2 years ago

Acceleration is displacement divided by time interval True Or False

Physics

1 answer:

Marat540 [252]2 years ago

8 0

It’s true because it doesn’t did use time in interval

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2. What is the mass of an object that has 50000 J

muminat

Answer:

m= 666.67 or 667

Explanation:

KE= 50000 J

v= 75 m/s

m= ?

M= 50000 / 75

m= 666.67 or 667

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2 years ago

How do you calculate acceleration

AfilCa [17]

A=f/m
Example a=10/2
A=5

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3 years ago

Read 2 more answers

What is the resulting velocity of the launcher if the net force on the launcher is equal to the reaction force?

xz_007 [3.2K]

Answer:

according to this question best answer is C

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2 years ago

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

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3 years ago

When a car stops, the brakes heat up because of friction. What is this an example of?

Umnica [9.8K]

Kinetic Energy would have to be the answer. Pretty sure.

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3 years ago