Answer:
200 KJ
Explanation:
h = 20 m, m = 1000 kg, g = 10 m/s^2
As the rock is sliding and it starts from rest. it falls from a f=height of 20 m
So, by using the energy conservation law
Potential energy at the top of hill = Kinetic energy at the bottom of hill
So, Kinetic energy at the bottom of hill = Potential energy at the top of hill
K.E = m g h
K. E = 1000 x 10 x 20 = 200,000 J
K.E = 200 KJ
A reservoir has a surface area of 30 km^2 and an average depth of 49 m. What mass of water is held behind the dam?
Answer:
Mass of water, [tex]m=1.47\times 10^{12}\ kg[/tex]
Explanation:
Given that,
Surface area of the reservoir, A = 30 km² = 3 × 10⁷ m²
Average depth, d = 49 m
The volume of the reservoir is V such that,
[tex]V=A\times d[/tex]
[tex]V=3\times 10^7\ m^2\times 49\ m[/tex]
[tex]V=1.47\times 10^9\ m^3[/tex]
We have to find the mass of water is held behind the dam. It can be calculated using the expression for density. We know that density of water, d = 1000 kg/m³
Density, [tex]d=\dfrac{m}{V}[/tex]
[tex]m=d\times V[/tex]
[tex]m=1000\ kg/m^3\times 1.47\times 10^9\ m^3[/tex]
[tex]m=1.47\times 10^{12}\ kg[/tex]
Hence, this is the required solution.
An ideal transformer has 60 turns on its primary coil and 300 turns on its secondary coil. If 120 V at 2.0 A is applied to the primary, what voltage and current are present in the secondary?
Explanation:
It is given that,
Number of turns in primary coil, [tex]N_p=60[/tex]
Number of turns in secondary coil, [tex]N_s=300[/tex]
Voltage in primary coil, [tex]V_p=120\ V[/tex]
Current in primary coil, [tex]I_p=2\ A[/tex]
We have to find the voltage and current in the secondary coli. Firstly calculating the voltage in secondary coil as :
[tex]\dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}[/tex]
[tex]\dfrac{60}{300}=\dfrac{120}{V_s}[/tex]
[tex]V_s=720\ Volts[/tex]
Now, calculating the current present in the secondary coil as :
[tex]\dfrac{N_p}{N_s}=\dfrac{I_s}{I_p}[/tex]
[tex]\dfrac{60}{300}=\dfrac{I_s}{2\ A}[/tex]
[tex]I_s=0.4\ A[/tex]
Hence, this is the required solution.
Final answer:
The voltage present in the secondary is 600 V and the current is 0.4 A.
Explanation:
The question involves an ideal transformer where the primary coil has 60 turns and the secondary coil has 300 turns, with 120 V at 2.0 A applied to the primary. To find the voltage and current present in the secondary, we can use the transformer equations:
Vp/Vs = Np/Ns (where V is voltage and N is the number of turns in the primary (p) and secondary (s) coils)
According to the conservation of energy in an ideal transformer, Pp = Ps (where P is power), and since P = VI, we get:
VpIp = VsIs (where I is current)
Using these equations:
Is = (Np/Ns) \\times Ip = (60/300) \\times 2.0 A = 0.4 A
Thus, the voltage present in the secondary is 600 V and the current is 0.4 A.
A system consists of two objects that collide in a partially elastic collision. Which two properties of the system are conserved by the collision? mass and kinetic energy momentum and mechanical energy momentum and kinetic energy momentum and mass
Answer:
Mass , momentum and mechanical energy
Explanation:
There are two types of collision.
1. Elastic collision: in case of elastic collision no energy is loss in form of heat and sound .
The momentum and the kinetic energy is conserved.
2. Inelastic collision: in case of inelastic or partially elastic collision, some of energy is lost in form of heat or sound. Here momentum and mechanical energy is conserved.
Mass of a system is always conserved.
A very hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. The ball rebounds back at essentially the same speed. If the collision with the floor lasts 0.05 s, what is the average force exerted by the floor on the ball?
The average force exerted by the floor on the ball during the collision is zero.
Explanation:To find the average force exerted by the floor on the ball during the collision, we can use the impulse-momentum principle. The impulse experienced by the ball is equal to the change in momentum. Since the ball rebounds back at the same speed, the change in momentum is zero. Therefore, the average force exerted by the floor on the ball is also zero.
Final answer:
The average force exerted by the floor on a 0.5 kg ball bouncing off the floor with a change in velocity from -4 m/s to 4 m/s, over a time of 0.05 s, is 40 N directed upwards.
Explanation:
To calculate the average force exerted by the floor on the ball, you can use the concept of impulse, which is the change in momentum. The change in momentum is equal to the final momentum minus the initial momentum. Since the speed of the ball is the same before and after the bounce, but the direction has changed, the change in velocity is twice the speed of the ball.
The formula for impulse (I) is:
I = Δp = m(v_f - v_i)
Where:
m is the mass of the ball
v_i is the initial velocity of the ball (before the bounce)
v_f is the final velocity of the ball (after the bounce)
Δp is the change in momentum
The impulse given by the floor can also be described by the average force (F_avg) multiplied by the time (t) the force was applied:
I = F_avg ⋅ t
Therefore, we can equate the two:
m(v_f - v_i) = F_avg ⋅ t
By plugging in the values:
(0.5 kg)(4 m/s + 4 m/s) = F_avg ⋅ (0.05 s)
Solving for F_avg gives:
F_avg = ±8.0 N
The positive value indicates that the average force direction is upwards.
