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Alenkinab
2 months ago
13

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

ation α. The flywheel is assumed to be at rest at time t=0 in Parts A and B of this problem.
Part A

Find the time t1 it takes to accelerate the flywheel to ω1 if the angular acceleration is α.

Express your answer in terms of ω1 and α.

Part B

Find the angle θ1 through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity ω1.

Express your answer in terms of some or all of the following: ω1, α, and t1.

Part C

Assume that the motor has accelerated the wheel up to an angular velocity ω1 with angular acceleration α in time t1. At this point, the motor is turned off and a brake is applied that decelerates the wheel with a constant angular acceleration of −5α. Find t2, the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero angular velocity).

Express your answer in terms of some or all of the following: ω1, α, and t1.
Physics
1 answer:
Maru [3.3K]2 months ago
6 0

Answer:

Part a)

t_1 = \frac{\omega_1}{\alpha}

Part b)

\theta = \frac{1}{2}\alpha t_1^2

Part c)

t = \frac{t_1}{5}

Explanation:

Part a)

Since there is a consistent torque, the time required to achieve acceleration is expressed as

\omega_f = \omega_i + \alpha t

\omega_1 = 0 + \alpha t_1

t_1 = \frac{\omega_1}{\alpha}

Part b)

Angular displacement can be determined as

\theta = \omega_i t_1 + \frac{1}{2}(\alpha) t_1^2

\theta = 0 + \frac{1}{2}\alpha t_1^2

\theta = \frac{1}{2}\alpha t_1^2

Part c)

The brakes create a certain angular deceleration that is calculated as

\alpha_d = - 5\alpha

Now we find

\omega_f = \omega _i + \alpha t

0 = \omega_1 - 5 \alpha_1 t

t = \frac{\omega_1}{5 \alpha}

Moreover, we know that

t_1 = \frac{\omega_1}{\alpha}

therefore, we derive

t = \frac{t_1}{5}

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The colored lines in the figure represent paths taken by different people walking around in a city. Assume that each city block
serg [3582]

Answer: 592.37m

Explanation:

Person D is represented by the blue line.

The total displacement is calculated by subtracting the initial position from the final position. Starting at (0,0), the path consists of moving down two blocks, then right six blocks, followed by moving up four blocks, and finally left one block.

Here, I consider the positive direction of the x-axis to the right and the positive direction of the y-axis as upward.

Thus, the new coordinates will be, with B representing a block:

P =(6*B - 1*B, -2*B + 4*B) = (5*B, 2*B)

Given that B = 110m

P = (550m, 220m)

The displacement corresponds to the length of the vector, since the change from the initial position (0,0) to P is simply P:

P = √(550^2 + 220^2) = 592.37m

4 0
2 months ago
Two infinite parallel surfaces carry uniform charge densities of 0.20 nC/m2 and -0.60 nC/m2. What is the magnitude of the electr
Yuliya22 [3333]

Answer:

The electric field strength, E = 45.19 N/C

Explanation:

It is indicated that,

Surface charge density on the first surface, \sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2

Surface charge density on the second surface, \sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2

The electric field at a location between the two surfaces can be calculated as:

E=\dfrac{\sigma}{2\epsilon_o}

E=\dfrac{\sigma1-\sigma_2}{2\epsilon_o}

E=\dfrac{0.2\times 10^{-9}-(-0.6\times 10^{-9})}{2\times 8.85\times 10^{-12}}

Consequently, E = 45.19 N/C

Therefore, the electric field's magnitude at a point between both surfaces is 45.19 N/C.

6 0
1 month ago
A 30 cm wrench is used to loosen a bolt with a force applied 0.3 mm from the bolt. It takes 60 N to loosen the bolt when the for
Yuliya22 [3333]

Answer:

the torque necessary to loosen the bolt when at a right angle was:

t = 60N x 0.3 = 18 Nm

To achieve the equivalent torque applied at a 30-degree angle:

18 Nm = Force x 0.3cos(30 degrees)

Simplifying yields:

18 Nm = Force x 0.2598

Thus, Force = 18Nm / 0.2598

this results in Force = 69.28 N

Adjust the answer as necessary.

6 0
1 month ago
If you accidentally touch the "hot" wire connected to the 120 V line, how much current will pass through your body?
serg [3582]

Complete Question:

Picture yourself on an aluminum ladder on the ground, attempting to fix an electrical connection with a metal screwdriver featuring a metallic handle. Since you are sweating profusely, your body has a resistance of 1.60 kΩ.

(a) If you accidentally contact the "hot" wire from the 120 V power line, what current will flow through your body?

(b) What is the amount of electrical power transferred to your body?

Answer:

(a) 0.075A

(b) 9W

Explanation:

According to Ohm's law, the voltage (V) applied to or passing through a body corresponds to the current (I) via the relationship:

V ∝ I

=> V = I x R                 ----------------------(i)

Where;

R denotes the resistance of the body

(a) As mentioned;

Due to wet conditions, the body will conduct electricity, and possesses the following values;

V = supplied voltage = 120V

R = resistance of the wet body = 1.60kΩ = 1.6 x 1000Ω = 1600Ω

Substituting these values into equation(i):

120 = I x 1600

To find I;

I = \frac{120}{1600}

I = 0.075A

Thus the current passing through your body is 0.075A

(b) Electrical power (P), which is expressed in Watts (W), delivered to the body is the product of current (I) and voltage (V) received. Thus:

P = I x V           ---------------------(ii)

Where;

I equates to 0.075A   [as derived above]

V is 120V     [as outlined in the question]

Plugging these values into equation (ii):

P = 0.075 x 120

P = 9W

Hence, the electrical power received by your body is 9W

7 0
1 month ago
A balloon was filled to a volume of 2.50 l when the temperature was 30.0∘c. What would the volume become if the temperature drop
Keith_Richards [3271]

Answer:

2.34 L

Explanation:

Assuming the pressure within the balloon is held steady, we can apply Charle's law which states that for a gas at constant pressure, the ratio of its volume to temperature remains unchanged:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

In our scenario:

V_1 = 2.50 L is the initial volume

V_2 is the final volume

T_1 = 30.0^{\circ}C+273 = 303 K is the starting temperature

T_2 = 11.0^{\circ}C+273 = 284 K is the concluding temperature

By plugging values into the equation and resolving for V2, we can calculate the final volume:

V_2 = \frac{V_1 T_2}{T_1}=\frac{(2.50 L)(284 K)}{303 K}=2.34 L

4 0
2 months ago
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