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Ksju [112]
3 years ago
6

A 310-N child is in a swing that is attached to a pair of ropes 2.10 m long. Find the gravitational potential energy of the chil

d–Earth system relative to the child's lowest position at the following times.
Physics
1 answer:
Softa [21]3 years ago
7 0

Explanation:

Given that,

Force acting on the child, F = 310 N

Length of the ropes, d = 2.1 m

We need to find the gravitational potential energy of the child–Earth system relative to the child's lowest position when the ropes are horizontal. The potential energy is simply given by :

E=(mg)\times d\\\\E=F\times d\\\\E=310\ N\times 2.1\ m\\\\E=651\ J

Hence, this is the required solution.

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Two children are riding on a merry-go-round that is rotating with a constant angular speed. Abbie is one meter from the center o
Ainat [17]

Answer:

  • <em>Abbie’s acceleration is (1/2) Zak’s acceleration.</em>

Explanation

1. <u>Data</u>:

a) ω = constant

b) Abbie: r₁ = 1 m

c) Zak: r₂ = 2 m

d) Ac₁ = ? Ac₂

2. <u>Formulae</u>

  • Ac = ω² r

3. <u>Solution</u>:

a) Abbie:

  • Ac₁ = ω² r₁  =  ω² (1m)

b) Zack:

  • Ac₂ = ω² r₂  = ω² (2m)

c) Divide Ac₁ / Ac₂

  • Ac₁ / Ac₂ =  ω² (1m) / [ω² (2m) ] = 1/2

⇒      Ac₁ = (1/2) Ac₂ = Ac₂ / 2 = 0.5 Ac₂

5 0
3 years ago
the speed of light in a certain medium is 0.6c. find critical angle , if the index of refraction is 1.67​
baherus [9]

Answer:

\theta_c = 36.78^o

Explanation:

The relationship between the refractive index and the critical angle is given as follows:

\eta = \frac{1}{Sin\ \theta_c} \\\\Sin\ \theta_c = \frac{1}{\eta}\\\\\theta_c = Sin^{-1}(\frac{1}{\eta} )

where,

η = refractive index = 1.67

θc = critical angle =?

Therefore,

\theta_c = Sin^{-1}(\frac{1}{1.67} )

\theta_c = 36.78^o

4 0
3 years ago
Suppose we hang a heavy ball with a mass 13 kg (so the weight is ) from a steel wire 3.9 m long that is 3.1 mm in diameter (radi
Katena32 [7]

Answer:

1.635×10^-3m

Explanation:

Young modulus is the ratio of the tensile stress of a material to its tensile strain.

Young modulus = Tensile stress/tensile strain

Tensile stress = Force/Area

Given force = 130N

Area = Πr² = Π×(1.55×10^-3)²

Area = 4.87×10^-6m²

Tensile stress = 130/4.87×10^-6 = 8.39×10^7N/m²

Tensile strain = extension/original length

Tensile strain = e/3.9

Substituting in the young modulus formula given young modulus to be 2×10¹¹N/m²

2×10¹¹N/m² = 8.39×10^7/{e/3.9)}

2×10¹¹ = (8.39×10^7×3.9)/e

2×10¹¹e = 3.27×10^8

e = 3.27×10^8/2×10¹¹

e = 1.635×10^-3m

The stretch of the steel wire will be

1.635×10^-3m

7 0
3 years ago
1. An object in free fall will have an initial velocity equal to zero when: a. It is thrown vertically down
Ne4ueva [31]

Answer:

b. It is dropped

Explanation:

If the initial velocity is zero, the object move from rest. That happens if the object is dropped

6 0
3 years ago
One horsepower (hp) is the amount of power required to lift a 75-kg mass a vertical distance of 1 m in 1 s. What is 2 hp equival
Vladimir [108]

Answer:

1470 W

Explanation:

Power: This can be defined as the rate at which work is done or energy is used up. The S.I unit of power is Watt (W).

The expression for power is given as,

P = Energy/time

P = mgh/t ...................... Equation 1

Where P = power, m = mass, h = height, t = time, g = acceleration due to gravity.

Given: m = 75 kg, g =9.8 m/s², h = 1 m, t = 1 s.

Substitute into equation 1

P = (75×1×9.8)/1

P = 735 W.

From the above,

1 hp = 735 W

2 hp = (2×735) W

2 hp = 1470 W.

Hence 2 hp = 1470 W

8 0
3 years ago
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