Answer:
The acceleration due to gravity is 
times the value of g at the Earth’s surface.
(D) is correct option.
Explanation:
Given that,
Radius = 4000 miles
We need to calculate the gravitational force at surface
Gravitational force on the mass m on the surface of the earth
At r = R
....(I)
We need to calculate the gravitational force at height
Gravitational force on a mass m from the center of the earth,
At r = R + R = 2 R
....(II)
Dividing equation (II) by equation (I)
Hence, The acceleration due to gravity is times the value of g at the Earth’s surface.
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Answer:
the volume decreases at the rate of 500cm³ in 1 min
Explanation:
given
v = 1000cm³, p = 80kPa, Δp/t= 40kPa/min
PV=C
vΔp + pΔv = 0
differentiate with respect to time
v(Δp/t) + p(Δv/t) = 0
(1000cm³)(40kPa/min) + 80kPa(Δv/t) = 0
40000 + 80kPa(Δv/t) = 0
Δv/t = -40000/80
= -500cm³/min
the volume decreases at the rate of 500cm³ in 1 min
Answer:
A constant value everywhere in the universe.
Explanation:
The speed of light in a vacuum is a constant value. It is not affected by change in frequency or wavelength of the light.
Mathematically the speed of light is given as:
c = λf
where λ = wavelength and f - frequency
The speed of light is the constant of proportionality between frequency and wavelength. In order words, wavelength and frequency are inversely proportional. As the wavelength increases, frequency decreases and vice versa.
While the change in wavelength and frequency of light affect the energy of the light, its speed is a constant value as long as the medium is a vacuum.
The speed of light is also not dependent on the manner with which the light wave is moving.
Answer:
Correct answer: P₂ = 25 W
Explanation:
Given: voltage V₁ = 220 V, power P₁ = 100 W, V₂ = 110 V, P₂ = ?
The formula for calculating power is:
P = V · I
We will include in the story and ohm's law:
I = V/R
We will replace the current in the expression for power
P = V · V/R = V²/R ⇒ R = V²/P
We will first calculate the electrical resistance of the bulb which is a constant in the electrical circuit
R = V₁²/P₁ = 220²/ 100 = 48,400 / 100 = 484 Ω
Power consumption of bulb connected to 110 V is:
P₂ = V₂²/R = 110²/484 = 12,100/484 = 25 W
P₂ = 25 W
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