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matrenka [14]
3 years ago
8

If the sprinter accelerates at that rate for a distance of 15 m, and then maintains the velocity he has at that point for the re

mainder of the 100-m dash, what will be his time for the entire race?
Physics
1 answer:
ahrayia [7]3 years ago
4 0

Answer:

The time for the entire race is 11.39 sec.

Explanation:

Given that,

Distance = 15 m

Remainder distance = 100 m

Suppose A sprinter begins a race with an acceleration of 3.4 m/s².

We need to calculate the time

Using equation of motion

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

Put the value in the equation

15=0+\dfrac{1}{2}\times3.4\times t^2

t^2=\dfrac{30}{3.4}

t=2.97\ sec

We need to calculate the final velocity of sprinter

Using equation of motion again

v=u+at

Put the value into the formula

v=0+3.4\times2.97

v=10.09\ m/s

We need to calculate the distance covers by sprinter

d=100-15=85\ m

The sprinter need to covers only 85 m.

We need to calculate the time

Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t'=\dfrac{85}{10.09}

t'=8.42\ sec

We need to calculate the time for the entire race

t''=t+t'

Put the value into the formula

t''=2.97+8.42

t''=11.39\ sec

Hence, The time for the entire race is 11.39 sec.

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By a factor of 1/4.

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If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by \small F_1, the following manipulation confirms the answer to this question.

\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}

Here \small F is the force that was applied to the object previously.

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What is the maximum possible resultant of two vectors with magnitudes of 2 and 4 units? what is the minimum possible resultant?
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Why filament lamp is filled with nitrogen and inert gas​
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Read 2 more answers
An 18.8 kg block is dragged over a rough, horizontal surface by a constant force of 156 N acting at an angle of angle 31.9◦ abov
Sav [38]

Answer:

a) W = 2635.56 J

b) Wf = 423.27 J

c) c)  The Sign of the work done by the frictional force (Wf) is negative (-)

d) W=0

Explanation:

Work (W) is defined as the Scalar product of the force (F) by the distance (d) that the body travels due to this force .  

The formula for calculate the work is :

W = F*d*cosα  

Where:

W : work in Joules (J)

F : force in Newtons (N)

d: displacement in meters (m)

α  :angle that form the force (F) and displacement (d)

Known data

m =  18.8 kg : mass of the block

F= 156 N,acting at an angle θ = 31.9◦°: angle  above the horizontal

μk= 0.209 : coefficient of kinetic friction between the cart and the surface

g = 9.8 m/s²: acceleration due to gravity

d = 19.9 m : displacement of the block

Forces acting on the block

We define the x-axis in the direction parallel to the movement of the cart on the floor  and the y-axis in the direction perpendicular to it.

W: Weight of the cart  : In vertical direction  downaward

N : Normal force :  In vertical direction the upaward

F : Force applied to the block

f : Friction force: In horizontal direction

Calculated of the weight  of the block

W= m*g  =  ( 18.8 kg)*(9.8 m/s²)= 184.24 N

x-y components of the force F

Fx = Fcosθ = 156 N*cos(31.9)° = 132.44 N

Fy = Fsinθ = 156 N*sin(31.9)°  = 82.44 n

Calculated of the Normal force

Newton's second law for the  block in y direction  :

∑Fy = m*ay    ay = 0

N-W+Fy= 0

N-184.24+82,44= 0

N = 184.24-82,44 

N = 101.8 N

Calculated of the kinetic friction force (fk):

fk = μk*N = (0.209)*( 101.8)

fk = 21.27 N

a) Work done by the F=156N.

W = (Fx) *d  *cosα

W = (132.44 )*(19.9)(cos0°) (N*m)

W = 2635.56 J

b) Work done by the force of friction

Wf = (fk) *d *cos(180°)

Wf = (21.27 )*(19.9) (-1) (N*m)

Wf = - 423.27 J

Wf = 423.27 J  :magnitude

c)  The Sign of the work done by the frictional force is negative (-)

d) Work done by the Normal force

W = (N) *d *cos(90°)

W = (101.8 )*(19.9) (0) (N*m)

W = 0

4 0
3 years ago
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