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2 years ago

If a 50 N force was applied to a really massive object as well as a tiny object, what can you say

Physics

1 answer:

frozen [14]2 years ago

6 0

Answer:

The tiny object's acceleration will be much greater.

Explanation:

Because of Newton's second law of motion, (which is F=ma or Force= Mass*acceleration), then if the force maintains the same (which in this case it does, because it says 50 N to both obejcts) and one mass was much greater than the other, then the ould be less for the more massive object and much greater for the lighter object.

For example: If 50 N were applied to a 500 kg object and a 50 kg object, then theformulas for each (respectively) are:

50 = 500*acceleration

and

50 = 50*acceleration

(Because of Newtons Second Law of Motion)

Then, solving for the equations, we get for equation 1:

Acceleration = .1 m/s^2

And for equation 2:

Accleration = 1 m/s^2

Thus, you can see that more massive objects (when applied he same amount of force as the smaller object) clearly have less accleration than the smaller objects.

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3 years ago

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Arrange the stars based on their temperature. Begin with the coolest star, and end with the hottest star.

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Answer:

Uranus, Pluto, Neptune, Saturn , Jupiter, mars, Venus ,mercury and sun

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2 years ago

A smooth circular hoop with a radius of 0.800 m is placed flat on the floor. A 0.300-kg particle slides around the inside edge o

Marrrta [24]

Answer:

a)11.25 J

b)Number of revolution = 1

Explanation:

Given that

Radius ,r= 0.8 m

m= 0.3 kg

Initial speed ,u= 10 m/s

final speed ,v= 5 m/s

Initial energy

$KE_i=\dfrac{1}{2}mu^2$

$KE_i=\dfrac{1}{2}0.3\times 10^2$

KEi= 15 J

Final kinetic energy

$KE_f=\dfrac{1}{2}mv^2$

$KE_f=\dfrac{1}{2}0.3\times 5^2$

KEf=3.75 J

The energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J

The minimum value to complete the circular arc

$V=\sqrt{r.g}$

Now by putting the values

$V=\sqrt{0.8\times 10}$

V= 2.82 m/s

So kinetic energy KE

$KE=\dfrac{1}{2}mV^2$

$KE=\dfrac{1}{2}0.3\times 2.82^2$

KE=1.19 J

ΔKE= KEi - KE

ΔKE= 15- 1.19 J

ΔKE=13.80 J

The minimum energy required to complete 2 revolutions = 2 x 11.25 J

= 22.5 J

Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.

Number of revolution = 1

4 0

2 years ago

A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i

Effectus [21]

Answer:

s = 1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}

• s = $\sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}$

s = $\sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}$

s = 1.7 m

7 0

3 years ago