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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
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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]
An electron moves at a speed of 1.0 x 104 m/s in a circular path of radius 2 cm inside a solenoid. The magnetic field of the solenoid is perpendicular to the plane of the electron’s path. Calculate (a) the strength of the magnetic field inside the solenoid and (b) the current in the solenoid if it has 25 turns per centimeter.
Answer:
(a) B = 2.85 × [tex]10^{-6}[/tex] Tesla
(b) I = I = 0.285 A
Explanation:
a. The strength of magnetic field, B, in a solenoid is determined by;
r = [tex]\frac{mv}{qB}[/tex]
⇒ B = [tex]\frac{mv}{qr}[/tex]
Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field
B = [tex]\frac{9.11*10^{-31*1.0*10^{4} } }{1.6*10^{-19}*0.02 }[/tex]
= [tex]\frac{9.11*10^{-27} }{3.2*10^{-21} }[/tex]
B = 2.85 × [tex]10^{-6}[/tex] Tesla
b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;
B = μ I N/L
⇒ I = B ÷ μN/L
where B is the magnetic field, μ is the permeability of free space = 4.0 ×[tex]10^{-7}[/tex]Tm/A, N/L is the number of turns per length.
I = B ÷ μN/L
= [tex]\frac{2.85*10^{-6} }{4*10^{-7} *25}[/tex]
I = 0.285 A
A rock falls from a vertical cliff that is 4.0 m tall and experiences no significant air resistance as it falls. At what speed will its gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy
Answer:
About 6.26m/s
Explanation:
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
Divide both sides by mass:
[tex]gh=\dfrac{1}{2}v^2[/tex]
Since the point of equality of kinetic and potential energy will be halfway down the cliff, height will be 4/2=2 meters.
[tex](9.8)(2)=\dfrac{1}{2}v^2 \\\\v^2=39.4 \\\\v\approx 6.26m/s[/tex]
Hope this helps!
The gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.
Given data:
The height of vertical cliff is, h = 4.0 m.
Since, we are asked for speed by giving the condition for gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy. Then we can apply the conservation of energy as,
Kinetic energy = Gravitational potential energy
[tex]\dfrac{1}{2}mv^{2}=mgh[/tex]
Here,
m is the mass of rock.
v is the speed of rock.
g is the gravitational acceleration.
Solving as,
[tex]v=\sqrt{2gh}\\\\v=\sqrt{2 \times 9.8 \times 4.0}\\\\v =8.85 \;\rm m/s[/tex]
Thus, we can conclude that the gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.
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In cricket how bowler and batsman use acceleration?
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.
An ideal, or Carnot, heat pump is used to heat a house to a temperature of 294 K (21 oC). How much work must the pump do to deliver 3000 J of heat into the house (a) on a day when the outdoor temperature is 273 K (0 oC) and (b) on another day when the outdoor temperature is 252 K (-21 oC)
Answer:
a) [tex]W_{in} = 214.286\,J[/tex], b) [tex]W_{in} = 428.571\,J[/tex]
Explanation:
a) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:
[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]
Where:
[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.
[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.
Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 273\,K[/tex]. The Coefficient of Performance is:
[tex]COP_{HP} = \frac{294\,K}{294\,K-273\,K}[/tex]
[tex]COP_{HP} = 14[/tex]
Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.
[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]
The input work to deliver a determined amount of heat to the house:
[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]
If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 14[/tex], the input work that is needed is:
[tex]W_{in} = \frac{3000\,J}{14}[/tex]
[tex]W_{in} = 214.286\,J[/tex]
b) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:
[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]
Where:
[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.
[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.
Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 252\,K[/tex]. The Coefficient of Performance is:
[tex]COP_{HP} = \frac{294\,K}{294\,K-252\,K}[/tex]
[tex]COP_{HP} = 7[/tex]
Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.
[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]
The input work to deliver a determined amount of heat to the house:
[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]
If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 7[/tex], the input work that is needed is:
[tex]W_{in} = \frac{3000\,J}{7}[/tex]
[tex]W_{in} = 428.571\,J[/tex]
New evidence increasingly emphasizes that __________.
