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3 years ago

If the mass of an object is 100 and Net force is 400, what would be the Acceleration?

Physics

1 answer:

ira [324]3 years ago

8 0

Answer:

$4m/s^{2}$

Explanation:

I'm assuming the units for force and mass are Newtons and kilograms, respectively.

Rearranging Newton's first law (F=m*a) to solve for acceleration:

F=m*a

$a=\frac{F}{m}=\frac{400 N}{100 kg}=4m/s^{2}$

The acceleration is 4 meters per second squared and was found by rearranging Newton's first law in order to solve for acceleration.

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What is the magnitude of velocity for a 2,000 kg car possessing 3,000 kg.m/s of<br> momentum?

yanalaym [24]

Answer:

The product of the mass and the volume is known as momentum.

According to the law of momentum, it is stated that the two or more bodies remain in a constant state unless an external force is applied in an isolated room.

Momentum depends on the following:-

Mass

Velocity

Momentum = MASS \ X \ VOLUMEMomentum=MASS X VOLUME

\begin{gathered}momentum = 3000kgm/s\\\\mass = 2000kg\\\\velocity =\frac{momentum}{mass}\\\\v= \frac{3000}{2000} \\\\v= 1.5m/s\end{gathered}

momentum=3000kgm/s

mass=2000kg

velocity=

mass

momentum

2000

3000

v=1.5m/s

3 0

3 years ago

What happened to the speed of light if it travels from air into glass?

vovikov84 [41]

Answer:

it slows down

Explanation:

the answer

8 0

3 years ago

If a system's internal energy increases by 250 kJ after the addition of375 kJ of energy as heat, what was the value ofthe work i

nata0808 [166]

Answer:

-125 kj

Explanation:

A system internal energy increases by 250kj after an additional 375kj

Therefore the value of the work in process can be calculated as follows

= 250kj-375kj

= - 125 kj

Hence the value of the work in process is - 125 kj

7 0

3 years ago

A car speeds up from 10 m/s to 24 m/s. Change in velocity

JulsSmile [24]

Answer:

14m/s

Explanation:

subtract initial velocity by final velocity

5 0

2 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Juliette [100K]

(a) 0.473

The potential energy of a satellite orbiting around Earth is given by

$U=-\frac{GMm}{R+h}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

R is the Earth's radius

h is the altitude of the satellite above the Earth's surface

So the potential energy of satellite A is

$U_A=-\frac{GMm}{R+h_A}$

while potential energy of satellite B is

$U_B=-\frac{GMm}{R+h_B}$

Therefore the ratio of the potential energy of satellite B to that of satellite A is

$\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}$

and using

hA = 5920 km

hB = 19600 km

R = 6370 km

we find

$\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473$

(b) 0.473

The kinetic energy of a satellite orbiting around Earth instead is given by

$K=\frac{GMm}{2(R+h)}$

So the kinetic energy of satellite A is

$K_A=\frac{GMm}{2(R+h_A)}$

while kinetic energy of satellite B is

$K_B=\frac{GMm}{2(R+h_B)}$

Therefore the ratio of the kinetic energy of satellite B to that of satellite A is

$\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}$

which is identical to before, so it gives

$\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473$

(c) Satellite B

The total energy of a satellite in orbit is given by

$E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}$

We see that the total energy is:

1) negative (because the satellite is on a bound orbit)

2) inversely proportional to the distance of the satellite from the Earth's center, R+h

So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.

(d) $1.03\cdot 10^7 J$