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3 years ago

Question 11(Multiple Choice Worth 3 points)

Physics

1 answer:

melomori [17]3 years ago

8 0

Answer:

Explanation:

The only thing that decreases the gravitational pull between 2 objects is if you<u> increase the distance.</u>

The formula is F = G*m*m/r^2

As r^2 gets larger and G and m1 and m2 remain the same, F has to get smaller because r^2 divides into G*m1*m2

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Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict

GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = $m$

Mass of the heavier cart = $2m$

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ $KE = \frac{mv^2}{2}$

⇒ $KE = \frac{mv^2}{2}\times \frac{m}{m}$

⇒ $KE = \frac{m^2v^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(mv)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}$

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ $K=\frac{(F\times t)^2}{2m}$ ...equation (i)

KE (K1) of mass 2m.

⇒ $K_1=\frac{(F\times t)^2}{2\times 2m}$

⇒ $K_1=\frac{(F\times t)^2}{4m}$ ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ $\frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}$

⇒ $\frac{K_1}{K} =\frac{2m}{4m}$

⇒ $\frac{K_1}{K} =\frac{2}{4}$

⇒ $\frac{K_1}{K} =\frac{1}{2}$

⇒ $K_1=\frac{K}{2}$

⇒ $K_1=\frac{1}{2}K$

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0

3 years ago

Carry's car has a mass of 1000 kg and its brakes can apply 8000 N of force. If she is driving at 24 m/s and sees something in th

-Dominant- [34]

Explanation:

Given parameters:

Mass of car = 1000kg

Force applied = 8000N

speed = 24m/s

Unknown:

time taken for the car to stop = ?

Solution:

According to newton's second law of motion; "the force on a body is the product of its mass and acceleration".

Force = mass x acceleration

let us find the acceleration of the car;

a = $\frac{F}{M}$ = $\frac{8000}{1000}$ = 8m/s²

since the car is accelerating at a rate of 8m/s², when the brakes are applied, it will start decelerating at the constant rate, - 8m/s²

Applying the appropriate equation of motion;

V = U + at

V is the final velocity

U is the initial velocity

a is the acceleration

t is the time taken

final velocity = 0

0 = U + at

-U = at

-24 = -8t

t = 3s

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

8 0

4 years ago

Which statement correctly compares infrared light to ultra violet light

mrs_skeptik [129]

What are the statements?

3 0

3 years ago

A kangaroo kicks downward with a 1000N force. According to Newton's Law the kangaroo is propelled into the air by:

bogdanovich [222]

<h3>Answer: B) his muscles</h3>

Explanation:

Specifically his leg muscles. As the leg muscles expand, they push down on the ground. Newton's 3rd law says that for any action, there's an opposite and equal reaction. That means a downward push into the ground will have the ground push back, more or less, and that's why the kangaroo will jump. The ground (and the earth entirely) being much more massive compared to the animal means that the ground doesn't move while the kangaroo does move. Perhaps on a very microscopic tiny level the ground/earth does move but it's so small that we practically consider it 0.

This experiment can be done with a wall as well. Go up to a wall and lean against it with your hands. Then do a pushup to move further away from the wall, but you don't necessarily need to lose contact with the wall's surface. As you push against the wall, the wall pushes back, and that causes you to move backward. If the wall was something flimsy like cardboard, then you could easily push the wall over and you wouldn't move back very much. It all depends how much mass is in the object you're pushing on.

5 0

3 years ago

URGENT!!!

andre [41]

Answer:

6.98 × 10^(14) Hz

Explanation:

wavelength in air; λ_o = 430nm = 430 × 10^(-9) m

Speed of violet light in the prism; v = 2 × 10^(8) m/s

Speed of light in air = 3 × 10^(8) m/s

To get the frequency, we will use the relationship;

f = c/λ_o

Plugging in the relevant values;

f = (3 × 10^(8))/(430 × 10^(-9))

f = 6.98 × 10^(14) Hz

6 0

3 years ago