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ycow [4]
3 years ago
7

How far from the base of the platform does she land?

Physics
1 answer:
seropon [69]3 years ago
3 0

When Janet leaves the platform, she's moving horizontally at 1.92 m/s.  We assume that there's no air resistance, and gravity has no effect on horizontal motion.  There's no horizontal force acting on Janet to make her move horizontally any faster or slower than 1.92 m/s.

She's in the air for 1.1 second before she hits the water.

Moving horizontally at 1.92 m/s for 1.1 second, she sails out away from the platform

(1.92 m/s) x (1.1 sec) = <em>2.112 meters</em>

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Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birm
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Where do people get their energy to turn a wheel??
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from food eaten (i.e contain carnlbohyadrates)

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The amount of heat energy required to raise the temperature of a unit mass of a material one degree is
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5 0
3 years ago
A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i
erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

\mu_k = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem,  we have

Work done by the friction force = change in kinetic energy of the mass 2

or

0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring =  Kinetic energy of the mass 2

or

\frac{1}{2}kx^2=\frac{1}{2}mv_3^2

on substituting the values, we get

\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2

or

x = 2.86 m

8 0
3 years ago
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