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Crazy boy [7]
3 years ago
6

What is the magnitude of the displacement of a car that travels half a lap along a circle of radius of 150m?

Physics
1 answer:
Lisa [10]3 years ago
6 0
It would be equal to 2r = 2* 150 = 300m
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1. A 0.40 kg ball is launched at a speed of 16 m/s from a 22 m cliff.
Masja [62]

Answer:

51.2 J, 86.2 J, 137.4 J

Explanation:

The kinetic energy of the ball is given by:

K=\frac{1}{2}mv^2

where

m = 0.40 kg is its mass

v = 16 m/s is its speed

Substituting,

K=\frac{1}{2}(0.40)(16)^2=51.2  J

The potential energy of the ball is given by

U=mgh

where

m = 0.40 kg

g=9.8 m/s^2 is the acceleration of gravity

h = 22 m is the heigth of the cliff

Substituting,

U=(0.40)(9.8)(22)=86.2 J

Finally, the total mechanical energy is the sum of the kinetic energy and the potential energy:

E=K+U=51.2 + 86.2=137.4 J

4 0
3 years ago
Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.7 m tall window in 0.129 sec. From how high above th
vladimir1956 [14]

Answer:

h = 22.35 m

Explanation:

given,

initial speed of the rock,u = 0 m/s

length of the window,l = 2.7 m

time taken to cross the window,t = 0.129 s

Speed of the rock when it crosses the window

v = \dfrac{l}{t}

v = \dfrac{2.7}{0.129}

  v = 20.93 m/s

height of the building above the window

using equation of motion

v² = u² + 2 g h

20.93² = 0² + 2 x 9.8 x h

h = 22.35 m

Hence, the height of the building above the top of window is equal to h = 22.35 m

8 0
3 years ago
Explain why theories may be changed or replaced overnight time.
vlabodo [156]
They may be changed because they may find evidence of some thing that will change their perspective on things.
5 0
3 years ago
A cyclist turns a corner with a radius of 50m at a speed of 10m/s. What is the cyclist's acceleration?
garri49 [273]

Answer:

2 m/s^2

Explanation:

a = v^2/r

a = (10m/s)^2 / 50m

a = 2 m/s^2

Leave a like and mark brainliest if this helped

Leave a like and mark brainliest if this helped

5 0
3 years ago
A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus
babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

I = m\cdot (\vec{v}_{2} - \vec{v_{1}}) (1)

Where:

I - Impulse, in kilogram-meters per second.

m - Mass, in kilograms.

\vec{v_{1}} - Initial velocity of the hockey park, in meters per second.

\vec{v_{2}} - Final velocity of the hockey park, in meters per second.

If we know that m = 0.2\,kg, \vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right] and \vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right], then the impulse applied by the stick to the park is approximately:

I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]

I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

8 0
3 years ago
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