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Flura [38]
3 years ago
13

To test the resilience of its bumper during low-speed collisions, a 4 060-kg automobile is driven into a brick wall. the car's b

umper behaves like a spring with a force constant 8.00 106 n/m and compresses 3.72 cm as the car is brought to rest. what was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall
Physics
1 answer:
Simora [160]3 years ago
4 0
Mass of car, m = 4060 kg
Spring constant, k = 8.00 x 10⁶ N/m
Spring compression, x = 3.72 cm = 3.72 x 10⁻² m

Let the car strike the wall with speed v m/s.

The kinetic energy of the car is released into the stored energy of the spring (if losses are ignored), so that
(1/2)mv² = (1/2)kx²
(4060 kg)*(v m/s)² = (8 x 10⁶ N/m)(3.72 x 10⁻² m)²
4060 v² = 1.1701 x 10⁴
v = 1.6513 m/s

Answer:  1.65 m/s (nearest hundredth)

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V = - 0.5 [m/s]

Explanation:

In order to solve this problem, we must use the principle of relative speeds. This is for an observer who is on the edge of the river he can see how the river moves to the left and the woman tries to move to the right but can not since:

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What happens when the crests of two waves overlap
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4 0
3 years ago
Can the tangent line to a velocity vs. time graph ever be vertical? Explain.
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Explanation:

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What is the weight of a person who has a mass of 76 kg?​
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5 0
3 years ago
A wire of radius 5 x 10⁻⁴ m is needed to prepare a coil of resistance 40 Ω. The resistivity of the material of the wire is 3.14x
SIZIF [17.4K]

Answer:

100 m

Explanation:

From the question,

R = Lρ/A.................... Equation 1

Where R = resistance of the wire, L = length of the wire, ρ = resistivity of the wire, A = cross sectional area of the wire.

But,

A = πr².................... Equation 2

Where r = radius of the wire.

Substitute equation 2 into equation 1

R = Lρ/πr²

Make L the subject of the equation

L = Rπr²/ρ...................... Equation 3

Given: R = 40 Ω, r = 5×10⁻⁴ m, ρ = 3.14×10⁻⁷ Ωm

Constant: π = 3.14

Substitute these values into equation 3

L = [40×3.14×( 5×10⁻⁴)²]/ (3.14×10⁻⁷)

L = 40×3.14×25×10⁻⁸/(3.14×10⁻⁷)

L = 100 m

Hence the length of the wire is 100 m

4 0
2 years ago
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