass tube with a narrow end and a wide end is connected to a battery at the narrow end. In the center of the tube there is a negatively charged plate above the tube, a positively charged plate below the tube, and a magnet with the field turned off. A beam originating at the narrow end of the tube travels toward the wide end of the tube. With the magnetic field turned off, the beam path bends toward the positively charged plate and ends at the lower half of the wide end of the tube. A bell-shaped evacuated glass tube with a narrow end and a wide end is connected to a battery at the narrow end. In the center of the tube there is a negatively charged plate above the tube, a positively charged plate below the tube, and a magnet with the field turned n. A beam originating at the narrow end of the tube travels toward the wide end of the tube. With the magnetic field turned on, the beam path travels in a straight path to the center of the wide end of the tube. What type of beam was used in this experiment?
Answer:
The beam used is a negatively charged electron beam with a velocity of
v = E / B
Explanation:
After reading this long statement we can extract the data to work on the problem.
* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.
* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced
[tex]F_{e} = F_{m}[/tex]
q E = qv B
v = E / B
this configuration is called speed selector
They ask us what type of beam was used.
The beam used is a negatively charged electron beam with a velocity of v = E / B
Two red blood cells each have a mass of 9.0 x 10-14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion from the excess charge prevents the cells from clumping together. One cell carries -2.5pC and the other -3.30 pC, and each cell can be modeled as a sphere 3.75 × 10-6 m in radius. If the red blood cells start very far apart and move directly toward each other with the same speed.
1. What initial speed would each need so that they get close enough to just barely touch?
2. What is the maximum acceleration of the cells as they move toward each other and just barely touch?
Answer:
Explanation:
Given that:
The mass of the cell is 9.0 x 10^-14 kg
The charges of the cell is -2.5pC and the other -3.30 pC
[tex]q_1=-2.5\times10^{-12}C \ \ and \ \ q_2=-3.75\times10^{-12}C[/tex]
Radius is 3.75 × 10-6 m
The final distance is twice the radius
i.e [tex]2*(3.75 \times 10^{-6}) = 7.5*10^{-6}m[/tex]
The formula for the velocity of the cell is
[tex]mv^2=\frac{q_1q_2}{4\pi \epsilon 2 r} \\[/tex]
[tex]v=\sqrt{\frac{q_1q_2}{4\pi \epsilon 2 r} }[/tex]
[tex]=\sqrt{\frac{(-2.5\times10^{-12})(-3.3\times10^{-12}}{4(3.14)(8.85\times10^{-112}(2\times3.75\times10^{-6})(9\times10^{-14})} } \\\\=\sqrt{\frac{(-8.25\times10^{-24})}{(7503.03\times10^{-32})} } \\\\=\sqrt{109955.5779} \\\\=331.60m/s[/tex]
The maximum acceleration of the cells as they move toward each other and just barely touch is
[tex]ma= \frac{q_1q_2}{4\pi \epsilon (2r)^2} \\\\a= \frac{q_1q_2}{4\pi \epsilon (2r)^2(m)}[/tex]
[tex]=\frac{(-2.5\times10^{-12})(-3.3\times10^{-12})}{4(3.14)(8.85\times10^{-12})(2\times3.75\times10^{-6})^2(9\times10^{-14})}[/tex]
[tex]=\frac{(-8.25\times10^{-24})}{(56272.725\times10^{-38})} \\\\=1.47\times10^{10}m/s^2[/tex]
The answers obtained are;
1. The initial speed of each of the red blood cells is [tex]v= 331.66\,m/s[/tex].
2. The maximum acceleration of the cells is [tex]a=1.47\times 10^{10}\,m/s^2[/tex].
The answer is explained as shown below.
We have, the mass of the red blood cell;
[tex]m=9\times 10^{-14}\,kg[/tex]Also, the charges of the cells are;
[tex]q_1=-2.5\times 10^{-12}\,C[/tex] and[tex]q_2=-3.30\times 10^{-12}\,C[/tex]The distance between the charges when they barely touch will be two times the radius of each charge.
[tex]r=2\times r\,'=2\times3.75\times10^{-6}\,m=7.5\times10^{-6}\,m[/tex]Kinetic Energy of moving charges1. As both the cells are negatively charged they will repel each other.
