Ask question

Ask question

All categories

11111nata11111 [884]

3 years ago

11

For this trajectory, what would the vertical component of acceleration for the module be at time tm=t0−σ=325s? Recall that accel

eration is the derivative of velocity with respect to time.

1 answer:

Karo-lina-s [1.5K]3 years ago

4 0

I think its half of pie so you divid that by 3.14 .

You might be interested in

Which method of separation would work on a homogeneous mixture salt water

Butoxors [25]

To separate a mixture of salt and water, you can try first by using filter paper hen with the extra water part set it out to the window so that the salt water evaporates and only the salt is remaining.

5 0

3 years ago

Read 2 more answers

Timmy drove 2/5 of a journey at an average speed of 20 mph.

mixer [17]

Answer:

4hr

Explanation:

5 0

3 years ago

Read 2 more answers

A bus is moving at a constant speed of 40 m/s. How many hours will it takes to travel 260 miles?

lawyer [7]

Explanation:

1 mile = 1609m

Distance=260 miles= 418340 m

Speed= 40 m/s

Time = distance/speed

= 418340/40

=10458.5 seconds

= 2.9 hours

8 0

3 years ago

TRUE OR FALSE <br> The nearby galaxy in the spectrum below is moving toward the close star.

katen-ka-za [31]

Answer:

i think its false tell me if im wrong

Explanation:

5 0

4 years ago

Two cars, one of mass 1100 kg, and the second of mass 2500 kg, are moving at right angles to each other when they collide and st

Pie

Answer:

$v_{f}=17.47 m/s$

Explanation:

Let's use the conservation of momentum to solve it.

$p_{initial}= p_{final}$ (1)

The total initial momentum will be: $m_{1}v_{1i}+m_{2}v_{2i}$
The total final momentum will be: $m_{1}v_{1f}+m_{2}v_{2f}$ , but as they stick together after the collision, v1f = v2f = vf.

So we can rewrite (1), using the above information:

$m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{f}+m_{2}v_{f}$

$m_{1}v_{1i}+m_{2}v_{2i}=v_{f}(m_{1}+m_{2})$

$v_{f}=\frac{m_{1}v_{1i}+m_{2}v_{2i}}{m_{1}+m_{2}}$

$v_{f}=\frac{1100\cdot 14+2500\cdot 19}{1100+2500}$

Finally, the magnitude of the velocity of the wreckage of the two cars immediately after the collision is:

$v_{f}=17.47 m/s$

I hope it helps you!

5 0

3 years ago