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Anettt [7]
3 years ago
8

A box falls off of a tailgate and slides along the street for a distance of 62.5 m. Friction slows the box at –5.0 m/s2. At what

speed was the truck traveling when the box fell?
Physics
1 answer:
Mila [183]3 years ago
3 0

Answer:

25 m/s

Explanation:

This question can be solved using equation of motion

v^2 = u^2 + 2as

where

v is the final velocity

u is the initial velocity

s is the distance covered while moving from initial to final velocity

a is the acceleration

_____________________________________________

Given

box moved for distance of 62.5 m

Friction slows the box at –5.0 m/s2----> this statement means that there is deceleration , speed of truck decreases by 5 m/s in every second until the box comes to rest. Friction causes this deceleration.

thus in this problem

a = -5.0 m/s2

V = 0   as body came to rest due to friction deceleration

u the initial velocity we have to find

the initial velocity of box will be the same as speed of truck, as the box was in the truck and hence box will pick the speed of truck.

so if we find speed of box, we will be able get sped of truck as well.

using equation of motion

v^2 = u^2 + 2as\\0^2 = u^2 + 2*-5* 62.5\\0 = u^2 - 625\\u^2 = 625\\\sqrt{u^2} = \sqrt{625} \\u = 25

Thus, initial speed with the truck was travelling was 25 m/s.

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A person's prescription for her new bifocal glasses calls for a refractive power of -0.450 diopters in the distance-vision part,
Angelina_Jolie [31]

Answer:

Far point of the eye is 22.24 m

Far point of the eye is 0.4 m

Explanation:

\frac{1}{f}=-0.045

Object distance = u

Image distance = v

Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=-0.045-\frac{1}{\infty}\\\Rightarrow \frac{1}{v}=\frac{1}{-0.045}\\\Rightarrow v=-22.22\ m

Far point

|v|+\text{Position from eye}\\ =|-22.22|+0.02\\ =22.24\ m

Far point of the eye is 22.24 m

Object distance = u = 0.25-0.02 = 0.23 m

\frac{1}{f}=1.75

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=1.75-\frac{1}{0.23}\\\Rightarrow v=-0.38\ m

Near point

|v|+\text{Position from eye}\\= |-0.38|+0.02\\ =0.4\ m

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7 0
3 years ago
Transverse waves travel with a speed of 20.0 m/s on a string under a tension of 6.00 N. What tension is required for a wave spee
marta [7]

Based on the sped of the waves and the tension as well as the needed wave speed, the required tension is 13.5 N.

<h3>What is the required tension?</h3>

Given the initial tension and speed, the tension that is required can be found by the formula:

= Initial tension x (Required speed / Initial speed)²

Solving gives:

= 6 x (30 / 20)²

= 6 x 9/4

= 13.5 N

In conclusion, the tension required is 13.5N.

Find out more on the tension on a wire at brainly.com/question/14290894.

#SPJ4

3 0
1 year ago
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