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MariettaO [177]
3 years ago
12

A child with mass of 25 kg gets into a toy car with mass of 80 kg on a playground, causing it to sink on its springs (with effec

tive spring constant 791 N/m). An adult walks by and gives the top of the car a shove, causing it to undergo oscillations with amplitude 31 cm in the vertical direction. Assuming the oscillations are simple harmonic, what is the frequency of the oscillations?
Physics
1 answer:
Bess [88]3 years ago
4 0

Answer:

The frequency of the oscillation is 0.9Hz

Explanation:

This problem bothers on simple harmonic motion of a spring

Given data

Mass of the child m= 25kg

Spring constant k=791 N/m

Amplitude a= 31cm

But the period of the motion as a result of the adults sholve is expressed as

T=2π√m/k

T=2*3.142√25/791

T=6.284√0.031

T=6.284*0.176

T=1.11 sec

But frequency F=1/T

F=1/1.11

F=0.9Hz

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Answer:

A

Explanation:

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n₁ sin θ₁ = n₂ sin θ₂

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The index of refraction of air is approximately 1.  So:

1 sin 30° = 1.52 sin θ

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5 0
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Consider two massless springs connected in series. Spring 1 has a spring constant k1, and spring 2 has a spring constant k2. A c
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Answer:

a. k = (1/k₁ + 1/k₂)⁻¹ b. k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹

Explanation:

Since only one force F acts, the force on spring with spring constant k₁ is F = k₁x₁ where x₁ is its extension

the force on spring with spring constant k₂ is F = k₂x₂ where x₁ is its extension

Let F = kx be the force on the equivalent spring with spring constant k and extension x.

The total extension , x = x₁ + x₂

x = F/k = F/k₁ + F/k₂

1/k = 1/k₁ + 1/k₂

k = (1/k₁ + 1/k₂)⁻¹

B

The force on spring with spring constant k₃ is F = k₃x₃ where x₃ is its extension

Let F = kx be the force on the equivalent spring with spring constant k and extension x.

The total extension , x = x₁ + x₂ + x₃

x = F/k = F/k₁ + F/k₂ + F/k₃

1/k = 1/k₁ + 1/k₂ + 1/k₃

k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹

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What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x
lyudmila [28]

The question is incomplete. Here is the complete question.

A uniform electric field of 2kN/C points in the +x-direction.

(a) What is the change in potential energy of a +2.00nC test charge, U_{electric,b} - U_{electric,a} as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?

(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?

(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?

Answer: (a) ΔU = 3.2×10^{-6} J

(b) KE = 2×10^{-6} J

Explanation: <u>Potential</u> <u>Energy</u> (U) is the amount of work done due to its position or condition and its unit is Joule (J). <u>Kinetic</u> <u>Energy</u> (KE) is the ability to do work by virtue of velocity and the unit is also (J). <u>Mechanical</u> <u>Energy</u> is the sum of Potential and Kinetic Energies of a system.

(a) Related to electricity, Potential Energy can be calculated as:

ΔU = Eqd

where E is the electric field (in N/C);

q is the charge (in C);

d is the distance between plaques (in m);

For a at x = - 30cm and b at x = 50 cm:

E = 2×10^{3} N/C

q = 2×10^{-9} C

d = 50 - (-30) = 80×10^{-2} = 8×10^{-1}m

ΔU = U_{electric,b} - U_{electric,a} = Eqd

U_{electric,b} - U_{electric,a} = 2×10^{3} .  2×10^{-9} . 8×10^{-1}

ΔU = 3.2×10^{-6} J

(b) Mechanical Energy is constant, so:

KE_{i} + U_{i} = KE_{f} + U_{f}

Since the initial position is zero and there is no initial kinetic energy:

KE_{f} = - U{f}

KE_{f} = - (2×10^{3}. 2×10^{-9} . 5×10^{-1})

KE_{f} = - 2.10^{-6} J

(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.

8 0
3 years ago
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