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ANEK [815]
3 years ago
11

a bicyclist is moving at a speed of 5.5 m/s. she squeezes the brakes, giving her an acceleration of -2.1 m/s². how fast is the b

icyclist going after 0.3 seconds of applying the brakes?
Physics
1 answer:
Blizzard [7]3 years ago
4 0
Formula relating time, velocity and acceleration:

Vf = Vi + at (Vf = Final velocity, Vi = initial velocity)

What you're given:
Initial speed, Vi = 5.5 m/s
Acceleration, a = -2.1 m/s^2
Time, t = 0.3 s

Required:
Vf = ?

Plugging in the given information:

Vf = (5.5) + (-2.1)(0.3)
Vf = 4.87 m/s

The bicyclist's speed is 4.87 m/s after 0.3 s.
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(b) both the temperature and pressure of the gas decrease.

Explanation:

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2 years ago
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You push a 85 kg shopping cart from rest with a net force of 250 n for 5 seconds,at which point it flies off a cliff that is 100
Vikki [24]

m = mass of the cart = 85 kg

F = net force on the cart = 250 N

a = acceleration of the cart

acceleration of the cart is given as

a = F/m

a = 250/85

a = 2.94 m/s²

t = time for which the force is applied = 5 sec

v₀ = initial velocity of the cart = 0 m/s

v = final velocity of the cart just before  it flies off the cliff = ?

using the equation

v = v₀ + a t

inserting the values

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v = 14.7 m/s

consider the motion of cart after it flies off the cliff in vertical direction :

v' = initial velocity in vertical direction = 0 m/s

a' = acceleration in vertical direction = g = acceleration due to gravity = 9.8 m/s²

t' = time taken for the cart to land = ?

Y' = vertical displacement of the cart = height of cliff = 100 m

using the kinematics equation

Y' = v' t' + (0.5) a' t'²

100 = (0) t' + (0.5) (9.8) t'²

t' = 4.52 sec


consider the motion of cart along the horizontal direction after it flies off the cliff

X = distance traveled from the base of cliff = ?

t' = time of travel = 4.52 sec

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distance traveled from the base of cliff is given as

X = v t'

X = 14.7 x 4.52

X  = 66.4 m


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