A bag is gently pushed off the top of a wall at A and swings in a vertical plane at the end of a rope of length l. Determine the angle θ for which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the bag.
Answer:
Dear user,
Answer to your query is provided below
The angle for which the rope will break θ = 41.8°
Explanation:
Explanation of the same is attached in image
A bag is gently pushed off the top of a wall at A and swings in a vertical plane at the end of a rope of length l. The angle θ for which the rope will break, is 41.81°
What is tension?The tension is a kind of force which acts on linear objects when subjected to pull.
The maximum tension Tmax =2W
From the work energy principle,
T₂ = 1/2 mv²
Total energy before and after pushing off
0+mglsinθ = 1/2 mv²
v² = 2gflsinθ..............(1)
From the equilibrium of forces, we have
T= ma +mgsinθ = mv²/l +mgsinθ
2mg = mv²/l +mgsinθ
2g = v²/l +gsinθ
Substitute the value of v² ,we get the expression for θ
θ = sin⁻¹(2/3)
θ =41.81°
Hence, the angle θ for which the rope will break, is 41.81°
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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
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The uniform dresser has a weight of 91 lb and rests on a tile floor for which μs = 0.25. If the man pushes on it in the horizontal direction θ = 0∘, determine the smallest magnitude of force F needed to move the dresser. Also, if the man has a weight of 151 lb , determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip.
Answer:
F = 22.75 lb
μ₁ = 0.15
Explanation:
The smallest force required to move the dresser must be equal to the force of friction between the man and the dresser. Therefore,
F = μR
F = μW
where,
F = Smallest force needed to move dresser = ?
μ = coefficient of static friction = 0.25
W = Weight of dresser = 91 lb
Therefore,
F = (0.25)(91 lb)
F = 22.75 lb
Now, for the coefficient of static friction between shoes and floor, we use the same formula but with the mas of the man:
F = μ₁W₁
where,
μ₁ = coefficient of static friction between shoes and floor
W₁ = Weight of man = 151 lb
Therefore,
22.75 lb = μ₁ (151 lb)
μ₁ = 22.75 lb/151 lb
μ₁ = 0.15
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
A car traveling with velocity v is decelerated by a constant acceleration of magnitude a. It takes a time t to come to rest. If its initial velocity were doubled, the time required to stop would
Answer:
If the initial speed is doubled the time is also doubled
Explanation:
You have that a car with velocity v is decelerated by a constant acceleration a in a time t.
You use the following equation to establish the previous situation:
[tex]v'=v-at[/tex] (1)
v': final speed of the car = 0m/s
v: initial speed of the car
From the equation (1) you solve for t and obtain:
[tex]t=\frac{v-v'}{a}=\frac{v}{a}[/tex] (2)
To find the new time that car takesto stop with the new initial velocity you use again the equation (1), as follow:
[tex]v'=v_1-at'[/tex] (3)
v' = 0m/s
v1: new initial speed = 2v
t': new time
You solve the equation (3) for t':
[tex]0=2v-at'\\\\t'=\frac{2v}{a}=2t[/tex]
If the initial speed is doubled the time is also doubled
when a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N.m^2/c. when the paper is turned 25 degree with respect to the field the flux through it is:
Answer:
22.66Nm²/C
Explanation:
Flux through an electric field is expressed as ϕ = EAcosθ
When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N.m^2/c. If the paper is turned 25 degree with respect to the field the flux through it can be calculated using the formula.
From the formula above where:
EA = 25N.m^2/C
θ = 25°
ϕ = 25cos 25°
ϕ = 22.66Nm²/C
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
An electron moving in a direction perpendicular to a uniform magnetic field at a speed of 1.6 107 m/s undergoes an acceleration of 7.0 1016 m/s2 to the right (the positive x-direction) when its velocity is upward (the positive y-direction). Determine the magnitude and direction of the field.
