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sesenic [268]
3 years ago
14

a person starts at a position of 2 meters and finishes at a postion of 25 meters. the trip takes 4.5 seconds. what is the person

's average velocity? what is the person's average speed?​
Physics
1 answer:
Nadya [2.5K]3 years ago
3 0

Explanation:

Average velocity = displacement / time

v = (25 m − 2 m) / 4.5 s

v = 5.11 m/s

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50 c should be right
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A 0.50-kilogram frog is at rest on the bank surrounding a pond of water. As the frog leaps from the bank, the magnitude of the a
marysya [2.9K]

Complete question:

A 0.50 kilogram frog is at rest on the bank surrounding a pond of water. As the frog leaps from the bank, the magnitude of the acceleration of the frog is 4.0 meters per second^2. Calculate The magnitude of the net force exerted on the frog as it leaps.

Answer:

2.0N

Explanation:

Given that,

Mass, m of the frog = 0.5 kg

The acceleration of the frog = 4.0 m/s².

We have been asked To find,

The magnitude of the net force exerted on the frog as it leaps.

So

We calculate this using the formula below :

F = ma

When we insert the values into the formula, we have:

F = 0.5 kg × 4 m/s²

F = 2.0 N

Therefore, the magnitude of net force is 2.0 N.

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Which of the following is an example of the electrostatic force acting in an atom?
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Read 2 more answers
The rear window in a car is approximately a rectangle, 1.3 m wide and 0.30 m high. The inside rear-view mirror is 0.50 m from th
maria [59]

Answer:

Height of mirror 0.075 m

width of mirror 0.325 m  

Explanation:

given data

wide = 1.3 m

high = 0.30 m

driver’s eyes = 0.50 m

rear window = 1.50 m

solution

we take here height / width of the mirror  is

height / width   = \frac{h}{w}   .................1

and

height /width of the window is

height /width  = \frac{h_w}{w_w}   .................2

and

distance of eye / window by the mirror is

distance of eye / window = \frac{x_e}{x_w}      .................3

so here

θ = θi  = θr    ....................4

and  tanθ for vertical is

tanθ  = \frac{h}{x_e}  

tanθ  =  \frac{h_w}{(x_e + x_w)}       ....................5

so

h =   h_w \times  \frac{x_e}{(x_e + x_e)}     ....................6

put  here value and we get

h = 0.30 \times  \frac{0.50}{(0.50 + 1.50)}  

h = 0.075 m

and

when we take here tanθ for horizontal than it will be

tanθ = \frac{w}{x_e}    

tanθ = \frac{w_w}{(x_e + x_w)}       .......................7

so

w = w_w \times  \frac{x_e}{(x_e + x_w)}         ....................8

put here value and we get

w = 1.3 \times  \frac{0.50}{(0.50 + 1.50)}  

w = 0.325 m

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If a car is traveling on the highway at a constant velocity, the force that pushes the car forward must be A. equal to the weigh
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the correct answer is c

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