A boxcar traveling at 12 m/s approaches a string of 5 identical boxcars sitting stationary on the track. The moving boxcar collides and links with the stationary cars and they all move off together along the track. What is the final speed of the cars immediately after the collision? (You may take the mass of each boxcar to be 18,537 kg.)
Answer:
2 m/s
Explanation:
Momentum is conserved:
mv = MV
where m is mass of the boxcar, v is its initial velocity, M is the mass of all six box cars, and V is the final velocity.
v = 12 m/s, and M = 6m, so:
m (12 m/s) = 6m V
12 m/s = 6V
V = 2 m/s
A 33.1 g copper object is launched from a 1.5 m 30° steel incline positioned on the floor by being pulled up a string attached to a 50.0 g mass suspended vertically over a pulley. The object is projected towards a glass table where it lands when it is at the point along its trajectory with the lowest speed. It comes to a halt when it clears the opposite edge of the table. It then falls and lands on 9 physics texts each 5 cm thick that are stacked on the floor on the opposite side of the table. Assume that the table does not have a ledge and the rectangular object experiences 0.05 N of air resistance as it falls towards the books. How far and how long did the object travel and how fast does it hit the books. Assume that at the moment the copper object leaves the incline, the massless string and ideal pulley break off
Explanation:
To answer how far and how long the object travels, we need to know either the length of the table or the kinetic coefficient of friction between copper and glass. We also need to know the kinetic coefficient of friction between the copper object and the steel incline. This information wasn't provided, so for sake of illustration, I'll assume both coefficients are the same, and that μk = 0.35.
Let's divide the path into four sections to stay organized:
a) object moves up incline
b) object is in free fall
c) object slides across table
d) object falls onto books
a)
First, the object slides up the incline. There are four forces acting on the object. Weight pulling down, normal force perpendicular to the incline, tension up the incline, and friction down the incline.
Sum of the forces normal to the incline:
∑F = ma
N - W cos θ = 0
N = mg cos θ
Sum of the forces parallel to the incline:
∑F = ma
T - F = ma
T - Nμ = ma
T - mgμ cos θ = ma
Now sum of the forces in the y direction on the hanging mass:
∑F = ma
T - W = M(-a)
T = Mg - Ma
Substituting:
Mg - Ma - mgμ cos θ = ma
Mg - mgμ cos θ = (m + M) a
a = g (M - mμ cos θ) / (m + M)
Given m = 33.1 g, M = 50.0 g, θ = 30°, μ = 0.35, and g = 9.8 m/s²:
a = 4.71 m/s²
So the velocity it reaches at the top of the incline is:
v² = v₀² + 2a(x - x₀)
v² = 0² + 2(4.71)(1.5)
v = 3.76 m/s
The time to reach the top of the incline is:
x = x₀ + v₀ t + ½ at²
1.5 = 0 + (0) t + ½ (4.71) t²
t = 0.798 s
The horizontal distance traveled is:
x = 1.5 cos 30°
x = 1.299 m
And the vertical distance traveled is:
y = 1.5 sin 30°
y = 0.750 m
b)
In the second stage, the object is in free fall. v₀ = 3.76 m/s and θ = 30°. The object lands on the table at the point where its speed is a minimum. This is at the highest point of the trajectory, when the vertical velocity is 0.
v = at + v₀
0 = (-9.8) t + (3.76 sin 30°)
t = 0.192 s
The horizontal distance traveled is:
x = x₀ + v₀ t + ½ at²
x = 0 + (3.76 cos 30°) (0.192) + ½ (0) (0.192)²
x = 0.625 m
The height it reaches is:
v² = v₀² + 2a(y - y₀)
(0)² = (3.76 sin 30°)² + 2(-9.8)(y - 0)
y = 0.180 m
c)
The object is now on the glass table. As it slides, there are three forces acting on it. Normal force up, gravity down, and friction to the left.
In the y direction:
∑F = ma
N - W = 0
N = mg
In the x direction:
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
The time it takes to reach the end of the table is:
v = at + v₀
0 = (-9.8×0.35) t + (3.76 cos 30°)
t = 0.949 s
And the distance it travels is:
v² = v₀² + 2a(x - x₀)
(0)² = (3.76 cos 30°)² + 2(-9.8×0.35) (x - 0)
x = 1.546 m
d)
Finally, the object falls onto the books. There are 9 books, each 5 cm thick, so the height is 45 cm. As the book falls, there are two forces acting on it: drag up and gravity down.
∑F = ma
D - W = ma
0.05 - (0.0331)(9.8) = 0.0331 a
a = -8.29 m/s²
Adding the vertical distances from parts a and b, we know the height of the table is 0.930 m. So the time it takes the object to land is:
y = y₀ + v₀ t + ½ at²
0.45 = 0.93 + (0) t + ½ (-9.8) t²
t = 0.313 s
Adding up the times from all four parts, the total time is:
t = 0.798 + 0.192 + 0.949 + 0.313
t = 2.25 s
Adding up the horizontal distances, the total distance traveled is:
x = 1.299 + 0.625 + 1.546
x = 3.47 m
A 93.5 kg snowboarder starts from rest and goes down a 60 degree slope with a 45.7 m height to a rough horizontal surface that is 10.0 m long and a coefficient of kinetic friction of 0.102. The then snowboarder goes up a slope with angle of 30 degrees up coming to a stop and then sits to stay in the same place. be considered frictionless. a. What is the speed of the snowboarder at the bottom of the 60 degree slope? b. What is the speed of the snowboarder at the bottom of the 30 degree slope? c. What is the final height of the snowboarder above the horizontal surface? until Both slopes are smooth and can
Answer:
a. 29.9 m/s, b. 29.6 m/s, c. 44.7 m
Explanation:
This can be answered with either force analysis and kinematics, or work and energy.
a) Using force analysis, we can draw a free body diagram for the snowboarder. There are two forces: normal force perpendicular to the slope and gravity down.
