3 years ago

An object with a mass of 5.0 kg is accelerating toward the surface of Earth at a rate of 9.8 m/s^2

1 answer:

4 0

The answer is A. Force is equal to mass x acceleration. The acceleration due to gravity is a constant so multiplying the mass by 2 would double the force.

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Que capacidade física consiste em deslocar o corpo no espaço o mais rápido possível, mudando o centro de gravidade de posição, s

grin007 [14]

Answer:ur mom

Explanation:because

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2 years ago

Pleasee help me pwease

Sedbober [7]

I think the answer is (C.) but im not sure

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2 years ago

Read 2 more answers

Hail comes straight down at a mass rate of m/Δt = 0.050 kg/s and an initial velocity of v0 = -15 m/s and strikes the roof perpen

cupoosta [38]

Answer:

$F = 1.5 N$

Explanation:

Let the hails are coming down with speed "v" and then rebound with the same speed in opposite direction

So here the change in the momentum of the hails is given as

$\Delta P = mv - (-mv)$

$\Delta P = 2mv$

now we know by Newton's 2nd Law

$F = \frac{\Delta P}{\Delta t}$

here we have

$F = \frac{\Delta m}{\Delta t} (2v)$

now plug in all data

$F = 0.050(2\times 15)$

$F = 1.5 N$

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2 years ago

Name some elements that have a symbol that is entirely different from the spelling of the world

Alenkasestr [34]

Some elements that have a symbol entirely different from the spelling are..

1.Fe. Iron
2.Na.Sodium
3.K. Pottasium
4.Ag.Silver
5.Sn.Tin
6.Sb. Antimony
7,Pb.Lead

7 0

2 years ago

A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressin

BARSIC [14]

Answer:

$U_k = 0.113$

Explanation:

using the law of the conservation of energy:

$E_i -E_f=W_f$

$\frac{1}{2}Kx^2=NU_kd$

where K is the spring constant, x is the spring compression, N is the normal force of the block, $U_k$ is the coefficiet of kinetic friction and d is the distance.

Also, by laws of newton, N is calculated by:

N = mg

N = 3.35 kg * 9.81 m/s

N = 32.8635

So, Replacing values on the first equation, we get:

$\frac{1}{2}(138)(0.123)^2= (32.8635)U_k(0.281m)$

solving for $U_k$ :

$U_k = 0.113$

8 0

2 years ago