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dybincka [34]
3 years ago
13

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff ideal sprin

gs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do 80.0 J of work when you compress the springs 0.180 m from their uncompressed length.
What magnitude of force must you apply to hold the platform in this position?
Physics
1 answer:
shepuryov [24]3 years ago
6 0

Answer:

The magnitude of force must you apply to hold the platform in this position = 888.89 N

Explanation:

Given that :

Workdone (W) = 80.0 J

length x = 0.180 m

The equation  for this  work done by the spring is expressed as:

W = \frac{1}{2}k_{eq}x^2

Making the spring constant k_{eq} the subject of the formula; we have:

k_{eq} = \frac{2W}{x^2}

Substituting our given values, we have:

k_{eq} = \frac{2*80}{0.180^2}

k_{eq} = 4938.27 \ N.m^{-1}

The magnitude of the force that must be apply  to the hold platform in this position is given by the formula :

F = k_{eq}x

F = 4938.27*0.180

F = 888.89 N

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From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up
NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

7 0
3 years ago
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Answer:

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7 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

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How could two waves on a rope interfere so the rope does not move at all?
coldgirl [10]

Answer:

If a crest formed by one wave interferes with a trough formed by the other wave then the rope will not move at all.

Explanation:

Assume a straight rope tied to both ends is at rest. When a wave is created at one end of the rope, it travels to the other end of the rope through formation of alternative crest and trough. Due to these crest and trough the rope shifts up and down.

But when there are two waves travelling through the rope and both have opposite direction (directed towards one another) in such a way that crest formed by one wave is interfering with the trough formed by the other wave then due to this interference the waves will cancel the effects of each other on the rope and rope will be stable.

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