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3 years ago

What is the actual depth of rainfall shown in this rain gauge?

Physics

2 answers:

ehidna [41]3 years ago

6 0

Answer:

The answer is D.

Explanation:

As shown in the picture, the measuring tube is collecting 10 cm of rain.

tensa zangetsu [6.8K]3 years ago

5 0

Answer:

D. 10cm

Explanation:

If you look closely, you can see the water is at the 10cm mark.

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8. (Liquids, Gas) have no definite shape, and no definite volume. A)Liquid B)Gas

Fynjy0 [20]

B. gas has no definite shape or volume. liquid has definite volume but no definite shape

4 0

2 years ago

4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two

Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=mV_{o}+MU_{o}$ (2)

$p_{f}=(m+M)V_{f}$ (3)

$m=1200 kg$ is the mass of the first car

$V_{o}=20 m/s$ is the velocity of the first car, to the North

$M=1400 kg$ is the mass of the second car

$U_{o}=-22 m/s$ is the mass of the second car, to the South

$V_{f}$ is the final velocity of both cars after the collision

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg}$ (6)

Finally:

$V_{f}=-2.61 m/s$ (7) This is the resulting velocity of the wreckage, to the south

7 0

4 years ago

Notice that all the initial spring potential energy was transformed into gravitational potential energy. If you compressed the s

Nostrana [21]

The question doesn't provide enough data to be solved, but I'm assuming some magnitudes to help you to solve your own problem

Answer:

The maximum height is 0.10 meters

Explanation:

Energy Transformation

It's referred to as the change of one energy from one form to another or others. If we compress a spring and then release it with an object being launched on top of it, all the spring (elastic) potential energy is transformed into kinetic and gravitational energies. When the object stops in the air, all the initial energy is now gravitational potential energy.

If a spring of constant K is compressed a distance x, its potential energy is

$\displaystyle P_E=\frac{Kx^2}{2}$

When the launched object (mass m) reaches its max height h, all that energy is now gravitational, which is computed as

$U=mgh$

We have then,

$U=P_E$

$\displaystyle mgh=\frac{Kx^2}{2}$

Solving for h

$\displaystyle h=\frac{Kx^2}{2mg}$

We have little data to work on the problem, so we'll assume some values to answer the question and help to solve the problem at hand

Let's say: x=0.2 m (given), K=100 N/m, m=2 kg

Computing the maximum height

$\displaystyle h=\frac{(100)0.2^2}{2(2)(9.8)}$

$\displaystyle h=\frac{4}{39.2}=0,10\ m$

The maximum height is 0.10 meters

8 0

3 years ago

What is the frame of reference for a plane moving at 500 km/h?

RUDIKE [14]

The frame of reference is the rotating earth underneath the flight path of the plane, the rate of rotation of the earth is 1036 miles per hour, meaning that the pilot has to compensate this fact when landing the aircraft.

3 0

3 years ago

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Natural selection allows populations to (2 points)

Dahasolnce [82]

adapt to a changing environment

organisms that possess heritable traits that enable them to better adapt to their environment

3 0

3 years ago