Answer:
The pressure is [tex]p_1 = 4051.4 \ Pa[/tex]
Explanation:
From the question we are told that
The gauge pressure at the mouth is [tex]p_1[/tex]
The radius of the column is [tex]r_2 = 4 \ mm = 0.004 \ m[/tex]
The speed of the liquid outside the body is [tex]v_2 = 3.1 \ m/s[/tex]
The area of the column is [tex]A_2[/tex]
The area inside the mouth [tex]A_1 = 10 A_2[/tex]
Generally according to continuity equation
[tex]v_1 A_1 = v_2 A_2[/tex]
=> [tex]v_ 1 = v_2 * \frac{A_2}{A_1}[/tex]
=> [tex]v_ 1 = 3.1 * \frac{1}{10}[/tex]
=> [tex]v_ 1 = 0.31 \ m/s[/tex]
So
[tex]A_1 = 10A_2[/tex]
=> [tex]\pi * r_1^2 = 10(\pi * r_2^2)[/tex]
=> [tex]r_1 = 10 * r_2[/tex]
substituting values
[tex]r_1 = 10 * 0.004[/tex]
[tex]r_1 =0.04 \ m[/tex]
Now the height of inside the mouth is [tex]h_1 = d = 2r_1 = 2* 0.04 = 0.08\ m[/tex]
Now the height of the column is [tex]h_2 = d = 2r_2 = 2* 0.004 = 0.008\ m[/tex]
Generally according to Bernoulli's equation
[tex]p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ][/tex]
Now [tex]\rho = 1000 \ kg m^{-3}[/tex] which is the density of water
[tex]p_2[/tex] is the gauge pressure of the atmosphere which is zero
So
[tex]p_1 = [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-[/tex]
[tex][(0.5 * 1000 * 0.31^2) + (0.08*1000 * 9.8)][/tex]
[tex]p_1 = 4051.4 \ Pa[/tex]
The initial pressure of the gauge pressure at the mouth is 4757 pascals.
The principle of continuity equation asserts that in a steady flow state, the quantity of fluid flowing at the inlet is equivalent to the quantity of fluid at the outlet given that there is a constant mass flow rate.
It can be expressed by using the formula:
[tex]\mathbf{A_1v_1=A_2v_2}[/tex]
where;
The speed of the liquid inside the body is [tex]\mathbf{v_1}[/tex] = ???The speed of the liquid outside the body is [tex]\mathbf{v_2}[/tex] = 3.1 m/sArea outside the column is = A₂Area inside the column is [tex]\mathbf{A_1}[/tex] = 10A₂∴
Making v₁ the subject of the formula:
[tex]\mathbf{v_1 = \dfrac{A_2}{A_1}\times v_2}[/tex]
[tex]\mathbf{v_1 = \dfrac{1}{10}\times3.1}[/tex]
[tex]\mathbf{v_1 =0.31 \ m/s}[/tex]
Since the area are equivalent to each other
[tex]\mathbf{A_1 = 10A_2}[/tex]
[tex]\mathbf{\pi r^2_1 = 10\times \pi r^2_2}[/tex]
[tex]\mathbf{ r^2_1 = 10\times r^2_2}[/tex]
where;
r₂ = radius of the column. = 4mm = 0.004 m∴
[tex]\mathbf{ r_1 = 10\times 0.004}[/tex]
[tex]\mathbf{ r^2_1 = 0.04 \ m}[/tex]
However, in fluid dynamics, Bernoulli's equation can be applied for the estimation of the initial gauge pressure by using the expression:
[tex]\mathbf{\dfrac{1}{2} \rho v_1^2 + h_1 \rho g + p_1 = \dfrac{1}{2}\rhov_2^2 + h_2 \rho g +p_2}[/tex]
Since the gauge pressure of the atmosphere [tex]\mathbf{p_2}[/tex] = 0
∴
The initial gauge pressure applied at the mouth can be determined as:
[tex]\mathbf{p_1 = \dfrac{1}{2} \rho ( v_2^2-v_1^2)}[/tex]
here;
[tex]\mathbf{\rho = 1000 \ kg/m^3}[/tex]
[tex]\mathbf{p_1 = \dfrac{1}{2} \times 1000 \ kg/m^3 ( 3.1^2-0.31^2)}[/tex]
[tex]\mathbf{p_1 =500 \ kg/m^3 (9.5139)}[/tex]
[tex]\mathbf{p_1 =4756.95 \ Pa}[/tex]
[tex]\mathbf{p_1 \simeq4757 \ Pa}[/tex]
Learn more about the principle of continuity equation here:
https://brainly.com/question/14619396
One kind of baseball pitching machine works by rotating a light and stiff rigid rod about a horizontal axis until the ball is moving toward the target. Suppose a 144 g baseball is held 82 cm from the axis of rotation and released at the major league pitching speed of 87 mph.