Sum of the forces parallel to the slope:
∑F = ma
mg sin θ = ma
a = g sin θ
Therefore, the velocity at the bottom is:
v² = v₀² + 2a(x - x₀)
v² = (0)² + 2(9.8 sin 60°) (45.7 / sin 60° - 0)
v = 29.9 m/s
Alternatively, using energy:
PE = KE
mgh = 1/2 mv²
v = √(2gh)
v = √(2×9.8×45.7)
v = 29.9 m/s
b) Drawing a free body diagram, there are three forces on the snowboarder. Normal force up, gravity down, and friction to the left.
Sum of the forces in the y direction:
∑F = ma
N - mg = 0
N = mg
Sum of the forces in the x direction:
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
Therefore, the snowboarder's final speed is:
v² = v₀² + 2a(x - x₀)
v² = (29.9)² + 2(-9.8×.102) (10 - 0)
v = 29.6 m/s
Using energy instead:
KE = KE + W
1/2 mv² = 1/2 mv² + F d
1/2 mv² = 1/2 mv² + mgμ d
1/2 v² = 1/2 v² + gμ d
1/2 (29.9)² = 1/2 v² + (9.8)(0.102)(10)
v = 29.6 m/s
c) This is the same as part a, but this time, the weight component parallel to the incline is pointing left.
∑F = ma
-mg sin θ = ma
a = -g sin θ
Therefore, the final height reached is:
v² = v₀² + 2a(x - x₀)
(0)² = (29.6)² + 2(-9.8 sin 30°) (h / sin 30° - 0)
h = 44.7 m
Using energy:
KE = PE
1/2 mv² = mgh
h = v² / (2g)
h = (29.6)² / (2×9.8)
h = 44.7 m
Question Part Points Submissions Used The difference in potential between the accelerating plates of a TV set is about 218 V. If the distance between these plates is 1.29 cm, find the magnitude of the uniform electric field (N/C) in this region. Consider entering your answer using scientific notation.
Answer:
[tex]1.69\cdot 10^4 N/C[/tex]
Explanation:
The relationship between electric field strength and potential difference is:
[tex]E=\frac{V}{d}[/tex]
where
E is the electric field strength
V is the potential difference
d is the distance between the plates
Here we have
V = 218 V
d = 1.29 cm = 0.0129 m
So, the electric field is
[tex]E=\frac{218 V}{0.0129 m}=16900 N/C = 1.69\cdot 10^4 N/C[/tex]
Two resistors have resistances R1 and R2. When the resistors are connected in series to a 12.7-V battery, the current from the battery is 1.81 A. When the resistors are connected in parallel to the battery, the total current from the battery is 9.06 A. Determine R1 and R2. (Enter your answers from smallest to largest.)
In the case of resistors in series and in parallel to a battery, the resistances can be determined by applying Ohm's law and using equations for total resistance in both series and parallel configurations. The calculations yield the values R1 = 1.53 ohms and R2 = 5.48 ohms.
Explanation:
The problem involves applying Ohm's law and equations for series and parallel resistors. When the resistors are in series, we add the resistances so R1 + R2 = 12.7 V / 1.81 A = 7.01 ohms. In a parallel connection, the total resistance R is given by 1/R = 1/R1 + 1/R2 or 12.7 V / 9.06 A = 1.40 ohms. Solving this pair of equations we get R1 = 1.53 ohms and R2 = 5.48 ohms, giving the pair of resistances asked for.
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A boat is traveling upstream at 10 km/h with respect to the water of a river. The water is flowing at 8.0 km/h with respect to the ground. What are the (a) magnitude and (b) direction of the boat’s velocity with respect to the ground? A child on the boat walks from front to rear at 4.0 km/h with respect to the boat.What are the (c) magnitude and (d) direction of the child’s velocity with respect to the ground?
Final answer:
The magnitude of the boat’s velocity with respect to the ground is 18.0 km/h downstream. The child’s velocity with respect to the ground is 14.0 km/h upstream.
Explanation:
To find the magnitude and direction of the boat’s velocity with respect to the ground, we can use vector addition. The boat's velocity with respect to the ground is the sum of its velocity with respect to the water and the water's velocity with respect to the ground. The magnitude of the boat's velocity with respect to the ground is the sum of its speed upstream and the speed of the water downstream: 10 km/h + 8.0 km/h = 18.0 km/h. The direction of the boat's velocity with respect to the ground is the same as the direction of the water's velocity, which is downstream.
To find the magnitude and direction of the child’s velocity with respect to the ground, we can also use vector addition. The child's velocity with respect to the ground is the sum of their velocity with respect to the boat and the boat's velocity with respect to the ground. The magnitude of the child's velocity with respect to the ground is the difference between their speed on the boat and the boat's speed with respect to the ground: 4.0 km/h - 18.0 km/h = -14.0 km/h (negative sign indicating direction opposite to the boat's velocity with respect to the ground). Therefore, the magnitude of the child's velocity with respect to the ground is 14.0 km/h. The direction is upstream, opposite to the direction of the boat's velocity with respect to the ground.
