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Ne4ueva [31]
2 years ago
10

1. For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a tur

n b. a child swinging around a pole c. a person sitting on a bench facing the center of a carousel d. a rock swinging on a string e. the Earth orbiting the Sun.
Physics
1 answer:
GenaCL600 [577]2 years ago
6 0

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

d. a rock swinging on a string

e. the Earth orbiting the Sun.

Answer:

Considering a

    The force providing the centripetal force is the frictional force on the tires \

          i.e  \mu mg  =  \frac{mv^2}{r}

    where \mu is the coefficient of static friction

Considering b

   The force providing the centripetal force is the force experienced by the boys  hand on the pole

Considering c

     The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

     The force providing the centripetal force is the tension on the string

Considering e

      The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation:

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Artyom0805 [142]

Heat engines were developed during industrial revolution.

Generally a heat engine contains three parts i.e source, sink and working substance.

The source of a heat engine is present at a higher temperature as compared to the sink. Due to the temperature difference, the heat will flow from source to sink through working substance.

Let us consider  T_{1}\ and\ T_{2} are the temperature of source and sink.

As the source is at higher temperature as compared to sink, heat will flow from source to sink.

Let\ Q_{1}\ and\ Q_{2} are the heat provided by source and heat rejected to sink.

Hence, the work done by the working substance will be -

                                                W\ =\ Q_{1}-Q_{2}

The efficiency of a heat engine is defined as the ratio of output to the input energy.

Here output = workdone [W]

Hence, the efficiency of a heat engine is calculated as -

                     Efficiency\ [\eta]=\frac{W}{Q_{1}}

                                        \eta\ =\frac{Q_{1}- Q_{2}} {Q_{1}}

                                               =\ 1-\frac{Q_{2}} {Q_{1}}

This is the expression for the efficiency of heat engine.

Here, all the heat absorbed by the working substance can not be converted to desired output. The efficiency of a heat engine can not be 100 percent. Some amount of heat is lost in the form of sound and heat due to the friction which is produced due to the relative motion between various parts of the machine.

6 0
2 years ago
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A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe
Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

=  surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ /  2ε₀

= 6.78 X 10³ N/C

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2 years ago
The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of +Q is located at one of the cor
Sveta_85 [38]

Answer:

d) 12 V

Explanation:

Due to the symmetry of the problem, the potential (relative to infinity) at the midpoint of the square, is the same for all charges, provided they be of the same magnitude and sign, and be located at one of the corners of the square.

We can apply the superposition principle (as the potential is linear with the charge) and calculating the total potential due to the 4 charges, just adding the potential due to any of  them:

V = V(Q₁) + V(Q₂) +V(Q₃) + V(Q₄) = 4* 3.0 V = 12. 0 V

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2 years ago
Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
balandron [24]

Answer with Explanation:

We are given that

Radius of  solid core wire=r=2.28 mm=2.28\times 10^{-3} m

1mm=10^{-3} m

Radius of each strand  of thin wire=r'=0.456 mm=0.456\times 10^{-3} m

Current density of each wire=J=3750 A/m^2

a.Area =\pi r^2

Where \pi=3.14

Using the formula

Cross section area of copper wire has solid core =3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2

Current density =J=\frac{I}{A}

Using the formula

3750=\frac{I}{16.3\times 10^{-6}}

I=3750\times 16.3\times 10^{-6}=0.061 A

Total number of strands=19

Area of strand wire=A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2

J'=\frac{I'}{A'}

3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}

I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2

I'=0.047 A

b.Resistivity of copper wire=\rho=1.69\times 10^{-8}\Omega-m

Length of each wire =6.25 m

Resistance, R=\frac{\rho l}{A}

Using the formula

Resistance of solid core wire=R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega

Resistance of strand wire=R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega

7 0
2 years ago
The chart shows data for four moving objects. A 4 column table with 4 rows. The first column is labeled Object with entries, W,
KatRina [158]

Answer:

y

Explanation:

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3 0
2 years ago
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