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Anna35 [415]
3 years ago
15

A sprinter starts from rest and accelerates to her maximum speed of 10.5 m/s in a distance of 11.0 m. (a) What was her accelerat

ion, if you assume it to be constant? 0 Incorrect: Your answer is incorrect. Units are required for this answer. seenKey 5.01 m/s^2 (b) If this maximum speed is maintained for another 96.3 m, how long does it take her to run 107.3 m?
Physics
1 answer:
Anvisha [2.4K]3 years ago
8 0

Answer:

a)a=5.01m/s^2

b)t=11.26s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)

{Vf^{2}-Vo^2}/{2.a} =X(2)

X=Xo+ VoT+0.5at^{2}    (3)

X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

to solve the question a, we can use the ecuation number 2

Vo=0

Vf=10.5 m/s

x=11m

{Vf^{2}-Vo^2}/{2.a} =X

{Vf^{2}-Vo^2}/{2.x} =a

{10.5^{2}-0^2}/{2x11} =a

a=5.01m/s^2

to find the time we can use the ecuation number 1

Vf=Vo+a.t

t=(Vf-Vo)/a

t=(10.5-0)/5.01=2.09s

part b

in this case  the spees is constant, so the movement is defined by the following ecuation

X=VT

t=x/v

t=96.3/10.5=9.17s

to find the total time we sum the times when the speed is constant and when the acceleration is constan

t=9.17+2.09

t=11.26s

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Answer:

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F = N e E     where E is the value of the field and N e the charge Q

M g = N e E      and M g is the weight of the drop

N = M g / (e E)

N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13

.00011 kg is a very large drop

Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs

Check:     N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13   electrons

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50 kg of water at 75o C is cooled to 25o C. How much heat was given off?
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A 200kg bumper car moving 3 m/s collides head on with a 392kg bumper car moving 6m/s. After impact, the larger car continues tra
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Answer:

the final speed of the smaller car is 5.624 m/s

Explanation:

Given;

mass of the small car, m₁ = 200 kg

initial velocity of the small car, u₁ = 3 m/s

mass of the larger car, m₂ = 392 kg

initial velocity of the larger car, u₂ = 6 m/s

final velocity of the larger car, v₂ = 1.6 m/s

let the direction of the larger car be positive

let the direction of the smaller car be negative

Apply the principle of conservation of linear momentum to determine the final speed of the smaller car.

m₁u₁ + m₂u₂ = m₁v₁  +  m₂v₂

200(-3)  +  392(6)  = 200v₁   +  392 x 1.6

-600 + 2352  = 200v₁    +  627.2

1752   =  200v₁     +   627.2

1752  -  627.2     =  200v₁

1124.8  = 200v₁

v₁  =  1124.8/200

v₁ = 5.624 m/s

Therefore, the final speed of the smaller car is 5.624 m/s

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