All categories

4 years ago

Does mass affect how long it takes an object to reach the ground from a high drop height?

1 answer:

s2008m [1.1K]4 years ago

3 0

If there is some outside resistance, like air resistance then yes, but if it is in a vacuum with no resistance, then mass wouldn't matter.

You might be interested in

What water pressure must a pump that is located on the first floor supply to have water on the thirteenth of a building with a p

irga5000 [103]

The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

The given parameters;

<em>Pressure on the 13 th floor, P₁ = 35 PSI</em>
<em>Distance between each floor, d = 10 ft</em>

The vertical pressure of the water is calculated as follows;

$P = \rho gh\\\\\frac{P}{h} = \rho g\\\\\frac{P}{h} = k\\\\\frac{P_1}{h_1} = \frac{P_2}{h_2} \\\\$

The vertical height of the first floor from the 13th floor = 130 ft

The vertical height of the 13 ft floor = 10 ft

$P_1 = \frac{P_2 h_1}{h_2} \\\\P_1 = \frac{35 \times 130}{10} \\\\P_1 = 455 \ PSI$

Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

Learn more about vertical height and pressure here: brainly.com/question/15691554

3 0

3 years ago

irga5000 [103]

Answer:

Explanation:

solution is found below

6 0

3 years ago

Please help!?!?

Vanyuwa [196]

Since bulb is connected in the closed circuit at the position of D

as well as switch B is also closed in that position so the current will flow through the bulb and bulb will glow in that position

So the most appropriate correct option will be

D. The light bulb will be on

5 0

4 years ago

Read 2 more answers

What is an invasive species plant??

lianna [129]

Japanese Honeysuckle

6 0

3 years ago

A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio

Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

$F_c = m\frac{v^2}{r}$

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

$v = r\omega$

with ω denoting the angular velocity, which we are given. With that, the above becomes:

$F_c = m\frac{v^2}{r}=m\omega^2 r$

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

$F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}$

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

$\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}$

and so 45 rev/min = 4.71 rad/s.

$\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27$

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.

3 0

3 years ago