Required:
a. What is the ball's centripetal acceleration just before it is released?
b. What is the magnitude of the net force that is acting on the ball just before it is released?
Answer:
a. ac = 1844.66 m/s²
b. Fc = 265.63 N
Explanation:
a.
The centripetal acceleration of the ball is given as follows:
ac = v²/r
where,
ac = centripetal acceleration = ?
v = speed of ball = (87 mph)(1 h/ 3600 s)(1609.34 m / 1 mile) = 38.9 m/s
r = radius of path = 82 cm = 0.82 m
Therefore,
ac = (38.9 m/s)²/0.82 m
ac = 1844.66 m/s²
b.
The centripetal force is given as:
Fc = (m)(ac)
Fc = (0.144 kg)(1844.66 m/s²)
Fc = 265.63 N
At a pressure of one atmosphere oxygen boils at −182.9°C and freezes at −218.3°C. Consider a temperature scale where the boiling point of oxygen is 100.0°O and the freezing point is 0°O. Determine the temperature on the Oxygen scale that corresponds to the absolute zero point on the Kelvin scale.
Answer: -254.51°O
Explanation:
Ok, in our scale, we have:
-182.9°C corresponds to 100° O
-218.3°C corresponds to 0°
Then we can find the slope of this relation as:
S = (100° - 0°)/(-182.9°C - (-218.3°C)) = 2.82°O/°C
So we can have the linear relationship between the scales is:
Y = (2.82°O/°C)*X + B
in this relation, X is the temperature in Celcius and Y is the temperature in the new scale.
And we know that when X = -182.9°C, we must have Y = 0°O
then:
0 = (2.82°O/°C)*(-182.9°C) + B
B = ( 2.82°O/°C*189.9°C) = 515.778°O.
now, we want to find the 0 K in this scale, and we know that:
0 K = -273.15°C
So we can use X = -273.15°C in our previous equation and get:
Y = (2.82°O/°C)*(-273.15°C) + 515.778°O = -254.51°O
Consider two identical containers. Container A is filled with water to the top. Container B has a block of wood floating in it, but the level of the water is also at the top. Which container weighs more
Answer:
Container A will weigh more
Explanation:
Both containers are identical, so we assume that they weigh the same.
They both have the same volume, and will contain an equal volume of a material.
Since they both contain water to the top, this means that their volume is fully occupied. But container B contain a block of wood floating in it.
The fact that the block of wood floats in the water in container B shows that it is less dense than the water around it, and in the container A, this same space is completely filled with water.
What we derive from this is that the portion of space contained by the block of wood in container B is occupied by water in container A, but, in container B, the density of this space is lesser now, since the wood block floats.
Since density is mass per unit volume, and weight is proportional to mass, then we can see that the weight of this volume portion in container B is lesser than that of container A. The consequence is that container A will weigh more than container B because of this extra weight.
Two metal spheres are hanging from nylon threads. When you bring the spheres close to each other, they tend to attract. Based on this information alone, discuss all the possible ways that the spheres could be charged. Is it possible that after the spheres touch, they will cling together? Explain.
Explanation:
In the given question, the two metal spheres were hanged with the nylon thread.
When these two spheres were brought close together, they attracted each other. The attraction between these spheres is the result of the opposite charges between them.
The possible ways by which these two metal spheres can be charged are by induction that is touching the metal or by rubbing them.
During induction, the same charges are transferred to each sphere. In this case, either both the spheres will be negatively charged or positively charged.
It is not possible that after the sphere touch each other they will cling together because the same charge repels each other and during touching, if one sphere is neutral, then the charged one will transfer the same charge. And as we know that same charge repel each other therefore they will repel each other.
The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1 W. HooRU's surface area is 1.55 m2 and the emissivity of its surface is 0.287. Ignoring the radiation that HooRU absorbs from the cold universe, what is HooRU's temperature T?
