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LUCKY_DIMON [66]
3 years ago
8

A 762 kg car experiences a braking force of 9045 N and skids to a stop in 4.3 seconds. What is the speed of the car just before

the braking force was applied?
Physics
1 answer:
PilotLPTM [1.2K]3 years ago
6 0

Answer:

The solution for this problem is:

We will be using the formula for force which is F = ma 

=>10,000 = 2000 * a 

but we need to solve for acceleration so divide both sides by 2000, we will get:

=>a = 5 m/s^2 

Let the initial velocity was u m/s 

=>By v = u - at 

=>0 = u - 5 x 6 

Since acceleration is constant the velocity can be computed by multiplying the acceleration by 6 seconds. 

=>u = 30 m/s

Explanation:

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Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second
shepuryov [24]

Answer:

distance between both oasis ( 1 and 2) is  27.83 Km

Explanation:

let d is the distance between oasis1 and oasis 2

from figure

OC  = 25cos 30

OE = 25sin30

OE = CD

Therefore BC =  30-25sin30

distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem

in\Delta BCO

OB^2 = BC^2 + OC^2

PUTTING ALL VALUE IN ABOVE EQUATION

d^2 = 930-25sin30)^2 + (25cos30)^2

d^2 = 775

d = 27.83 Km

distance between both oasis ( 1 and 2) is  27.83 Km

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2 years ago
Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at
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Answer:

Δu=1300kJ/kg  

Explanation:

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Δu=u₁-u₃

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Δu=1300kJ/kg  

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