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kodGreya [7K]
3 years ago
10

Newton’s second law: Force equals mass times acceleration (__=___).

Physics
1 answer:
DIA [1.3K]3 years ago
7 0
F n = ma / unbalanced force - In the direction of the force.
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When light passes from a faster medium into a slower medium, which of the following explains what will occur?
beks73 [17]

When light passes from a faster medium into a slower medium, light will be refracted toward a line drawn perpendicular to the point of refraction. <em>(B)</em>

5 0
4 years ago
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PLZ HELP IM REALLY CONFUSED a boy is playing catch with his friend. He throws the ball straight up. When it leaves his hand, the
erastovalidia [21]

Answer:

K.E = 100 J

Final P.E = 100 J

Explanation:

The kinetic energy of any object can be given by the following formula:

K.E = (\frac{1}{2})mv^{2}

where,

K.E = Kinetic Energy

m = mass of ball = 2 kg

v = speed of ball

Initially, v = 10 m/s. Therefore, the initial K.E is given as:

K.E = (\frac{1}{2})(2\ kg)(10\ m/s)^{2}

<u>K.E = 100 J</u>

Now, at the highest point the K.E of the ball becomes zero. because the ball stops for a moment at the highest point and its velocity becomes zero. So, from Law of Conservation of energy:

Initial K.E + Initial P.E = Final K.E + Final P.E

Initial P.E is also zero due to zero height initially.

K.E + 0 = 0 + Final P.E

<u>Final P.E = 100 J</u>

3 0
3 years ago
As denizens of the surface of a spinning planet, we are always in uniform circular motion. Imagine you are in Nairobi (on the Ea
Klio2033 [76]

Answer:

6.02 radian

Explanation:

At 12 noon position is zero radian on Monday

at 12 midnight the position is π/2

at 11 am position on Tuesday is 2π/24 x 23 ( after 23 hours ) = 6..02 radian.

7 0
3 years ago
The planet Krypton has a mass of 8.8 × 1023 kg and radius of 2.5 × 106 m. What is the acceleration of an object in free fall nea
djverab [1.8K]

Answer:

Acceleration, a=9.39\ m/s^2

Explanation:

Given that,

Mass of the planet Krypton, m=8.8\times 10^{23}\ kg

Radius of the planet Krypton, r=2.5\times 10^{6}\ m

Value of gravitational constant, G=6.6726\times 10^{-11}\ Nm^2/kg^2

To find,

The acceleration of an object in free fall near the surface of Krypton.

Solution,

The relation for the acceleration of the object is given by the below formula as :

a=\dfrac{Gm}{r^2}

a=\dfrac{6.6726\times 10^{-11}\times 8.8\times 10^{23}}{(2.5\times 10^{6})^2}

a=9.39\ m/s^2

So, the value of acceleration of an object in free fall near the surface of Krypton is 9.39\ m/s^2

5 0
4 years ago
“What are three ways that a cat pushing on a cat-flap door can change the amount of torque applied to the door?”
mixer [17]
It depends on where the cat is applying the force and how much the force is .. after all where and at what is distance from the axis of rotation of the door
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