Answer:
The temperature is [tex]T = 168.44 \ K[/tex]
Explanation:
From the question ewe are told that
The rate of heat transferred is [tex]P = 13.1 \ W[/tex]
The surface area is [tex]A = 1.55 \ m^2[/tex]
The emissivity of its surface is [tex]e = 0.287[/tex]
Generally, the rate of heat transfer is mathematically represented as
[tex]H = A e \sigma T^{4}[/tex]
=> [tex]T = \sqrt[4]{\frac{P}{e* \sigma } }[/tex]
where [tex]\sigma[/tex] is the Boltzmann constant with value [tex]\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.[/tex]
substituting value
[tex]T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }[/tex]
[tex]T = 168.44 \ K[/tex]
Potential difference of a battery is 2.2 V when it is connected
across a resistance of 5 ohm, if suddenly the potential difference
falls to 1.8V, its internal resistance will be
Answer:
1.1ohms
Explanation:
According to ohms law E = IR
If potential difference of a battery is 2.2 V when it is connected across a resistance of 5 ohm and if suddenly the voltage Falls to 1.8V then the current in the 5ohms resistor I = V/R = 1.8/5
I = 0.36A (This will be the load current).
Before we can calculate the value of the internal resistance, we need to know the voltage drop across the internal resistance.
Voltage drop = 2.2V - 1.8V = 0.4V
Then we calculate the internal resistance using ohms law.
According to the law, V = Ir
V= voltage drop
I is the load current
r = internal resistance
0.4 = 0.36r
r = 0.4/0.36
r = 1.1 ohms
A 50 gram meterstick is placed on a fulcrum at its 50 cm mark. A 20 gram mass is attached at the 12 cm mark. Where should a 40 gram mass be attached so that the meterstick will be balanced in rotational equilibrium
Answer:
The 40g mass will be attached at 69 cm
Explanation:
First, make a sketch of the meterstick with the masses placed on it;
--------------------------------------------------------------------------
↓ Δ ↓
20 g.................50 cm.................40g
38 cm y cm
Apply principle of moment;
sum of clockwise moment = sum of anticlockwise moment
40y = 20 (38)
40y = 760
y = 760 / 40
y = 19 cm
Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm
12cm 50 cm 69cm
--------------------------------------------------------------------------
↓ Δ ↓
20 g.................50 cm.................40g
38 cm 19 cm
A 1.0-kg ball on the end of a string is whirled at a constant speed of 2.0 m/s in a horizontal circle of radius 1.5 m. What is the work done by the centripetal force during one revolution
Answer:
The work done by the centripetal force is always, zero.
Explanation:
The formula for the work done by a force on an object is given as follows:
W = F d Cos θ
where,
W = Magnitude of the Work Done
F = Force applied to the body
θ = Angle between the direction of force and direction of motion of the object
In case of the circular motion, the force is the centripetal force. The centripetal force is always directed towards the center of the circle. While, the object moves in a direction, which is tangential to the circle. Hence, the angle between them is always 90°. Therefore,
W = F d Cos 90°
W = F d (0)
W = 0 J
Hence, the work done by the centripetal force is always, zero.
What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?
(Given, the universal gas constant = 8.315 J/(mol.k))
Answer:
U = 12,205.5 J
Explanation:
In order to calculate the internal energy of an ideal gas, you take into account the following formula:
[tex]U=\frac{3}{2}nRT[/tex] (1)
U: internal energy
R: ideal gas constant = 8.135 J(mol.K)
n: number of moles = 10 mol
T: temperature of the gas = 100K
You replace the values of the parameters in the equation (1):
[tex]U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J[/tex]
The total internal energy of 10 mol of Oxygen at 100K is 12,205.5 J
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 27.9 m/s2 with a beam of length 5.21 m , what rotation frequency is required
Answer: the angular frequency is 2.31 rad/s
Explanation:
The data we have is:
Radial acceleration A = 27.9 m/s^2
Beam length r = 5.21m
The radial acceleration is equal to the velocity square divided the radius of the circle (the lenght of the beam in this case)
And we can write the velocity as:
v = w*r where r is the radius of the circle, and w is the angular frequency.
w = 2pi*f
where f is the "normal" frequency.
So we have:
A = (v^2)/r = (r*w)^2/r = r*w^2
We can replace the values and find w.
27.9m/s^2 = 5.21m*w^2
√(27.9/5.21) = w = 2.31 rad/s
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 4.50 m/s at point A and 5.00 m/s at point C. The cart takes 4.00 s to go from point A to point C, and the cart takes 1.60 s to go from point B to point C. What is the cart's speed at point B
Answer:
Vi = 4.8 m/s
Explanation:
First we need to find the magnitude of constant tangential acceleration. For that purpose we use the following formula between points A and C:
a = (Vf - Vi)/t
where,
a = constant tangential acceleration from A to C = ?
