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Lelechka [254]
3 years ago
13

A bubble at the bottom of lake rises to surface within 10.0 seconds with an acceleration of 10.0 meters per second. What is the

depth of the lake?
Physics
1 answer:
amm18123 years ago
5 0
If the bubble travels 10 meters per second and it takes 10 seconds, then just multiply the distance per second by the total seconds to get the total depth.
10 • 10 = 100
The lake is 100 meters deep.

Think of it this way to clarify the answer:
It takes a bubble traveling at a speed of 10 meters per second 10 seconds to travel 100 meters.
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4 0
2 years ago
A particle with a charge of 5 × 10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a sta
creativ13 [48]

Answer:

r = 0.22m

Explanation:

To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.

Then, you have:

F_c=F_e=ma_c      (1)

m: mass of the particle = 20g = 20*10-3 kg

ac: centripetal acceleration = ?

q: charge of the particle = 5*10^-6C

Fe: electric force between the charges

The electric force is given by:

F_e=k\frac{qq'}{r^2}             (2)

r: radius of the orbit

q': charge of the particle at the center of the orbit = -5*10^-6C

Furthermore, the centripetal acceleration is:

a_c=\frac{v^2}{r}                 (3)

v: speed of the particle = 7m/s

You replace the expressions (2) and (3) in the equation (1) and solve for r:

k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}

Finally, you replace the values of all parameters in the previous expression:

r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m

The radius of the circular trajectory is 0.22m

5 0
3 years ago
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