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3 years ago

Which of the following statements apply to electric charges?

Physics

1 answer:

Gre4nikov [31]3 years ago

6 0

Answer:

The statement "If a positively charged rod is brought close to a positively charged object, the two objects will repel " applies to electric charges.

Explanation:

There are only two types of electric charges. Both having own magnitude but different charge.

1. Positive charge

2. Negative charge

Like charges repel each other and opposite charges always attract each other.

When a positively charged rod is brought close to a positively charged object, the rod and the object will repel.

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A 2-kW electric heater takes 15 min to boil a quantity of water. If this is done once a day and power costs 10 cents per kWh, wh

frozen [14]

Answer:

Explanation:

Given data

power P= 2 kW

time t= 15 min to hours = 15/60= 1/4 h

cost of power consumption per kWh= 10 cent = $0.1

We are expected to compute the cost of operating the heater for 30 days

but let us computer the energy consumption for one day

Energy of heater for one day= 2* 1/4 = 0.5 kWh

the cost of operating the heater for 30 days= 0.5*0.1*30= $1.50

<u><em>Hence it will cost $1.50 for 30 days operation</em></u>

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3 years ago

(a) Calculate the acceleration due to gravity on the surface of the Sun.

Ira Lisetskai [31]

<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

a) Acceleration due to gravity

$g=\frac{GM}{r^2}$

Here we need to find acceleration due to gravity of Sun,

G = 6.67259 x 10⁻¹¹ N m²/kg²

Mass of sun, M = 1.989 × 10³⁰ kg

Radius of sun, r = 6.957 x 10⁸ m

Substituting,

$g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2$

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²

b) Acceleration due to gravity in earth = 9.81 m/s²

Ratio of gravity = 274.21/9.81 = 27.95

Weight = mg

Factor of increase in weight = 27.95

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Eddi Din [679]

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3 years ago

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