Answer:
B. the air pressure decreases
Explanation:
As elevation increases, there is less overlying atmospheric pressure mass, so that atmospheric pressure decreases with increasing elevaton.
Answer:
2.9 E14 Hz
Explanation:
As we know by Einstein's equation that energy incident on the photo sensitive surface will be used by the surface to eject electron out of the surface with some kinetic energy.
This is given by
now the threshold frequency is the minimum frequency of the incident photons due to which electrons are ejected out with minimum kinetic energy or least kinetic energy.
So here when KE = 0 in the graph then corresponding to that position the frequency will be given as threshold frequency
so here from graph when KE = 0
The first question's answer is :
If, F=ma
Then, 15N= 2.1kg (a)
15/2.1=a
7.14=a
Therefore, acceleration = 7.14m/s^2
sorry I am not sure about the second question :-(
Answer:
3.64×10⁸ m
3.34×10⁻³ m/s²
Explanation:
Let's define some variables:
M₁ = mass of the Earth
r₁ = r = distance from the Earth's center
M₂ = mass of the moon
r₂ = d − r = distance from the moon's center
d = distance between the Earth and the moon
When the gravitational fields become equal:
GM₁m / r₁² = GM₂m / r₂²
M₁ / r₁² = M₂ / r₂²
M₁ / r² = M₂ / (d − r)²
M₁ / r² = M₂ / (d² − 2dr + r²)
M₁ (d² − 2dr + r²) = M₂ r²
M₁d² − 2dM₁ r + M₁ r² = M₂ r²
M₁d² − 2dM₁ r + (M₁ − M₂) r² = 0
d² − 2d r + (1 − M₂/M₁) r² = 0
Solving with quadratic formula:
r = [ 2d ± √(4d² − 4 (1 − M₂/M₁) d²) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(1 − (1 − M₂/M₁)) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(1 − 1 + M₂/M₁) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(M₂/M₁) ] / 2 (1 − M₂/M₁)
When we plug in the values, we get:
r = 3.64×10⁸ m
If the moon wasn't there, the acceleration due to Earth's gravity would be:
g = GM / r²
g = (6.672×10⁻¹¹ N m²/kg²) (5.98×10²⁴ kg) / (3.64×10⁸ m)²
g = 3.34×10⁻³ m/s²
