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kirza4 [7]
3 years ago
5

On a far-away (hypothetical) planet, a student is conducting an experiment to study vertical motion. The student fires a cannon

directly upward, and then records the position of the cannonball every second until it falls back to the ground. Their data is shown in the table below. Based on their data below, answer the following two questions:
1) What is the acceleration due to gravity on this planet? Clearly show how you calculated this.

2) Is there significant air resistance affecting the experiment? Write an evidence-based argument to justify your claim.

- One of the pieces of evidence that you use must be a graph based on the given data. The quantities on the graph are up to you; they can be position and time themselves, or different quantities that you calculate from the given position and time data.

- If you'd like, you can also include outside evidence from your own lab or research. This is not required; only include it if it makes your argument stronger.

Answer both questions on one document. The document can be digital, hand-written, or a mix of both (copy and paste images of your handwritten work into a doc).

Time (s) Position (m)
0 0
1 95
2 170
3 225
4 260
5 275
6 270
7 245
8 200
9 135
10 50
Physics
1 answer:
IgorC [24]3 years ago
3 0
Yeah it’s pretty much positive
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Answer:

When the Earth and sun are perfectly lined up, then it will happen. They can tell when it's going to happen.

Explanation:

This is why it only happens in some places. Some days it's not sunny out, so it's not going to happen.

5 0
3 years ago
Determine the Mutual Inductance per unit length between two long solenoids, one inside the other, whose radii are r1 and r2 (r2
Triss [41]

Answer:

M' = μ₀n₁n₂πr₂²

Explanation:

Since r₂ < r₁ the mutual inductance M = N₂Ф₂₁/i₁ where N₂ = number of turns of solenoid 2 = n₂l where n₂ = number of turns per unit length of solenoid 2 and l = length of solenoid, Ф₂₁ = flux in solenoid 2 due to magnetic field in solenoid 1 = B₁A₂ where B₁ = magnetic field due to solenoid 1 = μ₀n₁i₁ where μ₀ = permeability of free space, n₁ = number of turns per unit length of solenoid 1 and i₁ = current in solenoid 1. A₂ = area of solenoid 2 = πr₂² where r₂ = radius of solenoid 2.

So, M = N₂Ф₂₁/i₁

substituting the values of the variables into the equation, we have

M = N₂Ф₂₁/i₁

M = N₂B₁A₂/i₁

M = n₂lμ₀n₁i₁πr₂²/i₁

M = lμ₀n₁n₂πr₂²

So, the mutual inductance per unit length is M' = M/l = μ₀n₁n₂πr₂²

M' = μ₀n₁n₂πr₂²

3 0
3 years ago
6. Sandy adds 3 sugar packets to her tea in order to make it sweeter. She stirs in the sugar and then takes a sip. She notices t
irina1246 [14]

Answer:

C.Supersaturated

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There are three types of solution:

<u>SATURATED SOLUTION</u>:

It is the solution that contains maximum amount of solute dissolved in a solution in normal conditions.  

<u>UNSATURATED SOLUTION</u>:

It is the solution that contains less than the maximum amount of solute dissolved in a solution in normal conditions. It has space for more solute to be dissolved in it.

<u>SUPERSATURATED SOLUTION:</u>

It contains more than the maximum amount of solute dissolved in it. Such a solution has no capacity to dissolve any more solute under any conditions.

Since the sugar is no more dissolving in the tea and has settled down. Therefore, the solution is:

<u>C.Supersaturated</u>

7 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

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For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

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