Ask question

Ask question

All categories

3 years ago

12

A disk initially rotated counterclockwise at 1.0 rad/s, but has a counterclockwise angular acceleration of 0.50 rad/s^2 for 2 se

conds. After this acceleration, the disk is at an angle of 6.0 rad.
What is the disk’s angular position when the acceleration started?

1 answer:

Mumz [18]3 years ago

8 0

Answer:

Initial angular position of the disc is 3 radian

Explanation:

As we know that the angular acceleration of the disc is given as

$\alpha = 0.50 rad/s^2$

initial angular speed is given as

$\omega = 1 rad/s$

now we know that angular displacement of the disc is given as

$\theta = \omega t + \frac{1}{2}\alpha t^2$

$\theta = 1(2) + \frac{1}{2}(0.5)(2^2)$

$\theta = 3 rad$

now we have

$\theta_f - \theta_i = 3 rad$

$6 rad - \theta_i = 3$

$\theta_i = 3 ard$

You might be interested in

What experimental evidence led to the development of this atomic model from the one before it?

Marina86 [1]

Answer:

A few of the positive particles aimed at a gold foil seemed to bounce back

Explanation:

7 0

3 years ago

Read 2 more answers

In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys

nydimaria [60]

Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J

Solution :

(1) The equation used is,

$\Delta U=q+w\\\\U_{final}-U_{initial}=q+w$

where,

$U_{final}$ = final internal energy

$U_{initial}$ = initial internal energy

q = heat energy

w = work done

(2) The known variables are, q, w and $U_{initial}$

initial internal energy = $U_{initial}$ = 2000 J

heat energy = q = 1000 J

work done = w = 500 J

(3) Now plug the numbers into the equation, we get

$U_{final}-(2000J)=(1000J)+(500J)$

(4) By solving the terms, we get

$U_{final}-(2000J)=(1000J)+(500J)$

$U_{final}-(2000J)=1500J$

$U_{final}=2000J+1500J$

$U_{final}=3500J$

(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J

5 0

3 years ago

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

$\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]$

$\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]$

$v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) }$ (2)

If we know that $P_{1}-P_{2} = 0\,Pa$ , $\rho=1000\,\frac{kg}{m^{3}}$ , $v_{1} = 15.5\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $z_{1}-z_{2} = -3.5\,m$ , then the speed of the water flow in the pipe on the second floor is:

$v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}$

$v_{2} \approx 13.100\,\frac{m}{s}$

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

4 0

3 years ago

A 2-kg wheel rolls down the road with a linear speed of 15m/s. Find its trwansitional and rotational kinetic energies.

Elza [17]

Answer:

The translational kinetic energy is 225 J

The rotational kinetic energy is 225 J

Explanation:

Given;

mass of the wheel, m = 2-kg

linear speed of the wheel, v = 15 m/s

Transnational kinetic energy is calculated as;

E = ¹/₂MV²

where;

M is mass of the moving object

V is the velocity of the object

E = ¹/₂ x 2 x (15)²

E = 225 J

Rotational kinetic energy is calculated as;

E = ¹/₂Iω²

where;

I is moment of inertia

ω is angular velocity

$E = \frac{1}{2} I \omega^2\\\\E = \frac{1}{2} *mr^2*(\frac{v}{r})^2\\\\E = \frac{1}{2} *mr^2*\frac{v^2}{r^2} \\\\E = \frac{1}{2}mv^2$

E = ¹/₂ x 2 x (15)²

E = 225 J

Thus, the translational kinetic energy is equal to rotational kinetic energy

6 0

3 years ago

You are pushing a wooden crate across the floor at constant speed. You decide to turn the crate on end, reducing by half the sur

borishaifa [10]

Answer:

D. Half as great

Explanation:

Since we know that the friction force between the surface of crate and ground is given as

$F_f = \mu F_n$

so here we know that

$\mu$ = friction coefficient between two surfaces which depends on the effective contact area between two surfaces

$F_n$ = normal force due to the object

So when we turn the object on another side such that the surface area is half then the friction coefficient will become also half

So here the friction force will also reduce to half

so correct answer will be

D. Half as great

3 0

3 years ago