Which of the following are voltage sourWhich of the following are voltage sources?

A bowling ball of 35.2 kg, generates 218 kg* m/s units of momentum. What is the velocity of the ball?

Aleonysh [2.5K]

Answer:

6.19 m/s

Explanation:

$p = mv \\ 218 = (35.2)(v ) \\ v = 6.19 \: ms {}^{ - 1}$

7 0

2 years ago

For an object to be in static equilibrium, the net force must be equal to zero and and The center of gravity must be above the s

iren [92.7K]

Answer:

The normal force must equal the weight.

8 0

3 years ago

Definition for compression and rarefraction

s344n2d4d5 [400]

Compression and rarefaction are two phenomenon occurs in longitudunal wave!

when there is denser particle gathering in that wave , there we called it compression and the rarer part of particles is rarefaction !

6 0

2 years ago

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A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$

v is the speed of cat, $v=\dfrac{2\pi r}{t}$

$\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29$

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0

2 years ago

The electric field everywhere on the surface of a thin, spherical shell of radius 0.770 m is of magnitude 860 N/C and points rad

11111nata11111 [884]

Answer:

(a) $Q = 7.28\times 10^{14}$

(b) The charge inside the shell is placed at the center of the sphere and negatively charged.

Explanation:

Gauss’ Law can be used to determine the system.

$\int{\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}\\(860)4\pi(0.77)^2 = \frac{Q_{enc}}{8.8\times 10^{-12}}\\Q_enc = 7.28\times 10^{14}$

This is the net charge inside the sphere which causes the Electric field at the surface of the shell. Since the E-field is constant over the shell, then this charge is at the center and negatively charged because the E-field is radially inward.

The negative charge at the center attracts the same amount of positive charge at the surface of the shell.

5 0

3 years ago