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vlada-n [284]
3 years ago
7

At the surface of a certain planet, the gravitational acceleration g has a magnitude of 12.0 m/s2 . A 24.0-kg brass ball is tran

sported to this planet. What is?
(a) the mass of the brass ball on the Earth and on the planet, and
(b) the weight of the brass ball on the Earth and on the planet?
Physics
1 answer:
max2010maxim [7]3 years ago
8 0

Answer:

The mass of the ball is 24kg both on the Earth and on that planet.

The weight on Earth is 240N and on that planet is 288N.

Explanation:

The mass of a body is the same everywhere but the weight of a body varies from place to place.

To calculate the weight on Earth:

g on earth=10m/s2

Weight = mass *g

W=24*10

=240N

To calculate the weight on the planet:

g on the planet=12.0m/s2

W=24*12

=288N

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Answer:

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Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:

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s = (Vf² - Vi²)/2a

where,  

Vf = Final Velocity of Car = 0 mi/h

Vi = Initial Velocity of Car = 50 mi/h

a = deceleration of car  

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Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.

So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:

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FOR SOBER DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 0.33 s

s = s₁

Therefore,

s₁ = (73.33 ft/s)(0.33 s)

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v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 1 s

s = s₂

Therefore,

s₂ = (73.33 ft/s)(1 s)

s₂ = 73.33 ft

Now, the distance traveled by drunk driver's car further than sober driver's car is given by:

ΔS = s₂ - s₁

ΔS = 73.33 ft - 24.2 ft

<u>ΔS = 49.13 ft</u>

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