what is the result of 6.2×10 to the fourth power times 3.3×10 to the second power express in scientific notation

A rocket's acceleration is 6.0 m/s2. Assuming it starts at 0 m/s, how long will it take for the rocket to reach a velocity of 42

elena-s [515]

You said "<span>A rocket's acceleration is 6.0 m/s2.".

That just means that its speed increases by 6 m/s every second.
Whenever you look at it, its speed is 6 m/s faster than it was
one second earlier.

If it starts out with zero speed, then its speed is 6 m/s after 1 second,
12 m/s after 2 seconds, 18 m/s after 3 seconds . . . etc.

How long does it take to reach 42 m/s ?

Well, how many times does it have to go 6 m/s FASTER
in order to build up to 42 m/s ?

That's just (42/6) = 7 times.

Writing it correctly, with the units and everything, it looks like this:

(42 m/s) / (6 m/s</span>²)

= (42/6) (m/s) / (m/s²)

= (42/6) (m/s · s²/m)

= 7 seconds

7 0

3 years ago

Which of the following terms corresponds to #2 on the image?

Mila [183]

Answer:

Trough

Explanation:

cuz physics you see

7 0

3 years ago

you are building a race car. Your goal is to have a car that can go from 0 miles per hour to 80 miles per hour in 2 seconds. You

Vitek1552 [10]

<span>You want to focus on the car's acceleration.
(Hope this helps you! :) Have a nice day!)</span>

6 0

3 years ago

Read 2 more answers

A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220

disa [49]

First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, $\mu m g$ . For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
$ma=-\mu m g$
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
$a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2$

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
$2aS=v_f^2-v_i^2$
where $v_f=0$ is the final speed of the skier and $v_i=4.75 m/s$ is the initial speed. Substituting numbers, we find:
$S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m$