You have landed on an unknown planet, Newtonia, and want to know what objects will weigh there. You find that when a certain tool is pushed on a frictionless horizontal surface by a 12.2 N force, it moves 16.1 m in the first 2.40 s , starting from rest. You next observe that if you release this tool from rest at 11.0 m above the ground, it takes 2.58 s to reach the ground. What does the tool weigh on Newtonia? What would it weigh on Earth?
Final answer:
To find the weight of the tool on Newtonia, calculate the acceleration using the given displacement and time. Multiply this acceleration by the mass of the tool to find its weight. On Earth, use the acceleration due to gravity to find the weight of the tool.
Explanation:
To determine the weight of the tool on Newtonia, we can use the equation for weight: w = mg. From the given information, we know that the tool moves 16.1 m in the first 2.40 s when a 12.2 N force is applied. Using this information, we can calculate the acceleration of the tool. By using the equation of motion s = ut + 0.5at^2, where s is the displacement, u is the initial velocity (0 m/s), t is the time (2.40 s), and a is the acceleration, we can solve for a. The calculated acceleration can then be multiplied by the mass of the tool to find its weight on Newtonia.
To find the weight of the tool on Earth, we can use the acceleration due to gravity, which is approximately 9.8 m/s². Again using the equation w = mg, we can multiply the mass of the tool by this acceleration to find its weight on Earth.
The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is the gravitational constant 6.67 × 10-11 m3 kg-1 s-2. What is the approximate escape speed, in km/s, from the Solar System starting from an orbit at 1.0 AU? In this case, the mass of the Sun, 2.3e+30 kg, can be used as the mass of the Solar system
The approximate escape speed from the Solar System at 1.0 AU is around 42 km/s, calculated using the formula that includes the mass of the Sun and the gravitational constant.
The question asks for the approximation of the escape speed from the Solar System starting from an orbit at 1.0 AU using the given mass of the Sun and radius of the Earth's orbit. To find the escape speed, we would use the formula vesc = \\sqrt{2GM/R}, where G is the gravitational constant, M is the mass of the object (in this case, the Sun), and R is the starting radius (1 AU).
The escape speed formula does not depend on the mass of the escaping object, assuming that the central object (Sun) does not move. The escape velocity from the surface of the Earth is approximately 11 km/s, but to escape the sun's gravity, starting from Earth's orbit, requires an escape speed of about 42 km/s.
The escape speed from the Solar System starting from an orbit at 1 AU is 42.1 km/s.
To find the escape speed from the Solar System from an orbit at 1 AU (Astronomical Unit), we use the given formula for escape velocity:
[tex]\[ v_e = \sqrt{\frac{2GM}{R}} \][/tex]
Plugging in these values:
[tex]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \times 2.3 \times 10^{30} \, \text{kg}}{1.496 \times 10^{11} \, \text{m}}} \][/tex]
[tex]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 2.3 \times 10^{19}}{1.496}} \][/tex]
[tex]\[ v_e = \sqrt{\frac{30.482 \times 10^{19}}{1.496}} \][/tex]
[tex]\[ v_e = \sqrt{20.38 \times 10^{19}} \, \text{m/s} \][/tex]
[tex]\[ v_e = \sqrt{2.038 \times 10^{19}} \times 10 \, \text{m/s} \][/tex]
[tex]\[ v_e = \sqrt{2.038 \times 10^{19}} \, \text{m/s} \][/tex]
Calculating [tex]\( \sqrt{2.038 \times 10^{19}} \)[/tex]:
[tex]\[ v_e = 4.51 \times 10^{9.5} \, \text{m/s} \][/tex]
To convert meters per second to kilometers per second:
[tex]\[ v_e \approx 42.1 \, \text{km/s} \][/tex]
Tom kicks a soccer ball on a flat, level field with initial speed 20 m/s at an angle 35 degrees above the horizontal. (1) How long will the ball stay in the air (2) Max height will ball attain (3) How far away from tom will the ball land (4) what speed will the ball have in the instant just before it lands
Explanation:
Given:
x₀ = 0 m
y₀ = 0 m
v₀ = 20 m/s
θ = 35°
aᵧ = -9.8 m/s²
1) Find t when y = 0.
y = y₀ + v₀ᵧ t + ½ aᵧ t²
0 = 0 + (20 sin 35°) t + ½ (-9.8) t²
0 = t (20 sin 35° - 4.9 t)
t = 0, t = 2.34
The ball stays in the air 2.34 seconds.
2) Find y when vᵧ = 0.
vᵧ² = v₀ᵧ² + 2aᵧ (y - y₀)
0² = (20 sin 35)² + 2(-9.8) (y - 0)
y = 6.71 m
The ball reaches a maximum height of 6.71 meters.
3) Find x when y = 0.
x = x₀ + v₀ₓ t + ½ aₓ t²
x = 0 + (20 cos 35°) (2.34) + ½ (0) (2.3)²
x = 38.4 m
The ball lands 38.4 meters from Tom.
4) Find v when y = 0.
vₓ = aₓ t + v₀ₓ
vₓ = (0) (2.34) + 20 cos 35°
vₓ = 16.4 m/s
vᵧ = aᵧ t + v₀ᵧ
vᵧ = (-9.8) (2.34) + 20 sin 35°
vᵧ = -11.5 m/s
v = √(vₓ² + vᵧ²)
v = √((16.4)² + (-11.5)²)
v = 20 m/s
The ball has a speed of 20 m/s just before it lands.
