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natta225 [31]
3 years ago
12

a 5.0 kg tool box is raised from the ground by a rope. if the upward acceleration of the bucket is 2.5 m/s^2 , find the force ex

erted by the rope on the tool box. acceleration due to gravity is 9.8 m/s^2
Physics
1 answer:
denpristay [2]3 years ago
3 0

Answer:

62 N

Explanation:

Sum of the forces on the toolbox:

∑F = ma

T − mg = ma

T = mg + ma

T = m (g + a)

T = (5.0 kg) (9.8 m/s² + 2.5 m/s²)

T = 61.5 N

Rounded to two significant figures, the force exerted by the rope is 62 N.

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du
ELEN [110]

With constant angular acceleration \alpha, the disk achieves an angular velocity \omega at time t according to

\omega=\alpha t

and angular displacement \theta according to

\theta=\dfrac12\alpha t^2

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}

b. Under constant acceleration, the average angular velocity is equivalent to

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2

where \omega_f and \omega_i are the final and initial angular velocities, respectively. Then

\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}

c. After 1.00 s, the disk has instantaneous angular velocity

\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}

d. During the next 1.00 s, the disk will start moving with the angular velocity \omega_0 equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle \theta according to

\theta=\omega_0t+\dfrac12\alpha t^2

which would be equal to

\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}

5 0
4 years ago
Help please I don't understand this
BARSIC [14]
It's either staying there or is going at the same pace
4 0
4 years ago
2. A person began running due east and covered 15 kilometers in 2.0 hours. What is the average velocity of the person?
Angelina_Jolie [31]

Answer:

7.5 km/h (2.1 m/s) due east

Explanation:

The average velocity of the person is given by:

v=\frac{d}{t}

where

d is the displacement

t is the time taken

In this problem,

d = 15 km is the displacement

t = 2.0 h is the time elapsed

so the average velocity is

v=\frac{15 km}{2.0 h}=7.5 km/h

and the direction is the same as the displacement (east).

We can also convert the velocity into SI units (m/s). We have:

d = 15 km = 15,000 m

t = 2.0 h * 3600 s/h = 7200 s

v=\frac{15,000 m}{7200 s}=2.1 m/s

4 0
3 years ago
Read 2 more answers
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