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3 years ago

8

2. Turn off the Parallel line and turn on the Line through focal point. Move the light bulb around. What do you notice about the

reflected ray in this situation? __________________________________________________

1 answer:

MArishka [77]3 years ago

8 0

Answer:

The group of light rays is reflected back towards the focal point thereby producing a magnifying effect.

Explanation:

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Forces always act in pairs? True or false

nasty-shy [4]

Answer:

True

Explanation:

Just as Isaac Newton says, "For every action, there is an equal and opposite reaction."

7 0

3 years ago

Two identical balls move directly toward each other with equal speeds. how will the balls move if they collide and stick togethe

Citrus2011 [14]

The momentum of both the identical balls would eventually be transferred to one another when it comes to a point wherein they will collide. In addition, the phenomenon is called an elastic collision wherein both the momentum and energy of the system would considered to be conserved.

7 0

3 years ago

Differences between Pressure and upthrust

Angelina_Jolie [31]

Answer:

Pressure is equal to the ratio of thrust to the area in contact. Upthrust is a force exerted by the fluids on an object placed in the fluid . Upthrust acts in upward direction.

4 0

3 years ago

A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.

NemiM [27]

Answer:

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, $\omega=2.8\ rad/s$

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

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

$K_t=\dfrac{1}{2}m(r\omega)^2$

$K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2$

The rotational kinetic energy of the ball is :

$K_r=\dfrac{1}{2}I \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2$

Ratio of translational to the rotational kinetic energy as :

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

4 0

3 years ago

a ball is tossed straight up in the air to a maximum height of 25 meters before returning to the point that it was thrown, where

boyakko [2]

Answer:

height=25

velocity=8

force of gravity=10

Answer =2000

5 0

2 years ago