Vf = Final Velocity at C = 5 m/s
Vi = Initial Velocity at A = 4.5 m/s
t = time taken to move from A to C = 4 s
Therefore,
a = (5 m/s - 4.5 m/s)/4 s
a = 0.125 m/s²
Now, applying the same equation between points B and C:
a = (Vf - Vi)/t
where,
a = constant tangential acceleration from A to B = 0.125 m/s²
Vf = Final Velocity at C = 5 m/s
Vi = Initial Velocity at B = ?
t = time taken to move from B to C = 1.6 s
Therefore,
0.125 m/s² = (5 m/s - Vi)/1.6 s
Vi = 5 m/s - (0.125 m/s²)(1.6 s)
Vi = 4.8 m/s
Four fixed point charges are at the corners of a square with sides of length L. Q1 is positive and at (OL) Q2 is positive and at (LL) Q3 is positive and at (4,0) Q4 is negative and at (0,0) A) Draw and label a diagram of the described arrangement described above (include a coordinate system). B) Determine the force that charge Q1 exerts on charge Qz. C) Determine the force that charge Q3 exerts on charge Q2. D) Determine the force that charge Q4 exerts on charge Q2. E) Now assume that all the charges have the same magnitude (Q) and determine the net force on charge Q2 due to the other three charges. Reduce this to the simplest form (but don't put in the numerical value for the force constant).
Answer:
A) See Annex
B) Fq₁₂ = K * Q₁*Q₂ /16 [N] (repulsion force)
C) Fq₃₂ = K * Q₃*Q₂ /16 [N] (repulsion force)
D) Fq₄₂ = K * Q₄*Q₂ /32 [N] (attraction force)
E) Net force (its components)
Fnx = (2,59/64 )* K*Q² [N] in direction of original Fq₃₂
Fny =(2,59/64 )* K*Q² [N] in direction of original Fq₁₂
Explanation:
For calculation of d (diagonal of the square, we apply Pythagoras Theorem)
d² = L² + L² ⇒ d² = 2*L² ⇒ d = √2*L² ⇒ d= (√2 )*L
d = 4√2 units of length (we will assume meters, to work with MKS system of units)
B) Force of Q₁ exerts on charge Q₂
Fq₁₂ = K * Q₁*Q₂ /(L)² Fq₁₂ = K * Q₁*Q₂ /16 (repulsion force in the direction indicated in annex)
C) Force of Q₃ exerts on charge Q₂
Fq₃₂ = K * Q₃*Q₂ /(L)² Fq₃₂ = K * Q₃*Q₂ /16 (repulsion force in the direction indicated in annex)
D) Force of -Q₄ exerts on charge Q₂
Fq₄₂ = K * Q₄*Q₂ / (d)² Fq₄₂ = K * Q₄*Q₂ /32 (Attraction force in the direction indicated in annex)
E) Net force in the case all charges have the same magnitude Q (keeping the negative sign in Q₄)
Let´s take the force that Q₄ exerts on Q₂ and Q₂ = Q ( magnitude) and
Q₄ = -Q
Then the force is:
F₄₂ = K * Q*Q / 32 F₄₂ = K* Q²/32 [N]
We should get its components
F₄₂(x) = [K*Q²/32 ]* √2/2 and so is F₄₂(y) = [K*Q²/32 ]* √2/2
Note that this components have opposite direction than forces Fq₁₂ and
Fq₃₂ respectively, and that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively
In new conditions
Fq₁₂ = K * Q₁*Q₂ /16 becomes Fq₁₂ = K * Q²/ 16 [N] and
Fq₃₂ = K* Q₃*Q₂ /16 becomes Fq₃₂ = K* Q² /16 [N]
Note that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively
Then over x-axis we subtract Fq₃₂ - F₄₂(x) = Fnx
and over y-axis, we subtract Fq₁₂ - F₄₂(y) = Fny
And we get:
Fnx = K* Q² /16 - [K*Q²/32 ]* √2/2 ⇒ Fnx = K*Q² [1/16 - √2/64]
Fnx = (2,59/64 )* K*Q²
Fny has the same magnitude then
Fny =(2,59/64 )* K*Q²
The fact that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively, means that Fnx and Fny remains as repulsion forces
A block slides down a ramp with friction. The friction experienced by the block is 21 N. The mass of the block is 8 kg. The ramp is 6 meters long (meaning, the block slides across 6 meters of ramp with friction). The block is originally 2 meters vertically above the ground (the bottom of the ramp). What is the change in energy of the block due to friction, expressed in Joules
Complete Question
The complete question is shown on the first uploaded image
Answer:
the change in energy of the block due to friction is [tex]E = -126 \ J[/tex]
Explanation:
From the question we are told that
