All categories

3 years ago

552000 in scientific notation

1 answer:

4 0

5.52 × 10 to the 5th power (100000) . In scientific notation you need to have a decimal numver times 10 to the power of something so you can divide 552000 by 10 5 times. So in order to get 552000 you need to have 10 to the 4th power and 5.52

You might be interested in

Please help on this one

Murljashka [212]

Just find the area of the graft

4m/s x 5s
=20m

7 0

3 years ago

Read 2 more answers

What are two processes that result in rocks being broken down into smaller pieces ?

meriva

Mechanical and Chemical. (Weathering and erosion)

5 0

4 years ago

Read 2 more answers

Consider taking a time-lapse video of an analog clock that is missing the hour hand. Assume that one frame of video is taken eve

Vlad [161]

Answer:

the frame rate of the time-lapse video if no aliasing occurs is 10min per frame

Explanation:

Given the data in the question;

No aliasing effect, meaning the clock with missing hour hand will not have any effect

Time required for one frame = 10 minutes

frame rate of time = Time required / number of frame

= 10 min / 1 frame

= 10min per frame

Therefore, the frame rate of the time-lapse video if no aliasing occurs is 10min per frame

4 0

3 years ago

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

What does increasing the impulse on an object due to its momentum?

allochka39001 [22]

increaskng the impulse also <u>increases </u><u>momentum</u>

7 0

2 years ago