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3 years ago

If the fly experienced a force of 100 N when it hits how big was the force on the vehicle in this collision

Physics

1 answer:

iogann1982 [59]3 years ago

8 0

Answer:

100N

Explanation:

because newton's third law of motion states that if body A exerts a force on the body B, then Body B will exert an equal force but opposite in direction force on body A

so if the fly experienced 100N, then the car will also experience 100N

hope you get it

please mark

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Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet

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Answer:

v = 8.09 m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

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Em₀ = U = m gh

final point. To go down the slope

Em_f = K = ½ m v²

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W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

N - W_y = 0

N = W_y

X axis

Wₓ - fr = ma

let's use trigonometry

sin θ = y / L

sin θ = 11/110 = 0.1

θ = sin⁻¹ 0.1

θ = 5.74º

sin 5.74 = Wₓ / W

cos 5.74 = W_y / W

Wₓ = W sin 5.74

W_y = W cos 5.74

the formula for the friction force is

fr = μ N

fr = μ W cos θ

Work is friction force is

W_fr = - μ W L cos θ

Let's use the relationship of work with energy

W + ΔU = ΔK

-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²

v² = - 2 μ g L cos 5.74 +2 (gh)

v² = 2gh - 2 μ gL cos 5.74

let's calculate

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v² = 215.6 -150.16

v = √65.44

v = 8.09 m/s

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2 years ago

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Answer:

Part a)

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Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

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Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

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Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

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Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

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3 years ago

A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density

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Answer:

a) K = 2/3 π G m ρ R₁³ / R₂ , b) U = - G m M / r

Explanation:

The law of universal gravitation is

F = G m M / r²

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F = m a

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a = v² / R₂

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v² = G M / R₂

They give us the density of the planet

ρ = M / V

V = 4/3 π R₁³

M = ρ V

M = ρ 4/3 π R₁³

v² = 4/3 π G ρ R₁³ / R₂

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K = ½ m (4/3 π G ρ R₁³ / R₂)

K = 2/3 π G m ρ R₁³ / R₂

Part B

Potential energy and strength are related

F = - dU / dr

∫ dU = - ∫ F. dr

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U - U₀ = G m M (- r⁻¹)

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U - U₀ = -G m M (1 / $r_{f}$ - 1 / $r_{i}$ )

They indicate that for ri = ∞ U₀ = 0

U = - G m M / r

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3 years ago