Answer:
B = 0.024T positive z-direction
Explanation:
In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.
The magnitude of the magnetic force exerted on the electron is given by the following formula:
[tex]F=qvB[/tex] (1)
q: charge of the electron = 1.6*10^-19 C
v: speed of the electron = 1.6*10^7 m/s
B: magnitude of the magnetic field = ?
By the Newton second law you also have that the magnetic force is equal to:
[tex]F=qvB=ma[/tex] (2)
m: mass of the electron = 9.1*10^-31 kg
a: acceleration of the electron = 7.0*10^16 m/s^2
You solve for B from the equation (2):
[tex]B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T[/tex]
The direction of the magnetic field is found by using the right hand rule.
The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:
-^j X ^i = ^k
Where the minus sign of the ^j is because of the negative charge of the electron.
Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction
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
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
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
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
Your roommate is working on his bicycle and has the bike upside down. He spins the 68.0 cm -diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. A. What is the pebble's speed? B. What is the pebble's acceleration?
Answer:
a. 6.41 m/s
b. 120.85 m/s^2
Explanation:
The computation is shown below:
a. Pebble speed is
As we know that according to the tangential speed,
[tex]v = r \times \omega[/tex]
[tex]= \frac{0.68}{2} \times 18.84[/tex]
= 6.41 m/s
The 18.84 come from
[tex]= 2 \times 3.14 \times 3[/tex]
= 18.84
b. The pebble acceleration is
[tex]a = \frac{v^2}{r}[/tex]
[tex]= \frac{6.41^2}{0.34}[/tex]
= 120.85 m/s^2
We simply applied the above formulas so that the pebble speed and the pebble acceleration could come and the same is to be considered
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
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 car travels 2500 m in 8 minutes. Calculate the speed at which the car travelled
Answer:
5.95m/s to 2 decimal places
Explanation:
In physics speed is measured in metres per second so convert 8mins to seconds
8x60=420 seconds
The formula needed:
Speed (m/s)= Distance (m)/Time (s)
2500/420=5.95m/s
A force in the negative x-direction is applied for 27 ms to a 0.4 kg mass initially moving at 14 m/s in the x-direction. The force varies in magnitude and delivers an impulse with a magnitude of 32.4 N-s. What is the mass's velocity in the x-direction
Answer:
-67 m/s
Explanation:
We are given that
Mass of ball,m=0.4 kg
Initial speed,u=14 m/s
Impulse,I=-32.4 N-s
Time,t=27 ms=[tex]27\times 10^{-3} s[/tex]
We have to find the mass's velocity in the x- direction.
We know that
[tex]Impulse=mv-mu[/tex]
Substitute the values
[tex]-32.4=0.4v-0.4(14)[/tex]
[tex]-32.4+0.4(14)=0.4 v[/tex]
[tex]-26.8=0.4v[/tex]
[tex]v=\frac{-26.8}{0.4}=-67m/s[/tex]
An airplane is flying on a bearing of N 400 W at 500 mph. A strong jet-stream speed wind of 100 mph is blowing at S 500 W.
Required:
a. Find the vector representation of the plane and of the wind.
b. Find the resultant vector that represents the actual course of the plane.
c. Give the resulting speed and bearing of the plane.
Answer:
A. a (-321.393, 383.022) b (-76.40, -64.278)
B. (-397.991, 318.744)
C. a. resulting speed 509.9mph b. bearing of the plane = 51.6°
Explanation:
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.
A 3.10-mm-long, 430 kgkg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 69.0 kgkg construction worker stands at the far end of the beam.What is the magnitude of the gravitational torque about the point where the beam is bolted into place?
