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2 years ago

An object of the same mass has three different weights at different times.

Physics

1 answer:

Sindrei [870]2 years ago

5 0

Answer:

See below

Explanation:

Possible.....mass is the amount of matter an object has ....weight is the result of gravity on an object.... different gravity ( Earth , Moon or Mars etc)

results in different weights .

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A vector is 14.4 m long and

MaRussiya [10]

Answer:

Explanation:

The x-component is found in the magnitude of the vector times the cosine of the angle.

$A_x=14.4cos133$ and, to 3 sig dig,

$A_x=-9.82m$

3 0

3 years ago

22. A ball is thrown horizontally from the roof of a building 12 m tall with a speed of 3.1 m/s.

zysi [14]

Answer:

a) t = 1.6 s

b) d = 4.9 m

c) v = 16 m/s

d) θ = 79°

Explanation:

time of fall

t = √(2h/g) = √(2(12)/9.8) = 1.5649... s

d = vt = 3.1(1.56) = 4.8512...

vertical velocity vy = at = 9.8(1.56) = 15.336... m/s

v = √(15.336² + 3.1²) = 15.6464... m/s

θ = arctan(15.336/3.1) = 78.5724...°

5 0

3 years ago

Find the mass of an object on planet F if its weight is 650 N (g = 13m/s^2)

Andrew [12]

Answer:

the object's mass is 50 kg

Explanation:

We use Newton's second law to solve for the mass:

F = m * a , then m = F / a

In our case, the acceleration is the gravitational acceleration on the planet, and the force is the weight of the object on the planet. So we get:

m = w / a = 650 N / 13 m/s^2 = 50 kg

Then, the object's mass is 50 kg.

5 0

3 years ago

What are the subzones for each main zone?

Makovka662 [10]

Answer:

The oceanic zone is subdivided into the epipelagic, mesopelagic, and bathypelagic zones on the basis of amount of light that reaches different depths. The mesopelagic (disphotic) zone, where only small amounts of light penetrate, lies below the Epipelagic zone.

Explanation:

5 0

3 years ago

Three small balls of the same size but different masses are hung side-by-side in parallel on the strings of same length. They to

andrey2020 [161]

Answer:

m1/6 ( c )

Explanation:

since all the balls starts having the same momentum after the two collisions we will apply the principal of conservation of energy

After first collision

m1v = m1v1 + m2v2 --- ( 1 )

After second collision

m2v2 = m2v2 + m3v3 ---- ( 2 )

combining equations 1 and 2

m1v = m1v1 + m2v2 + m3v3 ----- ( 3 )

All balls moving at the same momentum ( p ) = m1v1 = m2v2 = m3v3

note ; 3p = m1v ∴ m3 = $\frac{m1v}{3v3}$ ----- ( 4 )

applying conservation of energy

3v = v1 + v2 + v3 ------- ( 5 )

also 3m1v1 = m1v = v1 = v/3 =

v2 + v3 = 8/3 v ----- ( 6 )

next eliminate V3 for equation 6 by applying conservation of energy and momentum

m1 = 2m2 ------ ( 7 )

now using p1 = p2 = m1v1 = 1/2 m1v1 hence v2 = 2v1 where v1 = 1/3 v

hence ; v2 = 2/3 v ------- ( 8 )

solving with equation 6 and 8

v3 = 2v ------ ( 9 ) ∴ v/v3 = 1/2 ---- ( 10 )

solving with equation 9 and 10

m3 = m1/3 * 1/2 = m1/6

8 0

3 years ago