Answer:
g' = 13.5 m/s²
Explanation:
The acceleration due to gravity on surface of earth is given by the formula:
g = GMe/Re² --------------- euation 1
where,
g = acceleration due to gravity on surface of earth
G = Universal Gravitational Constant
Me = Mass of Earth
Re = Radius of Earth
Now, the the acceleration due to gravity on the surface of Kepler-62e is:
g' = GM'/R'² --------------- euation 1
where,
g' = acceleration due to gravity on surface of Kepler-62e
G = Universal Gravitational Constant
M' = Mass of Kepler-62e = 3.57 Me
R' = Radius of Kepler-62e = 1.61 Re
Therefore,
g' = G(3.57 Me)/(1.61 Re)²
g' = 1.38 GMe/Re²
using equation 1:
g' = 1.38 g
where,
g = 9.8 m/s²
Therefore,
g' = 1.38(9.8 m/s²)
<u>g' = 13.5 m/s²</u>
<h2>
Option A is the correct answer.</h2>
Explanation:
When an elevator moves upward with consonant acceleration a, the overall acceleration on the body is given by
a' = a + g
So acceleration of pendulum is a + g.
We have equation for period of simple pendulum
In normal case a' = g here a' is more.
From the equation we can see that period of simple pendulum is inversely proportional to square root of acceleration.
Since acceleration increases period decreases.
Option A is the correct answer.
Is D) Zeroes and One's :))))
Answer:
29.0 g
Explanation:
The mass of the piece of gold is given by:
m = dV
where
m is the mass
d is the density
V is the volume of the piece of gold
The density of gold is
d = 19.3 g/cm^3
while the volume of the sample is equal to the volume of displaced water, so
V = 64.5 mL - 63.0 mL = 1.5 mL
And since
1 mL = 1 cm^3
the volume is
V = 1.5 cm^3
So the mass of the piece of gold is:
m = (19.3 g/cm^3)(1.5 cm^3)=29.0 g
<span>When an electron in a hydrogen atom moves from a higher to a lower energy state, "Energy Emits" that difference is equal to quantum of light
Hope this helps!</span>