The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational attraction of a much larger body, such as a planet of total mass M. The escape velocity is a function of the distance of the object from the center of the planet R, but unless otherwise specified this distance is taken to be the radius of the planet because it addresses the question "How fast does my rocket have to go to escape from the surface of the planet?"
Answer: [tex]V=\sqrt{\frac{2GM}{R}}[/tex]
Explanation:
Taking into account what is stated in this problem and considering there is no friction during the takeoff of the rocket of the planet, the rocket will escape the gravitational attraction of the massive body when its kinetic energy [tex]K[/tex] and its potential energy [tex]P[/tex] are equal in magnitude.
Written mathematically is:
[tex]K=P[/tex] (1)
Where:
[tex]K=\frac{1}{2}mV^{2}[/tex] (2)
Being [tex]m[/tex] the mass of the rocket
And:
[tex]P=-\frac{GMm}{R}[/tex] (3)
Being [tex]M[/tex] the mass of the planet, [tex]G[/tex] the gravitational constant and [tex]R[/tex] the radius of the planet.
Substituting (2) and (3) in (1):
[tex]\frac{1}{2}mV^{2}=-\frac{GMm}{R}[/tex] (4)
Finding [tex]V[/tex], which is the escape velocity:
[tex]V=\sqrt{\frac{2GM}{R}}[/tex] this is the velocity the rocket must have in order to escape from the surface of the planet
Which one of the following statements concerning superconductors is false? A constant current can be maintained in a superconducting ring for several years without an emf. Superconductors are perfect conductors. Critical temperatures for some superconductors exceed 100 K. All materials are superconducting at temperatures near absolute zero kelvin. Below its critical temperature, the resistivity of a superconductor is zero Ω-m
Answer:
All materials are superconducting at temperatures near absolute zero kelvin.
Explanation:
All materials are superconducting at temperatures near absolute zero kelvin is false concerning superconductors.
c. All materials are superconducting at temperatures near absolute zero kelvin.
What is superconductors?A superconductor is a material that achieves superconductivity, which is a state of matter that has no electrical resistance and does not allow magnetic fields to penetrate.
An electric current in a superconductor can persist indefinitely. Superconductivity can only typically be achieved at very cold temperatures.
Superconductivity is a phenomenon observed in several metals and ceramic materials.
When these materials are cooled to temperatures ranging from near absolute zero ( 0 degrees Kelvin, -273 degrees Celsius) to liquid nitrogen temperatures ( 77 K, -196 C), their electrical resistance drops with a jump down to zero.
Therefore,
All materials are superconducting at temperatures near absolute zero kelvin.
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If you are designing a force multiplying system with two pneumatic cylinders. Your system should increase the force by a multiple of 4. What should you do? O decrease the area in the second cylinder O increase the area in the second cylinder
Answer:
decrease the area in the second cylinder
Explanation:
In a two-cylinders pneumatic system, the pressure exerted on the first cylinder is equal to the pressure on the second cylinder:
[tex]p_1 = p_2[/tex]
We can rewrite the pressure as product between force (F) and area (A) of the cylinder:
[tex]F_1 A_1 = F_2 A_2[/tex]
In this system, the output force should be 4 times the input force:
[tex]F_2 = 4 F_1[/tex]
Substituting into the previous equation, we get:
[tex]F_1 A_1 = 4 F_1 A_2\\A_2 = \frac{A_1}{4}[/tex]
This means that the area of the second cylinder must be 1/4 of the area of the first cylinder, so the correct answer is
decrease the area in the second cylinder
Which of the following would double the amount of current flowing through a piece of metal wire? Quadruple the voltage across it. Quarter the voltage across it. Halve the voltage across it. Double the voltage across it.
Answer:
Double the voltage across it.- last choice
A playground merry-go-round has a radius of 4.6 m and a moment of inertia of 200 kg-m2 and turns with negligible friction about a vertical axle through its center. A child applies a 26.0 N force tangentially to the edge of the merry-go-round for 15.0 seconds. If the merry-go-round is initially at rest, how much work did the child do on the merry-go-round?
Answer:
8050 J
Explanation:
Given:
r = 4.6 m
I = 200 kg m²
F = 26.0 N
t = 15.0 s
First, find the angular acceleration.
∑τ = Iα
Fr = Iα
α = Fr / I
α = (26.0 N) (4.6 m) / (200 kg m²)
α = 0.598 rad/s²
Now you can find the final angular velocity, then use that to find the rotational energy:
ω = αt
ω = (0.598 rad/s²) (15.0 s)
ω = 8.97 rad/s
W = ½ I ω²
W = ½ (200 kg m²) (8.97 rad/s)²
W = 8050 J
Or you can find the angular displacement and find the work done that way:
θ = θ₀ + ω₀ t + ½ αt²
θ = ½ (0.598 rad/s²) (15.0 s)²
θ = 67.3 rad
W = τθ
W = Frθ
W = (26.0 N) (4.6 m) (67.3 rad)
W = 8050 J
What is the escape speed (in km/s) from an Earth-like planet with mass 6.5e+24 kg and radius 55.0 × 105 m? Use the gravitational constant G = 6.67 × 10-11 m3 kg-1 s-2.
