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2 years ago

How much force is required to accelerate a 12 kg mass at 5 m/s 2

Physics

1 answer:

Savatey [412]2 years ago

4 0

Answer:

60 N

Explanation:

This is just Newton's Second Law

F = m*a

F = ?

m = 12 kg

a = 5 m/^2

F = 5*12 = 60 Newtons

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Una persona de 80 kg de masa corre a una velocidad cuya magnitud es de 9m/s. Cual es la magnitud de la cantidad de movimiento? Q

natka813 [3]

Responder:

13,01 m / s

Explicación:

Paso uno:

datos dados

masa de la persona 1 m = 80 kg

velocidad de la persona 1 v = 9 m / s

masa de la persona 2 M = 55kg

velocidad de la persona 2 v =?

Segundo paso:

la expresión del impulso se da como

P = mv

para la primera persona, el impulso es

P = 80 * 9

P = 720N

Paso tres:

queremos que la segunda persona tenga el mismo impulso que la primera, por lo que la velocidad debe ser

720 = 55v

v = 720/55

v = 13,09

v = 13,01 m / s

Por lo tanto, la magnitud de la velocidad debe ser 13.01 m / s.

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3 years ago

1. What is the pull that all objects exert on each other?

Andrei [34K]

Answer:

Im pretty sure 1 is gravity, 2 is force

Explanation:

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3 years ago

An atom with 4 protons, 5 neutrons, and 4 electrons has an atomic mass of _____ amu. (Enter a whole number.)

Oksi-84 [34.3K]

Atomic mass = number of protons + number of neutrons = 4+5 = 9 amu

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3 years ago

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.

From the above we see that F ∝ mm' and $F\alpha \frac{1}{r^{2} }$

Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

For planet B the orbital radius $r_{2} =2r_{1}$ and mass $m_{2} = 3 m_{1}$

Hence the gravitational force $f_{2} =G\frac{m m_{2} }{r^{2} }$

$f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }$

$= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }$

Hence the ratio is $\frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2} }{Gmm_{1}/r_{1} ^2 }$

$=\frac{3}{4}$ [ ans]

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3 years ago

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Which of the following would most likely cause a decrease in the quality supplied

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2 years ago