A potter's wheel moves uniformly from rest to an angular speed of 0.20 rev/s in 32.0 s. (a) Find its angular acceleration in radians per second per second. rad/s2 (b) Would doubling the angular acceleration during the given period have doubled final angular speed?

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.

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.

Answer:

[tex]\frac{dB}{dt} = 11.55 T/s[/tex]

Explanation:

As we know that flux of magnetic field from the closed loop is given by

[tex]\phi = BAcos30[tex]

now by Faraday's law we know

[tex]EMF = \frac{d\phi}{dt}[/tex]

now we will have

[tex]EMF = Acos30\frac{dB}{dt}[/tex]

now we have

[tex]A = 1 m^2[/tex]

EMF = 10 Volts

[tex]10 = 1 (cos30)\frac{dB}{dt}[/tex]

[tex]\frac{dB}{dt} = \frac{20}{\sqrt3}[/tex]

[tex]\frac{dB}{dt} = 11.55 T/s[/tex]

Answer with Explanations:

Given:

mass of block, m = 0.244 kg

spring constant, k = 5075 N/m

compression distance in spring, x = 0.106 m

Find how high mass reaches after release (measured from point of release)

Solution:

Use conservation of energy, assuming a light spring (of negligible mass)

stored spring potential energy = gravitational potential energy of block

PE1 = PE2

PE1 = kx^2/2 = 5075*0.106^2/2 = 28.512 J

PE2 = mgh = 0.244 * 9.8 * h = 2.3912 h J

Solve for h = 28.512 / 2.3912 = 11.923 m

To find the maximum height a block rises after being released from a compressed spring, we use the conservation of energy principle to equate the initial spring potential energy to the gravitational potential energy at the peak height, then solve for that height.

The problem involves applying the conservation of energy principle to a mass-spring system when the block is released from compression. Initially, all the energy is stored as potential energy in the compressed spring (spring potential energy), and when the spring is uncompressed, that energy will be converted into kinetic energy of the mass. Finally, when the block leaves the spring and moves upward, its kinetic energy will be converted into gravitational potential energy until the block comes to a stop at its maximum height.

Let's examine the steps needed to solve the problem:

Let's perform these calculations for the provided problem:

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]

(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.

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.

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.

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.

Answer:

Gravity on the moon, g = 1.69 m/s²

Explanation:

It is given that,

Length of pendulum, l = 1 m

Time period, T = 4.82 seconds

We have to find the gravity of the moon. The time period of the pendulum is given by :

[tex]T=2\pi\sqrt{\dfrac{l}{g}}[/tex]

g = acceleration due to gravity on moon

[tex]g=\dfrac{4\pi^2l}{T^2}[/tex]

[tex]g=\dfrac{4\pi^2\times 1\ m}{(4.82\ s)^2}[/tex]

g = 1.69 m/s²

Hence, the gravity on the moon is 1.69 m/s².

Answer:

Gravity on the moon, g = 1.69 m/s²

Explanation:

If the pendulum of length is 1.0 m is pulled to the side and released on the moon and It's period is measured to be 4.82 seconds, the gravity on the moon is 1.69 m/s².

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

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]

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 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

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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The average force exerted by the floor on the ball during the collision is zero.

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.

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.

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.

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

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.

The magnitude of an electric field refers to the force in each electric charge on the test charge.

From the parameters given:

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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Answer:

[tex]4.41\cdot 10^5 N[/tex]

Explanation:

First of all, let's calculate the total mass of the train+the engines:

[tex]m=2(8.00\cdot 10^4 kg) + 45(5.50\cdot 10^4 kg) = 2.64\cdot 10^6 kg[/tex]

Then we can apply Newton's second law, which states that the resultant of the forces is equal to the product between mass (m) and acceleration (a):

[tex]\sum F = ma[/tex] (1)

In this case there are two forces:

- The pushing force exerted by the engines, F

- The frictional force, [tex]F_f = 7.50 \cdot 10^5 N[/tex], in an opposite direction to the acceleration

So (1) becomes

[tex]F-F_f = ma[/tex]

Since the acceleration must be

[tex]a=5.00\cdot 10^{-2} m/s^2[/tex]

We  can solve the formula to find F:

[tex]F=ma+F_f = (2.64\cdot 10^6 kg)(5.00\cdot 10^{-2} m/s^2) + 7.50 \cdot 10^5 N = 8.82\cdot 10^5 N[/tex]

However, this is the force exerted by both engines. So the force exerted by each engine must be half this value:

[tex]F=\frac{8.82\cdot 10^5 N}{2}=4.41\cdot 10^5 N[/tex]

Explanation:

It is given that,

Mass of the car, m = 1200 kg

Radius of circular path, r = 84 m

The coefficient of static friction between tires and road is 0.60

Static friction is given by :

[tex]f=\mu N=\mu mg[/tex]..............(1)

When an object is moving in a circular path, the centripetal force will act on it. It is given by :

[tex]F=\dfrac{mv^2}{r}[/tex]

Let v is the maximum speed of the car. From equation (1) and (2) as :

[tex]v=\sqrt{\mu gr}[/tex]

g = acceleration due to gravity

[tex]v=\sqrt{0.6\times 9.8\ m/s^2\times 84\ m}[/tex]

v = 22.22 m/s

or

v = 22 m/s

Hence, this is the required solution.

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]

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.

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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Answer:

B. Remain the same

Explanation:

Answer:

Double the voltage across it.- last choice

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.

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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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

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

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

Answer:

g

Explanation:

gg