Answer:
Escape velocity, v = 12.6 km/s
Explanation:
It is given that,
Mass of earth like planet, [tex]m=6.5\times 10^{24}\ kg[/tex]
Radius of planet, [tex]r=55\times 10^5\ m[/tex]
The escape velocity of a planet is given by the following formula as:
[tex]v=\sqrt{\dfrac{2Gm}{r}}[/tex]
G = universal gravitational constant
[tex]v=\sqrt{\dfrac{2\times 6.67\times 10^{-11}\ m^3kg^{-1}s^{-2}\times 6.5\times 10^{24}\ kg}{55\times 10^5\ m}[/tex]
v = 12556.05 m/s
or
v = 12.6 km/s
Hence, this is the required solution.
An electric motor rotating a workshop grinding wheel at a rate of 1.31 ✕ 102 rev/min is switched off. Assume the wheel has a constant negative angular acceleration of magnitude 3.40 rad/s2. (a) How long does it take for the grinding wheel to stop? (b) Through how many radians has the wheel turned during the interval found in (a)?
(a) 4.03 s
The initial angular velocity of the wheel is
[tex]\omega_i = 1.31 \cdot 10^2 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=13.7 rad/s[/tex]
The angular acceleration of the wheel is
[tex]\alpha = -3.40 rad/s^2[/tex]
negative since it is a deceleration.
The angular acceleration can be also written as
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_f = 0[/tex] is the final angular velocity (the wheel comes to a stop)
t is the time it takes for the wheel to stop
Solving for t, we find
[tex]t=\frac{\omega_f - \omega_i }{\alpha}=\frac{0-13.7 rad/s}{-3.40 rad/s^2}=4.03 s[/tex]
(b) 27.6 rad
The angular displacement of the wheel in angular accelerated motion is given by
[tex]\theta= \omega_i t + \frac{1}{2}\alpha t^2[/tex]
where we have
[tex]\omega_i=13.7 rad/s[/tex] is the initial angular velocity
[tex]\alpha = -3.40 rad/s^2[/tex] is the angular acceleration
t = 4.03 s is the total time of the motion
Substituting numbers, we find
[tex]\theta= (13.7 rad/s)(4.03 s) + \frac{1}{2}(-3.40 rad/s^2)(4.03 s)^2=27.6 rad[/tex]
The grinding wheel takes 38.5 seconds to stop, covering an angular displacement of 105 radians.
(a) To calculate the time it takes for the grinding wheel to stop, we can use the equation: t = ωf / α where ωf is the final angular velocity (0), and α is the angular acceleration (-3.40 rad/s²). Solving gives t = 38.5 seconds.
(b) The total angular displacement can be found using the equation: θ = ωi*t + 0.5*α*t^2 where ωi is the initial angular velocity (1.31 × 10² rev/min converted to rad/s), t is the time found in part (a), and α is the angular acceleration. This gives θ = 105 radians.
A person walks first at a constant speed of 4.89 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 2.95 m/s. What is the average speed over the entire trip?
Answer:
approximately 3.68
Explanation:
Explained in the attached picture.
A wind turbine is rotating counterclockwise at 0.626 rev/s and slows to a stop in 12.9 s. Its blades are 17.9 m in length. What is the centripetal acceleration of the tip of the blades at t=0~\text{s}t=0 s?
Answer:
276.5 m/s^2
Explanation:
The initial angular velocity of the turbine is
[tex]\omega=0.626 rev/s \cdot 2\pi rad/rev =3.93 rad/s[/tex]
The length of the blade is
r = 17.9 m
So the centripetal acceleration is given by
[tex]a=\omega^2 r[/tex]
At the instant t = 0,
[tex]\omega=3.93 rad/s[/tex]
So the centripetal acceleration of the tip of the blades is
[tex]a=(3.93 rad/s)^2 (17.9 m)=276.5 m/s^2[/tex]
If the work required to stretch a spring 3 ft beyond its natural length is 15 ft-lb, how much work is needed to stretch it 27 in. beyond its natural length?
Answer:
8.4 ft-lb
Explanation:
Work = change in energy
W = ½ kx²
When x = 3 ft, W = 15 ft-lb:
15 ft-lb = ½ k (3 ft)²
k = 30/9 lb/ft
When x = 27 in = 2.25 ft:
W = ½ kx²
W = ½ (30/9 lb/ft) (2.25 ft)²
W = 8.4375 ft-lb
Rounding to 2 sig-figs, it takes 8.4 ft-lb of work.
Final answer:
The work required to stretch the spring 27 in. beyond its natural length is 12.728 N-m.
Explanation:
To find the work required to stretch the spring 27 in. beyond its natural length, we can use the work done formula:
Work = (1/2)kx²
Where k is the spring constant and x is the displacement of the spring. We know that the work required to stretch the spring 3 ft beyond its natural length is 15 ft-lb. Let's convert the given measurements into consistent units:
3 ft = 36 in.
15 ft-lb = 12.728 N-m
Now, we can solve for the work required to stretch the spring 27 in. beyond its natural length:
(1/2)k(27)² = (1/2)k(729) = 12.728 N-m
We can rearrange the equation to solve for k:
k = (2 * 12.728) / 729 = 0.0349 N/m
Vector A with arrow has a magnitude of 5.35 units, and vector B with arrow has a magnitude of 9.09 units. The two vectors make an angle of 47.0° with each other. Find A with arrow · B with arrow.
Answer:
33.167
Explanation:
The dot product of two vectors is given by
A . B = A B Cos Ф
Where, A and B be the magnitudes of vector A and vector B.