The frictional force is [tex]F_f = 21 \ N[/tex]
The mass of the block is [tex]m_b = 8 \ kg[/tex]
The length of the ramp is [tex]l = 6 \ m[/tex]
The height of the block is [tex]h = 2 \ m[/tex]
The change in energy of the block due to friction is mathematically represented as
[tex]\Delta E = - F_s * l[/tex]
The negative sign is to show that the frictional force is acting against the direction of the block movement
Now substituting values
[tex]\Delta E = -(21)* 6[/tex]
[tex]\Delta E = -126 \ J[/tex]
A swimmer of mass 64.38 kg is initially standing still at one end of a log of mass 237 kg which is floating at rest in water. He runs toward the other end of the log and dives off with a horizontal speed of 3.472 m/s relative to the water. What is the speed of the log relative to water after the swimmer jumps off
Answer:
0.9432 m/s
Explanation:
We are given;
Mass of swimmer;m_s = 64.38 kg
Mass of log; m_l = 237 kg
Velocity of swimmer; v_s = 3.472 m/s
Now, if we consider the first log and the swimmer as our system, then the force between the swimmer and the log and the log and the swimmer are internal forces. Thus, there are no external forces and therefore momentum must be conserved.
So;
Initial momentum = final momentum
m_l × v_l = m_s × v_s
Where v_l is speed of the log relative to water
Making v_l the subject, we have;
v_l = (m_s × v_s)/m_l
Plugging in the relevant values, we have;
v_l = (64.38 × 3.472)/237
v_l = 0.9432 m/s
A 4.5 kg ball swings from a string in a vertical circle such that it has constant sum of kinetic and gravitational potential energy. Ignore any friction forces from the air or in the string. What is the difference in the tension between the lowest and highest points on the circle
Answer:
88.29 N
Explanation:
mass of the ball = 4.5 kg
weight of the ball will be = mass x acceleration due to gravity(9.81 m/s^2)
weight W = 4.5 x 9.81 = 44.145 N
centrifugal forces Tc act on the ball as it swings.
At the top point of the vertical swing,
Tension on the rope = Tc - W.
At the bottom point of the vertical swing,
Tension on the rope = Tc + W
therefore,
difference in tension between these two points will be;
Net tension = tension at bottom minus tension at the top
= Tc + W - (Tc - W) = Tc + W -Tc + W
= 2W
imputing the value of the weight W, we have
2W = 2 x 44.145 = 88.29 N
An object spins in a horizontal circle with a radius of 15.0 cm. The rotations are timed and the amount of time it takes for it to go around once is 0.56 s. The centripetal force is measured to be 6.1 N.According to the experiment, the speed of the object is closest to:'
Answer:
1.7 m/s
Explanation:
Relevant Data provided as per the question below:-
Radius = 15.0 cm
Time = 0.56 s.
Based on the above information
The computation of the speed of the object is shown below:-
[tex]Velocity = \frac{2\times \pi \times Radius}{Time}[/tex]
[tex]Velocity = \frac{2\times \frac{22}{7} \times 0.15}{0.56}[/tex]
[tex]= \frac{0.942857}{0.56}[/tex]
= 1.683 m/s
or
= 1.7 m/s
Therefore for computing the speed of the object or velocity we simply applied the above formula by considering the pi and all other given data
A 25 kg box is 220 N pulled at constant speed up a frictionless inclined plane by a force that is parallel to the incline. If the plane is inclined at an angle of 25o above the horizontal, the magnitude of the applied force is
Answer:
F = 103.54N
Explanation:
In order to calculate the magnitude of the applied force, you take into account that the forces on the box are the applied force F and the weight of the box W.
The box moves with a constant velocity. By the Newton second law you have that the sum of forces must be equal to zero.
Furthermore, you have that the sum of forces are given by:
[tex]F-Wsin\theta=0[/tex] (1)
F: applied force = ?
