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Blababa [14]
3 years ago
7

Under which of the following conditions is the magnitude of the average velocity of a particle moving in one dimension smaller t

han the average speed over some time interval?
Physics
1 answer:
vesna_86 [32]3 years ago
6 0

Answer:

A particle moves in the +x direction and then reverses the direction of its velocity

Explanation:

This is illustrated in that when a particle moves in a straight route with no alterations in direction, this will lead to displacement and distance being equal at any point in time during the movement. Thereby the quantity of average speed and average velocity will equal.

On the other hand, should the particle reverses direction, the distance traveled will be greater than it's displacement, thereby, the average speed will be greater than the average velocity.

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A motorcycle is traveling up one side of a hill and down the other side. The crest of the hill is a circular arc with a radius o
Setler79 [48]

Answer:

21.8 m/s

Explanation:

At the top of the hill (crest), there are two forces acting on the motorcycle:

- The reaction force of the road, N (upward)

- The force of gravity, mg (downward)

Since the motorcycle is moving by circular motion, the resultant of these forces will give the centripetal force, so:

mg-N = m\frac{v^2}{r}

where the direction of the weight (mg) is equal to that of the centripetal force, and where

m is the mass of the cycle

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

v is the speed

r = 48.6 is the radius of the hill

The cycle loses contact with the road when the reaction force becomes zero:

N = 0

Substituting into the equation, we therefore find the maximum speed that is allowed for the cycle before losing constact:

mg = m\frac{v^2}{r}\\v=\sqrt{gr}=\sqrt{(9.8)(48.6)}=21.8 m/s

6 0
2 years ago
PHYSICAL SCIENCE
omeli [17]

Answer:

Option D

Explanation:

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7 0
3 years ago
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This graph shows the energy of a reaction over time. Which statement is
kirill115 [55]

Answer: D.

Explanation: Just took the quiz.

3 0
2 years ago
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What is the difference in KE between a 52.5 kg person running 3.50 m/s and a 0.0200 kg bullet flying 450 m/s?
rusak2 [61]

Answer:

Ek = 1705.28 [J]

Explanation:

In order to solve this problem, we must remember that kinetic energy can be calculated by means of the following equation.

E_{k}=\frac{1}{2} *m*v^{2}

where:

m = mass [kg]

v = velocity [m/s]

Ek = kinetic energy [J] (Units of Joules)

<u>For the person running</u>

<u />E_{k} =\frac{1}{2}*52.2*(3.5)^{2} \\ E_{k} =319.72[J]<u />

<u />

<u>For the bullet</u>

<u />E_{k} =\frac{1}{2} *m*v^{2}<u />

<u />E_{k} =\frac{1}{2} *0.02*(450)^{2} \\E_{k}=2025 [J]<u />

<u />

The difference in Kinetic energy is equal to:

Ek = 2025 - 319.72

Ek = 1705.28 [J]

8 0
2 years ago
Gravity is affected by
mrs_skeptik [129]

Gravity is affected by mass and distance

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3 years ago
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