So, for the cells to come nearly close, their kinetic energies must be equal to the electric potential between them.[tex]\frac{1}{2}mv^2+ \frac{1}{2}mv^2=k\frac{q_1 q_2}{r^2}[/tex]Where, [tex]k=9\times10^9\,Nm^2/C^2[/tex] is the Coulomb's constant.Now, substituting all the known values in the equation, we get;
[tex](9\times 10^{-14}\,kg)\times v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m}[/tex][tex]v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m\times(9\times 10^{-14}\,kg)} =110000\,m^2/s^2[/tex]
[tex]\implies v=\sqrt{110000\,m^2/s^2}=331.66\,m/s[/tex]Electrostatic force between two charges2. Also as the force between them is repulsive, there must be an acceleration to make them barely touch each other.
[tex]ma=k\frac{q_1 q_2}{r^2}[/tex]Substituting the known values, we get;
[tex](9\times 10^{-14}\,kg)\times a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2}[/tex]
[tex]\implies a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2\times(9\times 10^{-14}\,kg) }[/tex]
[tex]a=1.47\times 10^{10}\,m/s^2[/tex]Find out more information about moving charges here:
https://brainly.com/question/14632877
Which one of the following is closely related to the law of conservation of
energy, which states that energy can be transformed in different ways but can
never be created or destroyed?
O A. Charles's Law
B. Boyle's Law
C. Second law of thermodynamics
O D. First law of thermodynamics
Answer:
D
Explanation:
Answer:
It is D
Explanation: No cap
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands
Complete question is;
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.98 m long that pivots freely about the hip.
Answer:
Tangential speed of foot just before it lands is; v = 5.37m/s
Explanation:
Let U (potential energy) be zero on the ground.
So, initially, U = mgh
where, h = 0.98/2 = 0.49m (midpoint of the leg)
Now just before the leg hits the floor it would have kinetic energy as;
K = ½Iω²
where ω = v/r and I = ⅓mr²
So, K = ½(⅓mr²)(v/r)²
K = (1/6) × (mr²)/(v²/r²)
K = (1/6) × mv²
From principle of conservation of energy, we have;
Potential energy = Kinetic energy
Thus;
mgh = (1/6) × mv²
m will cancel out to give;
gh = (1/6)v²
Making v the subject, we have;
v = √6gh
v = √(6 × 9.81 × 0.49)
v = √28.8414
v = 5.37m/s
A 300 g bird flying along at 6.2 m/s sees a 10 g insect heading straight toward it with a speed of 35 m/s (as measured by an observer on the ground, not by the bird). The bird opens its mouth wide and enjoys a nice lunch.
Required:
What is the bird's speed immediately after swallowing?
Answer:
The velocity of the bird is [tex]v_f = 4.87 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the bird is [tex]m_1 = 300 \ g = 0.3 \ kg[/tex]
The initial speed of the bird is [tex]u_1 = 6.2 \ m/s[/tex]
The mass of the insect is [tex]m_2 = 10 \ g = 0.01 \ kg[/tex]
The speed of the insect is [tex]u_ 2 =-35 \ m/s[/tex]
The negative sign is because it is moving in opposite direction to the bird
According to the principle of linear momentum conservation
[tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2 )v_f[/tex]
substituting values
[tex](0.3 * 6.2 ) + (0.01 * (-35)) = (0.3 + 0.01 )v_f[/tex]
[tex]1.51 = 0.31 v_f[/tex]
[tex]v_f = 4.87 \ m/s[/tex]
The Final velocity of Bird = 4.87 m/s
Mass of the bird = 300 g = 0.3 kg
Velocity of bird = 6.2 m/s
Momentum of Bird = Mass of bird [tex]\times[/tex] Velocity of Bird = 0.3 [tex]\times[/tex] 6.2 = 1.86 kgm/s
Mass of the insect = 10 g = 0.01 kg
Velocity of insect = - 35 m/s
Momentum of the Insect = Mass of Insect [tex]\times[/tex] Velocity of Insect = - 0.35 kgm/s
According to the law of conservation of momentum We can write that
In the absence of external forces on the system , the momentum of system remains conserved in that particular direction.
The bird opens the mouth and enjoys the free lunch hence
Let the final velocity of bird is [tex]v_f[/tex]
Initial momentum of the system = Final momentum of the system
1.86 -0.35 = [tex]v_f[/tex] ( 0.01 + 0.3 )
1.51 = [tex]v_f[/tex] 0.31
[tex]v_f[/tex] = 4.87 m/s
The Final velocity of Bird = 4.87 m/s
For more information please refer to the link below
https://brainly.com/question/18066930
In cricket how bowler and batsman use acceleration?