W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N
θ: degree of the incline = 25°
You solve the equation (1) for F:
[tex]F=Wsin\theta=(245N)sin(25\°)=103.54N[/tex] (2)
The applied force on the box is 103.54N
A charged particle q moves at constant velocity through a crossed electric and magnetic fields (E and B, which are both constant in magnitude and direction). Write the magnitude of the electric force on the particle in terms of the variables given. Do the same for the magnetic force
Answer:
The magnitude of the electric force on the particle in terms of the variables given is, F = qE
The magnitude of the magnetic force on the particle in terms of the variables given is, F = q (v x B)
Explanation:
Given;
a charged particle, q
magnitude of electric field, E
magnitude of magnetic field, B
The magnitude of the electric force on the particle in terms of the variables given;
F = qE
The magnitude of the magnetic force on the particle in terms of the variables given;
F = q (v x B)
where;
v is the constant velocity of the charged particle
Answer:
The magnitude of the electric force acting on a charged particle moving through an electric field = |qE|
The magnitude of the magnetic force of a charged particle moving at a particular velocity through a magnetic field = |qv × B|
Explanation:
The electric force acting on a charged particle, q, moving through an electric field, E, is given as a product of the charge on the particle (a scalar quantity) and the electric field (a vector quantity).
Electric force = qE
The magnitude of the electric force = |qE|
That is, magnitude of the product of the charge and the electric field vector.
The magnetic force acting on a charged particle, q, moving with a velocity, v, through a magnetic field, B is a vector product of qv [a product of the charge of the particle (a scalar quantity) and the velocity of the particle (a vector quantity)] and B (a vector quantity).
It is given mathematically as (qv × B)
The magnitude of the magnetic force is the magnitude of the vector product obtained.
Magnitude of the magnetic force = |qv × B|
Hope this Helps!!!
what is the orbital speed for a satellite 3.5 x 10^8m from the center of mars? Mars mass is 6.4 x 10^23 kg
Answer:
v = 349.23 m/s
Explanation:
It is required to find the orbital speed for a satellite [tex]3.5\times 10^8\ m[/tex] from the center of mass.
Mass of Mars, [tex]M=6.4\times 10^{23}\ kg[/tex]
The orbital speed for a satellite is given by the formula as follows :
[tex]v=\sqrt{\dfrac{GM}{r}} \\\\v=\sqrt{\dfrac{6.67\times 10^{-11}\times 6.4\times 10^{23}}{3.5\times 10^8}} \\\\v=349.23\ m/s[/tex]
So, the orbital speed for a satellite is 349.23 m/s.
Cart A, with a mass of 0.20 kg, travels on a horizontal air trackat 3.0m/s and hits cart B, which has a mass of 0.40 kg and is initially traveling away from Aat 2.0m/s. After the collision the center of mass of the two cart system has a speed of:____________.A. zeroB. 0.33m/sC. 2.3m/sD. 2.5m/sE. 5.0m/s
Answer:
[tex]\large \boxed{\text{C. 2.3 m/s}}[/tex]
Explanation:
Data:
[tex]m_{\text{A} } = \text{0.20 kg};\,v_{\text{Ai}} = \text{3.0 m/s}\\m_{\text{B} } = \text{0.40 kg};\,v_{\text{Bi}} = \text{2.0 m/s}\\[/tex]
Calculation:
This is a perfectly inelastic collision. The two carts stick together after the collision and move with a common final velocity.
The conservation of momentum equation is
[tex]\begin{array}{rcl}m_{\text{A}}v_{\text{Ai}} +m_{\text{B}} v_{\text{Bi}}&=&(m_{\text{A}} + m_{\text{B}})v_{\text{f}}\\0.20\times 3.0 + 0.40\times 2.0 & = & (0.20 + 0.40)v_{\text{f}}\\0.60 + 0.80 & = & 0.60v_{\text{f}}\\1.40 & = & 0.60v_{\text{f}}\\v_{\text{f}}&=& \dfrac{1.40}{0.60}\\\\& = & \textbf{2.3 m/s}\\\end{array}\\\text{The centre of mass has a velocity of $\large \boxed{\textbf{2.3 m/s}}$}[/tex]
A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point circled A, at a potential of 1.50 103 V, to point circled B, at 4.00 103 V. What is its speed at point circled B?
Answer:
[tex]v_B=3.78\times 10^5\ m/s[/tex]
Explanation:
It is given that,
Charge on helium nucleus is 2e and its mass is [tex]6.63\times 10^{-27}\ kg[/tex]
Speed of nucleus at A is [tex]v_A=6.2\times 10^5\ m/s[/tex]
Potential at point A, [tex]V_A=1.5\times 10^3\ V[/tex]
Potential at point B, [tex]V_B=4\times 10^3\ V[/tex]
We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :
increase in kinetic energy = increase in potential×charge
[tex]\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s[/tex]
So, the speed at point B is [tex]3.78\times 10^5\ m/s[/tex].
