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1. In which direction can a force be exerted on an object?

A. Any Direction
This should be obvious, as you can push down on something, push it to the sides, and push something up. Hope this helped!

3 0

4 years ago

13. A lever does 5.0 J of work on a 0.10-kg ball bearing in a pinball machine. The ball's

shusha [124]

Answer:

The resulting speed of the ball is;

10 m/s

Explanation:

The given parameters are;

The amount of work done by the lever = 5.0 J

The mass of the ball on which the work is done, m = 0.10 kg

By the principle of conservation of energy, we have;

The work done by the lever = The kinetic energy. K.E., gained by the ball

The kinetic energy, K.E., is given by K.E. = 1/2·m·v²

Where, for the ball, we have;

m = The mass of the ball

v = The resulting speed of the ball

Therefore, by substituting the known values, we have;

The work done by the lever = 5.0 J = The kinetic energy. K.E., gained by the ball = 1/2 × m × v² = 1/2 × 0.10 kg × v²

∴ 5.0 J = 1/2 × 0.10 kg × v²

v² = 5.0 J/(1/2 × 0.10 kg) = 100 m²/s²

v = √(100 m²/s²) = 10 m/s

The resulting speed of the ball, v = 10 m/s.

6 0

3 years ago

What does the balloon of the air capacitor represent in an electrical capacitor?

bulgar [2K]

Answer:

The balloon prohibits the flow of air through the air capacitor.

Explanation:

Just like an electric capacitor has an insulator between the plates, the air capacitor has a balloon between the chambers.

6 0

3 years ago

Read 2 more answers

A bike race against the clock takes place on a straight road. Yan drives at 37 km / h and he starts the course 30s before Christ

joja [24]

Given data:

Yan speed;

$u_1=37\text{ km/h}$

Christopher speed;

$u_2=38.9\text{ km/h}$

Christophe starts 30 s later than Yan. Therefore, Christophe takes 30 s less than Yan to reach the same distance.

Part (A)

The distance is given as,

$d=ut$

Let both Yan and Christophe meet at d distance from the start position. Therefore,

$u_1t=u_2(t-30)$

Substituting all known values,

$\begin{gathered} (37\text{ km/h})t=(38.9\text{ km/h})\times(t-30) \\ \frac{(37\text{ km/h})}{(38.9\text{ km/h})}=\frac{(t-30)}{t} \\ 0.95=1-\frac{30}{t} \\ \frac{30}{t}=1-0.95 \\ \frac{30}{t}=0.05 \\ t=\frac{30}{0.05} \\ t=600\text{ s} \end{gathered}$

Therefore, 600 s after Yan's departure Christophe will join him.

Part (B)

The distance is given as,

$d=u_1t$

Substituting all known values,

$\begin{gathered} d=(37\text{ km/h})\times(600\text{ s}) \\ =(37\text{ km/h})\times(600\text{ s})\times(\frac{1\text{ hr}}{3600\text{ s}}) \\ \approx6.17\text{ km} \end{gathered}$

Therefore, Christophe joins Yan after 6.17 km from the start.

3 0

1 year ago

It takes 185 kj of work to accelerate a car from 23.0 m/s to 28.0 m/s. what is the car's mass?

Pachacha [2.7K]

<span>This question is based on conservation of energy as the work done would lead to change in kinetice energy of car change in KE = 1/2 mv(f)^2 - 1/2mv(i)^2 = 1/2m(v(f)^2-v(i)^2) where v(f) and v(i) are the final and initial speeds change in KE = 185kJ = 185,000J = 1/2 m((28m/s)^2-(23m/s)^2) 185,000=1/2 m(255m^2/s^2) solving for m m=1451kg</span>

6 0

3 years ago