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bagirrra123 [75]

3 years ago

5

What has more momentum a large truck stopped at a stop sign or a motorcycle moving down the road.

1 answer:

Alona [7]3 years ago

4 0

<h2>I'm pretty sure it's</h2><h2>"<u>a motorcycle moving down the road.</u>"</h2><h3></h3><h3>The truck isn't moving so the truck cannot have momentum.</h3><h3>Momentum is basically "mass in motion."</h3><h3></h3><h3><em>Please let me know if I am wrong.</em></h3>

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How does a compound differ from a mixture?

Mashcka [7]

A compound is the substances that are formed by combining two are more chemical elements. A mixture is a substance created from two or more matter that can be separate with the help of physical methods. ... Mainly pure water is part of the compound. Mixtures fall under impure water.

Hope this helps!

Have a great day!

4 0

3 years ago

Read 2 more answers

Can someone help me?

Leviafan [203]

Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.

In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

$V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t$

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

$V_y.t = \frac{2d}{t} .t = 2d$

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.

4 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:

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

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

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!

7 0

3 years ago

mezya [45]

Choose answer C.....

4 0

3 years ago

From an ethical standpoint, which best describes why new biotechnology products should be tested extensively before they are all

Gnoma [55]

Its actually A. I did the same question and A was the correct answer.

7 0

3 years ago