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Oksana_A [137]
3 years ago
10

A typical male sprinter can maintain his maximum acceleration for 2.0 s, and his maximum speed is 10 m> s. After he reaches t

his maximum speed, his acceleration becomes zero, and then he runs at constant speed. Assume that his acceleration is constant during the first 2.0 s of the race, that he starts from rest, and that he runs in a straight line.
(a) How far has the sprinter run when he reaches his maximum speed?
(b) What is the magni- tude of his average velocity for a race of these lengths:
(i) 50.0 m;
(ii) 100.0 m;
(iii) 200.0 m?
Physics
1 answer:
Elina [12.6K]3 years ago
6 0

Answer:

a) 10m

b)

i) v=8.33m/s

ii) v=9.09m/s

iii) v=9.52m/s

Explanation:

The first 2.0s the sprinter maintains a constant acceleration. So we have to use the Constant acceleration motion formulas to solve the first part of the movement, then we need to use the contant velocity motion formulas for the rest of the movement.

a)

x=Vo*t+\frac{1}{2}*a*t^2\\\\

x=Vo*t+\frac{1}{2}*(vf-vo)*t\\x=10m

b)

the sprinter run 10 meters with constant accelaration then it start a constant velocity movement, so:

i) x=50-10=40m

t=\frac{40m}{10m/s}=4s

v=\frac{50m}{2s+4s}=8.33m/s

ii) x=100-10=90m

t=\frac{90m}{10m/s}=9s

v=\frac{100m}{2s+9s}=9.09m/s

iii) x=200-10=190m

t=\frac{190m}{10m/s}=19s

v=\frac{200m}{2s+19s}=9.52m/s

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Answer:

option B

Explanation:

Radius of new planet,R' = 2R

Mass of the earth, M' = 2 M

R and M is the Radius and Mass of the earth.

Weight of the person on earth = 500 N

Weight of the person in the new planet = ?

We know acceleration due to gravity is calculated by using formula

   g = \dfrac{GM}{R^2}

now, acceleration due to gravity on the new planet

   g' = \dfrac{GM'}{R'^2}

   g' = \dfrac{G(2M)}{(2R)^2}

   g' =\dfrac{1}{2} \dfrac{GM}{R^2}

   g' =\dfrac{g}{2}

here acceleration due to gravity is half in new planet, so weight will also be half on the new planet.

Weight on the new plane is equal to \dfrac{500}{2} = 250 N

correct answer is option B

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3 years ago
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3 years ago
A competitive go-cart driver is traveling 32 m/s. He sees a caution flag go up, so he slows at a rate of -1.5 m/s^2 in 10.8 s. W
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Answer:

15.8 m/s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 32 m/s.

Acceleration (a) = – 1.5 m/s²

Time (t) = 10.8 s.

Final velocity (v) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration (a) = [final velocity (v) – initial velocity (u)] / time (t)

a = (v – u) /t

With the above formula, we can obtain the final velocity of go-cart driver as follow:

Initial velocity (u) = 32 m/s.

Acceleration (a) = – 1.5 m/s²

Time (t) = 10.8 s.

Final velocity (v) =?

a = (v – u) /t

– 1.5 = (v – 32) / 10.8

Cross multiply

(v – 32) = –1.5 × 10.8

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Collect like terms

v = – 16.2 + 32

v = 15.8 m/s

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3 years ago
A 2 kg and 2 meter long stick is held horizontally. A 4 kg mass is placed at 30 cm, and a 5 kg mass is placed at 75 cm. Determin
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Answer:

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