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vagabundo [1.1K]

3 years ago

8

Find the force needed to accelerate a .3 kg bullet at 2100 m / s / s.

1 answer:

klasskru [66]3 years ago

6 0

The force is 630 m/s hope this helps!

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Could it be hypothesis?

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myrzilka [38]

<em>Anything</em> that's dropped through air is somewhat affected by air resistance. But, out of that list, the leaf and the balloon are the items that will be affected by air resistance enough so that you can plainly see it.

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3 0

3 years ago

A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart

serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

$E_{initial} = E_{final}$

$PE_{initial}+KE_{initial} = PE_{final}+KE_{final}$

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

$KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}$

$KE_{final} = (5100+4200)-5700$

$KE_{final} = 3600MJ$

Therefore the final kinetic energy is 3600MJ

5 0

3 years ago

Consider a golf ball with a mass of 45.9 grams traveling at 200 km/hr. If an experiment is designed to measure the position of t

Kipish [7]

Answer:

speed of golf ball is 1.15 × $10^{-30}$ m/s

and % of uncertainty in speed = 2.07 × $10^{-30}$ %

Explanation:

given data

mass = 45.9 gram = 0.0459 kg

speed = 200 km/hr = 55.5 m/s

uncertainty position Δx = 1 mm = $10^{-3}$ m

to find out

speed of the golf ball and % of speed of the golf ball

solution

we will apply here heisenberg uncertainty principle that is

uncertainty position ×uncertainty momentum ≥ $\frac{h}{4\pi }$ ......1

Δx × ΔPx ≥ $\frac{h}{4\pi }$

here uncertainty momentum ΔPx = mΔVx

and uncertainty velocity = ΔVx

and h = 6.626 × $10^{-34}$ Js

so put here all these value in equation 1

$10^{-3}$ × 0.0459 × ΔVx = $\frac{6.626*10^{-34}}{4\pi }$

ΔVx = 1.15 × $10^{-30}$ m/s

and

so % of uncertainty in speed = ΔV / m

% of uncertainty in speed = 1.15 × $10^{-30}$ / 55.5

% of uncertainty in speed = 2.07 × $10^{-30}$ %

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3 years ago

What is the force of friction between the dragsters tires and the track if the dragster mass is 300kg and the coefficient of the

Nuetrik [128]

Ff=μN=μmg=0.8*300*10=2400 N

7 0

3 years ago