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spayn [35]
3 years ago
10

A runner accelerates to 4.2 m/s2 for 10 seconds before winning the race. How far did he/she run?

Physics
2 answers:
timofeeve [1]3 years ago
8 0

She run the distance of 42metres

hodyreva [135]3 years ago
6 0

Answer:

Runner will run for 210m

Step by Step Explanation:

According to Newton's first of motion,

Vf=Vi+at

Given is

the speed with which runner is running will be Vi, he is running with accelertion a=4.2m/s2 fot time t=10sec

Vf will be 0 because runner will stop after winning the race

we will find Vi from equation 1,

Vi=Vf-at

   = 0- 4.2*10

   =-42 m/s  (negative sign is showing decceleration)

So, Vi=42 m/s

According to Newton's third of motion,

2aS= Vf^2-Vi^2

S is distance, he will cover before stopping,

2(4.2)S= 0- 42^2

S=1764/8.4

S=210m

Runner will run for 210m

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A driver in a 2144 kg car traveling at 15 m/s hits the brakes, coming to a stop in 67 meters. How far would it take the car to s
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3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping:
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4 0
3 years ago
An iron anchor of density 7890.00 kg/m3 appears 299 N lighter in water than in air. (a) What is the volume of the anchor? (b) Ho
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Answer:

weigh is 2353.13 N

Explanation:

Given data

density = 7890.00 kg/m3

lighter =  299 N

to find out

the volume of the anchor and weigh in air

solution

from question we can say that

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we know weight = mg and buoyant force = water density × g

so volume of anchor is = actual weight - apparent weight / buoyant force

volume of anchor is = 299 / 1000 × 9.81

volume of anchor is = 0.0304791 m³

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weight of anchor is mg

here mass m = density Fe g

density Fe = 7870 from table 14-1

so weight = 7870 × 0.0304791  × 9.81

weigh is 2353.13 N

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