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WITCHER [35]
3 years ago
15

If a car manufacturer reduced the mass of a new car model, so that it took a force of 4000 N

Physics
1 answer:
Sauron [17]3 years ago
5 0

Answer:

1000 kg

Explanation:

Force = mass x acceleration, Hence, mass = Force/acceleration

mass = 4000/4 = 1000 kg

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Zeb was lifting a box onto a moving truck. He lifted with a net force of 2000N and the box had a mass of 100 kg. What was the re
Aleksandr [31]

Answer:

<h2>20 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{2000}{100}  = 20 \\

We have the final answer as

<h3>20 m/s²</h3>

Hope this helps you

8 0
3 years ago
What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne
irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance h= 1790 km = 1.790\times10^{6}\ m

Magnetic field B=4\times10^{-8}\ T

Mass of proton m=1.673\times10^{-21}\ Kg

Radius of earth R =6.38\times10^{6}\ m

Radius of orbit r=R+h

r=6.38\times10^{6}+1.790\times10^{6}

r=8170000\ m

We need to calculate the speed

Using formula of magnetic field

Bvq=\dfrac{mv^2}{r}

v=\dfrac{Bqr}{m}

Put the value into the formula

v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}

v=31.25\ m/s

Hence, The velocity is 31.25 m/s and direction is toward west.

6 0
3 years ago
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
The equation for the conservation of mechanical energy is:
marusya05 [52]

Answer:

The answer is A because the equation is  KEi+PEi=KEf+PEf

i means initial (before) and f means final (after)

7 0
3 years ago
How much do a steel BB weight?
pychu [463]
Each BB 0.349 grams and 5.386 grains. 349 mg & 5.386 grains each steel BBs.
6 0
3 years ago
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