Who proposed the idea that gravity could actually bend light?

A ball traveling with an initial momentum of 5.1 kgm/s bounces off a wall and comes back in the opposite direction with a moment

Elina [12.6K]

Answer: Change in momentum=9.4kgm/s

Impulse=9.4kgm/s

Explanation:

Change in momentum=5.1-(-4.3)=5.1+4.3=9.4kgm/s

Impulse=Change in momentum

There impulse=9.4kgm/s

4 0

2 years ago

Which of the following is not an example of energy transmission by waves? sound ripples on a pond an apple falling from a tree l

Bingel [31]

Answer:

an apple falling from a tree light.

3 0

3 years ago

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The space between galaxies is continually ________________, thus causing the light from galaxies to become ____________. A. expa

Vinil7 [7]

A. expanding, redder

8 0

2 years ago

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The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass

devlian [24]

Answer:

The velocity of the recoil is $v=1.001 \frac{m}{s}$

Explanation:

Kinetic Energy

$m_{bullet}*v_{bullet}=m_{gun}*v_{recoil}\\m_{gun}= 15.0kg+3.6kg$

The mass of th gun is the both mass the shotgun and the arm shoulder combination

$m_{bullet}=0.049kg\\v_{bullet}=380\frac{m}{s} \\m_{bullet}*v_{bullet}=m_{gun}*v_{recoil}\\0.049kg*380\frac{m}{s}=(15.kg+3.6kg)* v_{recoil}\\v_{recoil}=-\frac{18.62 kg \frac{m}{s} }{18.6 kg}\\ v_{recoil}=-1.0010 \frac{m}{s}$

The velocity is negative because is in opposite direction of the bullet

6 0

2 years ago

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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

2 years ago