A solenoidal coil with 23 turns of wire is wound tightly around another coil with 310 turns. The inner solenoid is 20.0 cm long and has a diameter of 2.20 cm. At a certain time, the current in the inner solenoid is 0.130 A and is increasing at a rate of 1800 A/s. For this time, calculate:
(a) the average magnetic flux through each turn of the inner solenoid;
(b) the mutual inductance of the two solenoids;
(c) the emf induced in the outer solenoid by the changing current in the inner solenoid.
Answer:
Explanation:
From the given information:
(a)
the average magnetic flux through each turn of the inner solenoid can be calculated by the formula:
[tex]\phi _ 1 = B_1 A[/tex]
[tex]\phi _ 1 = ( \mu_o \dfrac{N_i}{l} i_1)(\pi ( \dfrac{d}{2})^2)[/tex]
[tex]\phi _ 1 = ( 4 \pi *10^{-7} \ T. m/A ) ( \dfrac{310}{20*10^{-2} \ m }) (0.130 \ A) ( \pi ( \dfrac{2.20*10^{-2} \ m }{2})^ 2[/tex]
[tex]\phi_1 = 9.625 * 10^{-8} \ Wb[/tex]
(b)
The mutual inductance of the two solenoids is calculated by the formula:
[tex]M = 23 *\dfrac{9.625*10^{-8} \ Wb}{0.130 \ A}[/tex]
M = [tex]1.703 *10^{-5}[/tex] H
(c)
the emf induced in the outer solenoid by the changing current in the inner solenoid can be calculate by using the formula:
[tex]\varepsilon = -N_o \dfrac{d \phi_1}{dt}[/tex]
[tex]\varepsilon = -M \dfrac{d i_1}{dt}[/tex]
[tex]\varepsilon = -(1.703*10^{-5} \ H) * (1800 \ A/s)[/tex]
[tex]\varepsilon = -0.030654 \ V[/tex]
[tex]\varepsilon = -30.65 \ V[/tex]
How much force is needed to cause a 15 kilogram bicycle to accelerate at a rate of 10
meters per second per second?
O A. 15 newtons
OB. 1.5 newtons
C. 150 newtons
OD. 10 newtons
Question 4
3 pts
I am approaching a traffic light at a speed of 135 km/h when I suddenly notice that
the light is red. I slam on my brakes and come to a stop in 4.29 seconds. What is the
acceleration of the car as I screech to a complete stop? (Note that an object that slows down
simply has a negative acceleration.)
& show work please I want to also understand
Answer:
The deceleration of the car is [tex]\approx -0.065m/s^{2}[/tex]
Explanation:
to solve this, we will have to apply the knowledge that will be got from the equations of motion.
There are several equations of motion, and depending on the parameters given in the problem, we can choose the perfect equation that can best be used to solve the problem.
In this case, since we are given the velocity and time, and we are solving for the acceleration, we will use this formula
[tex]v = u +at[/tex]
where v= final velocity = 0
u = initial velocity = 135Km/h [tex]\approx 0.278 m/s[/tex]
t= time = 4.29 seconds.
[tex]a = \frac{v - u}{t}[/tex]
[tex]a =\frac{0-0.278}{4.29} \approx 0.065m/s^{2}[/tex]
Hence, the deceleration of the car is [tex]\approx -0.065m/s^{2}[/tex]
A long cylindrical rod of diameter 200 mm with thermal conductivity of 0.5 W/m⋅K experiences uniform volumetric heat generation of 24,000 W/m3. The rod is encapsulated by a circular seeve having an outer diameter of 400 mm and a thermal conductivity of 4 W/m⋅K. The outer surface of the sleeve is exposed to cross flow air at 27∘C with a convection coefficient of 25 W/m2⋅K.
(a) Find the temperature at the interface between the rod and sleeve and on the outer surface.
(b) What is the temperature at the center of the rod?
