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Stells [14]
2 years ago
7

If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?

Physics
1 answer:
Law Incorporation [45]2 years ago
5 0

a) 4.8 m/s² his average acceleration during the first 2.5 s

b) 15 m distance he cover during the first 2.5 seconds.

c) 31 m/s his speed as he finished the race.

<h3>(a) How to calculate the Average Acceleration ?</h3>

Average acceleration is calculated by using Newtons first law of motion

v = u + at

where

u is initial velocity

v is final velocity

a is constant acceleration

t is time

Given u = 0m/s , v = 12 m/s and t = 2.5 sec

Therefore average acceleration is given by

a=\frac{v}{t}

a=\frac{12}{2.5}

a = 4.8 m/sec²

b) Here we will use newtons second law of motion

s=ut+\frac{1}{2}at^{2}

On substituting value we get

s = 15m

c) Here we will use newtons third law of motion

v² = u² + 2as

Here u = 0 , a = 4.8 m/s² and s = 100 m

Therefore

v² = 960

v = 31 m/s

Disclaimer: the question was given incomplete in the portal. Here is the complete question.

Question: Usain Bolt's 100m sprint Runner set the world record for the100 meter sprint during the 2009World Championships in Berlin with a time of 9.58 s, reaching a top speed of 12 m/s in about 2.5 s.

a. What was his average acceleration during the first 2.5 s?

b. What distance did he cover during the first 2.5 seconds, assuming his acceleration was constant?

c. If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?

Learn more about Newtons law of motion here:

brainly.com/question/12525794

#SPJ4

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In diving to a depth of 248 m, an elephant seal also moves 296 m due east of his starting point. What is the magnitude of the se
Vinvika [58]

Answer:

The displacement is 386.16m

Explanation:

A seal dives to a depth of 248m. To find displacement, we must calculate the resultant vectors which will give us the displacement

R= sqrt(vector1+vector2)

Since this is a right angle triangle

R= sqrt(248^2 + 296^2)

R= sqrt(149120)

R= 386.16m

Displacement = 386.16m

4 0
3 years ago
At rest, hydrogen has a spectral line at 116 nm. if this line is observed at 107 nm for the star sirius, how fast is sirius movi
Citrus2011 [14]

2.3275862×10¹²km/s fast is sirius moving in km/s.

<h3>Briefing:</h3>

Hydrogen has a spectral line at = 116nm=116×10⁻⁹m

Line is observed at = 107 nm=107×10⁻⁹m

Now, from the Hubble's law

V=(\Delta \lambda / \lambda)×C

Where,

v is the velocity

Δλ = Change in wavelength = 116 - 107= 9nm=9×10⁻⁹m

λ = Actual wavelength=116nm=116×10⁻⁹m

C is the speed of the light=3×10⁸m/s

on substituting the respective values, we get

V=(9/116)×3×10⁸=23275862.069×10⁵m/s

V=2.3275862×10¹²km/s.

<h3>What is the wavelength?</h3>

A waveform signal's wavelength, which is the distance between two identical locations (adjacent crests) in the succeeding cycles, determines whether it is sent through space or via a wire. Typically, in wireless systems, this length is specified in meters (m), centimeters (cm), or millimeters (mm).

To know more about Wavelength visit:

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4 0
1 year ago
You hear the engine roaring on a race car at the starting line. Predict the changes in the sound as the race starts and the car
gogolik [260]

B. The sound of the engine will get louder and the pitch higher.

4 0
3 years ago
Read 2 more answers
Can someone help me?!!!!!
SOVA2 [1]

Answer:magnitude -5; angle 160°

Explanation:

Vector A is described as having magnitude 5 and angle -20°.

To get an equivalent vector, we either leave the magnitude at 5 and add 360° to the angle, or we reverse the magnitude to -5 and add 180° to the angle.

5 @ -20° = 5 @ 340°

5 @ -20° = -5 @ 160°

The third one is the answer.

8 0
3 years ago
A fire helicopter carries a 700-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee
rodikova [14]

Answer:

F = 50636.873 N

Explanation:

given,

bucket of water =  700-kg

length of cable = 20 m

Speed  = 40 m/s

angle of the cable = 38.0°

let air resistance be = F

tension in rope be = T

T cos 38° = m×g..................(1)

T sin 38^0= \dfrac{mv^2}{l} + F..........(2)

equation (1)/(2)

tan 38^0 =\dfrac{\dfrac{mv^2}{l} + F}{mg}

       0.781=\dfrac{\dfrac{700\times 40^2}{20} + F}{700\times 9.8}

           F = 50636.873 N

Hence the force exerted on the bucket is equal to F = 50636.873 N

5 0
3 years ago
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