Answer:
Explanation:
Given that,
The length of the beam L = 3.10m
The mass of the steam beam [tex]m_1[/tex] = 430kg
The mass of worker [tex]m_2[/tex] = 69.0kg
The distance from the fixed point to centre of gravity of beam = [tex]\frac{L}{2}[/tex]
and our length of beam is 3.10m
so the distance from the fixed point to centre of gravity of beam is
[tex]\frac{3.10}{2}=1.55m[/tex]
Then the net torque is
[tex]=-W_sL'-W_wL\\\\=-(W_sL'+W_wL)[/tex]
[tex]W_s[/tex] is the weight of steel rod
[tex]=430\times9.8=4214N[/tex]
[tex]W_w[/tex] is the weight of the worker
[tex]=69\times9.8\\\\=676.2N[/tex]
Torque can now be calculated
[tex]-(4214\times1.55+676.2\times3.9)Nm\\\\-(6531.7+2637.18)Nm\\\\-(9168.88)Nm[/tex]
≅ 9169Nm
Therefore,the magnitude of the torque is 9169NmA hockey puck slides off the edge of a horizontal platform with an initial velocity of 28.0 m/shorizontally in a city where the acceleration due to gravity is 9.81 m/s 2. The puck experiences no significant air resistance as it falls. The height of the platform above the ground is 2.00 m. What is the angle below the horizontal of the velocity of the puck just before it hits the ground
Answer:
θ = 12.60°
Explanation:
In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})[/tex] (1)
θ: angle below he horizontal
vy: y component of the velocity just after the puck hits the ground
vx: x component of the velocity
The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:
[tex]v_x=v_o[/tex]
vo: initial velocity = 28.0 m/s
The y component is calculated with the following equation:
[tex]v_y^2=v_{oy}^2+2gy[/tex] (2)
voy: vertical component of the initial velocity = 0m/s
g: gravitational acceleration = 9.8 m/s^2
y: height
You solve the equation (2) for vy and replace the values of the parameters:
[tex]v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}[/tex]
Finally, you use the equation (1) to find the angle:
[tex]\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°[/tex]
The angle below the horizontal is 12.60°
The angle below the horizontal of the velocity of the puck just before it hits the ground is 12.60°.
Given the following data:
Initial velocity = 28.0 m/s Acceleration due to gravity = 9.81 [tex]m/s^2[/tex]Displacement (height) = 2.00 meters.To find the angle below the horizontal of the velocity of the puck just before it hits the ground:
First of all, we would determine the horizontal and vertical components of the hockey puck.
For horizontal component:
[tex]V_y^2 = U_y^2 + 2aS\\\\V_y^2 = 0^2 + 2(9.81)(2)\\\\V_y^2 = 39.24\\\\V_y = \sqrt{39.24} \\\\V_y = 6.26 \; m/s[/tex]
For vertical component:
[tex]V_x = U_x\\\\V_x = 28.0 \;m/s[/tex]
Now, we can find the angle by using the formula:
[tex]\Theta = tan^{-1} (\frac{V_y}{V_x} )[/tex]
Substituting the values, we have:
[tex]\Theta = tan^{-1} (\frac{6.26}{28.0} )\\\\\Theta = tan^{-1} (0.2236)\\\\\Theta = 12.60[/tex]
Angle = 12.60 degrees.
Read more: https://brainly.com/question/8898885
One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the floor. A man weighing 700 N could climb up to 7.0 m before slipping. What is the coefficient of static friction between the floor and the ladder
Answer:
μ = 0.37
Explanation:
For this exercise we must use the translational and rotational equilibrium equations.
We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive
let's write the rotational equilibrium
W₁ x/2 + W₂ x₂ - fr y = 0
where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances
cos 60 = x / L
where L is the length of the ladder
x = L cos 60
sin 60 = y / L
y = L sin60
the horizontal distance of man is
cos 60 = x2 / 7.0
x2 = 7 cos 60
we substitute
m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0
fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60
let's calculate
fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)
fr = (735 + 2450) / 8.66
fr = 367.78 N
the friction force has the expression
fr = μ N
write the translational equilibrium equation
N - W₁ -W₂ = 0
N = m₁ g + W₂
N = 30 9.8 + 700
N = 994 N
we clear the friction force from the eucacion
μ = fr / N
μ = 367.78 / 994
μ = 0.37