Answer:
a, 71.8° C, 51° C
b, 191.8° C
Explanation:
Given that
D(i) = 200 mm
D(o) = 400 mm
q' = 24000 W/m³
k(r) = 0.5 W/m.K
k(s) = 4 W/m.K
k(h) = 25 W/m².K
The expression for heat generation is given by
q = πr²Lq'
q = π . 0.1² . L . 24000
q = 754L W/m
Thermal conduction resistance, R(cond) = 0.0276/L
Thermal conduction resistance, R(conv) = 0.0318/L
Using energy balance equation,
Energy going in = Energy coming out
Which is = q, which is 754L
From the attachment, we deduce that the temperature between the rod and the sleeve is 71.8° C
At the same time, we find out that the temperature on the outer surface is 51° C
Also, from the second attachment, the temperature at the center of the rod was calculated to be, 191.8° C
In a remote location near the North Pole, an iceberg floats in a lake. Next to the lake (assume it is not frozen) sits a comparably sized glacier sitting on land. If both chunks of ice should melt (and the melted ice all goes into the lake), which ice chunk would give the greatest increase in the level of the lake water, if any?
Answer:
The correct answer to the following question will be "Glacier".
Explanation:
A large mass of frozen ground that flows very steadily through both the valley as well as spreads out through the middle. Throughout several centuries, glaciers have formed from compacted snow in places whereby snow builds up quicker although this melts.The iceberg has been on the bay, so there's no adjustment in a size equivalent to the whole iceberg.So that the above seems to be the right answer.
Answer:
Glaciers not only transport material as they move, but they also sculpt and carve away the land beneath them. A glacier's weight, combined with its gradual movement, can drastically reshape the landscape over hundreds or even thousands of years. The ice erodes the land surface and carries the broken rocks and soil debris far from their original places, resulting in some interesting glacial landforms.
Explanation:
If the number of loops in a coil around a moving magnet doubles, the emf created:_________
a. Doubles
b. Halves
c. Remains the same
Answer is a. Doubles
when the loops are increased in the coil then the magnetic field created doubles
A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mass. It has a rotation rate of 8.0 rev/s. Two antennas deploy in the plane of rotation extending from the center of mass of the satellite. Each antenna can be approximated as a rod of mass 150.0 kg and length 6.6 m. What is the new rotation rate of the satellite (in rev/s)
Answer:
6.3 rev/s
Explanation:
The new rotation rate of the satellite can be found by conservation of the angular momentum (L):
[tex] L_{i} = L_{f} [/tex]
[tex] I_{i}*\omega_{i} = I_{f}*\omega_{f} [/tex]
The initial moment of inertia of the satellite (a solid sphere) is given by:
[tex] I_{i} = \frac{2}{5}m_{s}r^{2} [/tex]
Where [tex]m_{s}[/tex]: is the satellite mass and r: is the satellite's radium
[tex] I_{i} = \frac{2}{5}m_{s}r^{2} = \frac{2}{5}1900 kg*(4.6 m)^{2} = 1.61 \cdot 10^{4} kg*m^{2} [/tex]
Now, the final moment of inertia is given by the satellite and the antennas (rod):
[tex] I_{f} = I_{i} + 2*I_{a} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}m_{a}l^{2} [/tex]
Where [tex]m_{a}[/tex]: is the antenna's mass and l: is the lenght of the antenna
[tex] I_{f} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}150.0 kg*(6.6 m)^{2} = 2.05 \cdot 10^{4} kg*m^{2} [/tex]
So, the new rotation rate of the satellite is:
[tex] I_{i}*\omega_{i} = I_{f}*\omega_{f} [/tex]
[tex]\omega_{f} = \frac{I_{i}*\omega_{i}}{I_{f}} = \frac{1.61 \cdot 10^{4} kg*m^{2}*8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kg*m^{2}} = 6.3 rev/s[/tex]
Therefore, the new rotation rate of the satellite is 6.3 rev/s.
I hope it helps you!
If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is the magnitude of the magnetic field at a distance of 19.8 cm from the wire
Answer:
The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Explanation:
Given;
first magnetic field at first distance, B₁ = 2.50 mT
first distance, r₁ = 12.6 cm = 0.126 m
Second magnetic field at Second distance, B₂ = ?
Second distance, r₂ = ?
Magnetic field for a straight wire is given as;
[tex]B = \frac{\mu I}{2 \pi r}[/tex]
Where:
μ is permeability
B is magnetic field
I is current flowing in the wire
r distance to the wire
[tex]Let \ \frac{\mu I}{2\pi} \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT[/tex]
Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Light rays from stars bend toward smaller angles as they enter Earth's atmosphere. a. Explain why this happens using Snell's law and the speed of light. b. Where are the actual stars in relation to their apparent position as viewed from the Earth's surface?
Answer:
Following are the answer to this question:
Explanation:
In option (a):
The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle. Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.In option (b):
Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth. Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.