So,
A . B = 5.35 x 9.09 x Cos 47
A . B = 33.167
The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on an electron with kinetic energy of 1 keV? Compare with the gravitational force on the electron.
Answer:
[tex]F = 1.5 \times 10^{-16} N[/tex]
this force is [tex]1.68 \times 10^{13}[/tex] times more than the gravitational force
Explanation:
Kinetic Energy of the electron is given as
[tex]KE = 1 keV[/tex]
[tex]KE = 1 \times 10^3 (1.6 \times 10^{-19}) J[/tex]
[tex]KE = 1.6 \times 10^{-16} J[/tex]
now the speed of electron is given as
[tex]KE = \frac{1}{2}mv^2[/tex]
now we have
[tex]v = \sqrt{\frac{2 KE}{m}}[/tex]
[tex]v = 1.87 \times 10^7 m/s[/tex]
now the maximum force due to magnetic field is given as
[tex]F = qvB[/tex]
[tex]F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})[/tex]
[tex]F = 1.5 \times 10^{-16} N[/tex]
Now if this force is compared by the gravitational force on the electron then it is
[tex]\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}[/tex]
[tex]\frac{F}{F_g} = 1.68 \times 10^{13}[/tex]
so this force is [tex]1.68 \times 10^{13}[/tex] times more than the gravitational force
Final answer:
The maximum magnetic force on an electron in Earth's magnetic field is calculated using the charge of the electron, its velocity derived from kinetic energy, and the field strength. This force is much larger than the gravitational force acting on the electron due to its minuscule mass.
Explanation:
The maximum possible magnetic force on an electron moving through Earth's magnetic field can be found using the formula F = qvB sin(\theta), where q is the charge of the electron, v is its velocity, B is the magnetic field strength, and \theta is the angle between the velocity and the magnetic field. Since the maximum force occurs when the angle is 90 degrees (sin(\theta) = 1), we first need to calculate the velocity from its kinetic energy, 1 keV. Given the kinetic energy E = 1 keV = 1 × 103 eV and using the relation E = \frac{1}{2}mv2, we can find v. The electron charge is q = -1.6 × 10-19 C, and the magnetic field is B = 0.5 G = 5 × 10-5 T.
For the gravitational force, we use F = mg, where m is the mass of the electron and g is the acceleration due to gravity, approximately 9.8 N/kg on the surface of Earth. The mass of the electron is about 9.11 × 10-31 kg. Comparing these forces, the magnetic force is several orders of magnitude greater than the gravitational force acting on an electron due to its negligible mass.
A nuclear power plant operates at 40.0% efficiency with a continuous production of 1192 MW of usable power in 1.00 year and consumes 1.22×106 g of uranium-235 in this time period. What is the energy in joules released by the fission of a single uranium-235 atom?
Final answer:
To calculate the energy released by fission of a single U-235 atom, convert the power output to total energy in joules, adjust for plant efficiency, and divide by the number of fissioned U-235 atoms, using Avogadro's number and U-235's molar mass.
Explanation:
The question involves calculating the energy released from the fission of a single uranium-235 (U-235) atom, given the power output and efficiency of a nuclear power plant. First, determine the total energy produced by the power plant in a year by converting megawatts to joules. Since 1 watt equals 1 joule per second (J/s), and there are 3.1536×107 seconds in a year, the total energy output in joules is 1192 MW × 3.1536×107 s/year × 106 W/MW. Taking into consideration the 40.0% efficiency of the plant, the total energy from fission is obtained by dividing the calculated energy output by 0.40. Finally, to find the energy per fission event, divide the total fission energy by the number of grams of U-235 consumed, converted to number of atoms using Avogadro's number, which is approximately 6.022×1023 atoms/mol. Using the atomic mass of U-235, which is roughly 235 grams per mole, we can calculate the energy per fission event.
A block slides on a horizontal surface with an initial speed of 10 m/s. If the block takes 2s to come to a stop, what is the coefficient of kinetic friction between the block and the surface? (The answer does not depend on the mass of the block.) (A) 0.1 (B) 0.2 (C) 0.4 (D) 0.5
Answer:
The coefficient of kinetic friction between the block and the surface is 0.5
Explanation:
It is given that,
Initial velocity of the block, u = 10 m/s
Time taken by the block to come to rest, t = 2 s
So, final velocity, v = 0
We need to find the coefficient of kinetic friction between the block and the surface. According to second law of motion :
F = ma
And friction force F = -μmg
i.e.
[tex]\mu mg=ma[/tex]
[tex]\mu=\dfrac{a}{g}[/tex]...........(1)
Firstly, we will find the value of a i.e. acceleration
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{0-10\ m/s}{2\ s}[/tex]
a = -5 m/s²
So, equation (1) becomes :
[tex]\mu=\dfrac{5\ m/s^2}{9.8\ m/s^2}[/tex]
[tex]\mu=0.5[/tex]
So, the coefficient of kinetic friction between the block and the surface is 0.5. hence, this is the required solution.
Consider an electric field perpendicular to a work bench. When a small charged sphere of mass 3.62 g and charge −19.9 µC is carefully placed in the field, the sphere is in static equilibrium. What are the magnitude and direction of the electric field? (Give the magnitude in N/C.)
