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3 years ago

A 4 kg bowling ball rolls at a speed of 5 m/s on the roof of a building that is 30 meters tall. What is its kinetic energy?

Physics

1 answer:

klasskru [66]3 years ago

3 0

ans: B

Kinetic energy = $\frac{1}{2} mv^{2}$

= $\frac{1}{2} (4)(5^{2} )$

= 50J

ps. the height will only affect its potential energy, not kinetic.

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What force is needed to accelerate an object 5 m/s2 if the object has a mass of 10 kg?

Helga [31]

We know, F = m * a
F = 10 * 5
F = 50 N

In short, Your Answer would be 50 Newtons

Hope this helps!

5 0

3 years ago

An antelope moving with constant acceleration covers the distance 70.0 m between two points in time 6.60 s. Its speed as it pass

Veronika [31]

Answer:

A) The speed at the first point is 6.91 m/s.

B) The acceleration is 1.12 m/s²

Explanation:

The equations of velocity and position of the antelope will be as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the antelope at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A) If we place the origin of the frame of reference at the first point, we can say that at t = 6.60 the position of the antelope is 70.0 m and its velocity is 14.3 m/s. In this way, we will have 2 equations with 2 unknowns, the initial velocity (the velocity at the first point) and the acceleration.

Let´s start finding the speed at the first point:

v = v0 + a · t (solving for "a")

(v - v0)/t = a

Replacing a = (v - v0)/t in the equation of the position:

x = x0 + v0 · t + 1/2 · (v - v0)/t · t² (x0 = 0)

x = v0 · t + 1/2 · v · t - 1/2 · v0 · t

x - 1/2 · v · t = 1/2 · v0 · t

2/t · (x - 1/2 · v · t) = v0

2/6.60 s · (70.0 m - 1/2 · 14.3 m/s · 6.60s) = v0

v0 = 6.91 m/s

The speed at the first point is 6.91 m/s.

B) Using the equation of velocity

a = (v - v0)/t

a = (14,3 m/s - 6.91 m/s) / 6.60 s

a = 1.12 m/s²

The acceleration is 1.12 m/s²

4 0

3 years ago

What best describes the cosmic microwave background radiation? A. Existing at the 'red' end of the light spectrum. B. Moving tow

Monica [59]

Answer:

C: Low energy, a few degrees above 0 Kelvin

3 0

3 years ago

The relative density of oxygen and carbon dioxide are 16, 12 respectively. If 25cm3 of carbon dioxide effuse out in 75 sec what

Mrac [35]

Answer:

32 cm³

Explanation:

The given gas data are;

The relative density of oxygen = 16

The relative density of carbon dioxide = 12

The time it takes 25 cm³ of carbon dioxide to effuse out = 75 seconds'

The duration of effusion of the oxygen = 96 seconds

The rate of effusion of carbon dioxide, R1 = 25 cm³/(75 sec) = (1/3) cm³/sec

According to Graham's law of diffusion and effusion of a gas, we have;

$\dfrac{Rate \ of \ effudion \ of \ gas \ 1}{Rate \ of \ effudion \ of \ gas \ 2} =\dfrac{The \ relative \ density \ of \ gas \ 2}{The \ relative \ density \ of \ gas \ 1}$