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AlladinOne [14]
2 years ago
6

There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the differe

nce between these two quantities.
Physics
1 answer:
Usimov [2.4K]2 years ago
5 0

An example that illustrates the difference is the circular motion

Explanation:

Let's start by reminding the definition of the two quantities:

- Speed is a scalar quantity that tells "how fast" an object is moving, regardless of its direction of motion.

Speed can be  calculate as:

speed = \frac{d}{t}

where:

d is the distance travelled

t is the time taken

- Velocity is instead a vector quantity, given by:

velocity = \frac{d}{t}

where;

d is the displacement of the object (displacement is a a vector connecting the initial position to the final position of motion)

t is the time taken

Since it is a vector, velocity has both a magnitude and a direction, therefore it also takes into account the direction of motion of the object.

For an object in motion in a straight line, speed and velocity are the same. However, this is not always the case.

In fact, an example of motion in which the two quantities are different is the circular motion. Consider for example the object making one complete revolution along the circle. Therefore, its average speed is the ratio between the length of the perimeter (the distance) divided by the time taken:

speed = \frac{2\pi r}{t}

where r is the radius of the circle.

However, the displacement of the object is zero (because the object returns to the starting point), and so the average velocity is also zero:

velocity = \frac{0}{t}=0

Learn more about speed and velocity:

brainly.com/question/8893949

brainly.com/question/5063905

brainly.com/question/5248528

#LearnwithBrainly

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A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then
Elan Coil [88]

Answer:

A) W_{ff} =-744.12J

B) F_f=-W_{ff}*sin\theta /hy = 112.75N

C) F_{f2}=207.58N

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.  </em>

<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.  </em>

<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder.  </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

W_{ff}=\Delta E

W_{ff}=1/2*m*vf^2-m*g*hy

W_{ff}=-744.12J

For part B), we use our previous value for the work:

W_{ff}=-F_f*(hy/sin\theta)   Solving for friction force:

F_f=-W_{ff}*sin\theta /hy

F_f=112.75N

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

Vf^2=Vo^2+2*a*d

Solving for a:

a=-3.844m/s^2

Now, by dynamics:

|F_f|=|m*a|

|F_f|=207.58N

8 0
2 years ago
Joe wanted to experiment with different factors that affect the freezing rate of water. He put two cups of water into each of tw
creativ13 [48]

the answer is not D ....... the answer is {B} if you got it right give me a 5 stars and a hard

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2 years ago
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A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

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2 years ago
A 16-Ω loudspeaker, an 8.0-Ω loudspeaker, and a 4.0-Ω loudspeaker are connected in parallel across the terminals of an amplifier
kolbaska11 [484]

Answer:

2.286 ohm

Explanation:

R1 = 16 ohm

R2 = 8 ohm

R3 = 4 ohm

They all are connected in parallel combination

Let the equivalent resistance is R.

1/R = 1/R1 + 1/R2 + 1/R3

1/R = 1/16 + 1/8 + 1/4

1/R = (1 + 2 + 4) / 16

1/R = 7 / 16

R = 16/7 = 2.286 ohm

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2 years ago
A particularly scary roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radiu
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Answer:

v = 10.89\ m/s

Explanation:

given,                          

radius of loop = 12.1 m                              

to find the minimum speed transverse by the rider to not to fall out upside down                                                                

centripetal force = \dfrac{mv^2}{r}

gravitational force  = m g

computing both the equation]

mg = \dfrac{mv^2}{r}

v = \sqrt{rg}

v = \sqrt{12.1 \times 9.8}

v = \sqrt{118.58}

v = 10.89\ m/s

5 0
2 years ago
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