Answer:
1782.7 N/C, downward
Explanation:
Since the charged sphere is in static equilibrium, it means that the electric force acting is balanced with the weight of the sphere, so we can write:
[tex]F_G = F_E\\mg = qE[/tex]
where
m = 3.62 g = 0.00362 kg is the mass of the sphere
g = 9.8 m/s^2 is the acceleration of gravity
[tex]q=19.9 \mu C= 19.9\cdot 10^{-6}C[/tex] is the magnitude of the charge
E is the magnitude of the electric field
Solving for E,
[tex]E=\frac{mg}{q}=\frac{(0.00362 kg)(9.8 m/s^2)}{19.9\cdot 10^{-6} C}=1782.7 N/C[/tex]
In order for the sphere to be in equilibrium, the electric force must be in opposite direction to the weight, so the electric force must point upward. Since the sphere has a negative charge, the electric field has opposite direction to the electric force: so, the electric field direction is downward.
The magnitude of the electric field is 1782.7 N/C. The direction of the electric field is in the downward direction due to the negative charge on the charged sphere.
What is the magnitude of an electric field?The magnitude of an electric field refers to the force in each electric charge on the test charge.
From the parameters given:
The static equilibrium indicates that the electric field is in a balanced position.Thus, the magnitude of the electric field can be computed by using the formula:
[tex]\mathbf{E =\dfrac{F}{q}}[/tex]
[tex]\mathbf{E =\dfrac{m \times g}{q}}[/tex]
[tex]\mathbf{E =\dfrac{0.00362 \ kg \times 9.8 \ m/s^2}{19.9\times 10^{-6} C}}[/tex]
E = 1782.71 N/C
However, the direction of the electric field is in the downward direction due to the negative charge on the charged sphere.
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A soccer player kicks a ball horizontally off a 40.0-m-high cliff into a pool of water. If the player hears the sound of the splash 3.00 s later, what was the initial speed given to the rock? Assume the speed of sound in air is 343 m/s. Determine the speed of the ball immediately before it reaches the pool of water.
the speed of the ball before it reaches the pool of water would be 9.91 m/s.
The initial speed given to the ball is 32.35m/s.
The speed of the ball immediately before it reaches the pool of water is 28m/s
What is Speed ?Speed is a rate of change of distance with respect to time. i.e. v=dx÷dt. Speed can also be defined as distance over time i.e. speed= distance ÷ time it is denoted by v and its SI unit is m/s. it is a scalar quantity. Speed shows how much distance can be traveled in unit time.
To find dimension for speed is, from formula
Speed = Distance ÷ Time
Dimension for distance is [L¹] ,
Dimension for Time is [T¹],
Dividing dimension of distance by dimension of time gives,
[L¹] ÷ [T¹] = [L¹T⁻¹]
Dimension for speed is [L¹T⁻¹].
In this problem,
Given,
Hight of the cliff y = 40m
speed of the sound = 343 m/s
vertical direction,
y = u(y)t + 1/2 at² where u(y) is initial velocity of ball along y direction which is zero.
-40 = 0t + 1/2 (-9.8)*t²
80 = 9.8t²
t = 2.85s
time taken by sound to hear is = 3-2.85 = 0.15s
distance covered by the sound,
343m/s × 0.15s = 51.45m
By applying Pythagoras theorem,
AC = √(x² + y²)
51.45² = x²+40²
x = 32.35
The initial velocity of the ball is,
u = 32.35m / 2.85s = 11.35m/s
The speed of the ball immediately before it reaches the pool of water is
v² = u² + 2as
v² = 0 +2(-9.8)40m
here initial velocity of the ball along y direction is 0. and distance along it is 40m.
v² = 784
v = 28m/s ......this is velocity along verticle.
Hence initial velocity of ball is 32.35m/s.
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A ball is whirled on the end of a string in a horizontal circle of radius R at const v. By which one of the following means can the centripetal acceleration of the b increased by a factor of two? 13. Keep the radius fixed and increase the period by a factor of two. Keep the radius fixed and decrease the period by a factor of two. Keep the speed fixed and increase the radius by a factor of two. Keep the speed fixed and decrease the radius by a factor of two. A) B) C) D)
Answer:
D
Explanation:
Centripetal acceleration is the square of velocity divided by radius:
a = v² / r
Velocity is equal to angular velocity times radius, so this can also be written as:
a = ω² r
And angular velocity is 2π divided by the period, so:
a = (2π / T)² r
We can use these two equations to determine which scenario results in double the centripetal acceleration.
A) Keep the radius fixed and increase the period by a factor of two.
(2π / 2T)² r
1/4 (2π / T)² r
1/4 a
B) Keep the radius fixed and decrease the period by a factor of two.
(2π / ½T)² r
4 (2π / T)² r
4a
C) Keep the speed fixed and increase the radius by a factor of two.
v² / 2r
1/2 a
D) Keep the speed fixed and decrease the radius by a factor of two.
v² / ½r
2a
Only D doubles the centripetal acceleration.
To double the centripetal acceleration of a ball moving in a circular path at a constant velocity, halve the radius of the path while keeping the speed constant. This is because centripetal acceleration is inversely proportional to the radius.
Explanation:This is possible because centripetal acceleration is inversely proportional to the radius (as described by the formula ac = v²/r). Conversely, increasing the radius or altering the period would not have the desired effect on the centripetal acceleration.
If the radius is halved and the speed remains constant, the force required to keep the object moving in a circular path increases, which is reflected in an increase in the centripetal acceleration. Hence, if you were to plot centripetal acceleration against radius, you would see it decrease as the radius increases and vice versa.
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