All categories

2 years ago

A rocket ship leaves Earth's atmosphere. Its initial velocity is less than its final velocity. What is this an example of?

2 answers:

6 0

I believe this would be an example of positive acceleration as the initial velocity of the rocket is less than the final velocity, indicating that the rocket is accelerating and thus is positive.

mina [271]2 years ago

4 0

The correct answer among all the other choices is positive acceleration. When a rocket ship leaves Earth's atmosphere, its initial velocity is less than its final velocity, this is an example of positive acceleration. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help.

You might be interested in

Why does paper stand up when a van de graaf generator is on?

joja [24]

Answer:

because follicles are getting charged to the same potential

Explanation:

7 0

3 years ago

Which does not contain a lens?

denis23 [38]

Im pretty sure it’s A eye

8 0

3 years ago

A synthesis reaction occurs when a substance breaks down into two or more simple

max2010maxim [7]

True . this is called a complex substance

4 0

3 years ago

Read 2 more answers

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago

Particles 1 and 2 each mass m fixed to the ends of a rigid massless rod of length L1 + l2 with L1 = 20cm and l2 = 80 cm. The rod

Degger [83]

Answer:

Sorry bro I don't even know the answer

7 0

3 years ago