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3 years ago

Collision 2: Blue Cart Moving Slower than the Red Cart

Physics

1 answer:

pshichka [43]3 years ago

3 0

Answer:

this is what i always used

Explanation:

http://www.sjutsscience.com/uploads/3/7/4/5/37458459/phet_collision_lab_2020_key.pdf

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A 43-N crate is suspended from the left end of a plank. The plank weighs 21 N, but it is not uniform, so its center of gravity d

lesya [120]

Answer:

$d_{2}=0.585m$

Explanation:

The torque is the force by the distance so to determinate that both torque are the same magnitude so

$T_{1}-T_{2}=0$

$T_{1}=T_{2}$

$F_{1}*d_{1}=F_{2}*d_{2}$

$43N*0.30m=21N*d_{2}$

Solve to d2

$d_{2}=\frac{41N*0.30m}{21N}$

$d_{2}=0.585m$

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3 years ago

A car speeds up from 0 m/s to 14.0 m/s in 325 s. What is the acceleration?

Anton [14]

Answer: $0.04308 m/s^2$

Explanation: Plug the variables into the equation: $a=\frac{v-v_{0} }{t}$ | $v$ equals initial speed, $v_{0}$ equals final speed and $t$ equals time.

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3 years ago

Imagine that a tank is filled with water. The height of the liquid column is 7 meters and the area is 1.5 square meters (m2). Wh

Crazy boy [7]

Answer:

B. 102,900 N

I remember this question from one of my, test. Really hope this helps..

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3 years ago

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What did the moon of Titan contain that is of interest to scientists?

statuscvo [17]

The answer to this question is C. Lakes of hydrocarbons. Trust me this is the answer before i got the results i thought it was lakes of mercury.

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3 years ago

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the average power of the Sun is 3.79x1026 Watts, answer the following questions: 13. What is the average intensity of light at E

Lemur [1.5K]

Explanation:

Given that,

Average power of sun $P=3.79\times10^{26}\ Watt$

We need to calculate the intensity of light at Earth's position

Using formula of intensity

$I=\dfrac{P_{avg}}{4\pi r^2}$

Where, I = intensity

P = power

Put the value into the formula

$I=\dfrac{3.79\times10^{26}}{4\pi\times(1.496\times10^{11})^2}$

$I=1347.616\ W/m^2$

So, The intensity is 1347.616 W/m².

(A). We need to calculate the pressure on a solar sail due to the light of the sun if it's fully reflective

Using formula for fully reflective

$P = \dfrac{2I}{c}$

Put the value into the formula

$P=\dfrac{2\times1347.616}{3\times10^{8}}$

$P=8.984\times10^{-6}\ N/m$

(B). We need to calculate the pressure on a solar sail due to the light of the sun if it's fully reflective

Using formula for fully absorptive

$P=\dfrac{I}{c}$

$P=\dfrac{1347.616}{3\times10^{8}}$

$P=4.492\times10^{-6}\ N/m$

Hence, This is the required solution.

6 0

3 years ago