How are the surface of Mercury and Mars similar to the surface of the Earth?
1 answer:
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Answer:
See the explanation below.
Explanation:
Solving the first image question:
C ) The resulting force is defined by Newton's second law which tells us that the sum of the forces on a body is equal to the product of mass by acceleration. That is, there must be a force that acts on a body to produce an acceleration. If there is no acceleration it is because there are no external forces or developed by the body. And if there is no acceleration the body moves at a constant speed, in a straight line, so the response is C.
For the second image, we must remember that weight is defined as the product of mass by gravitational acceleration.
W = m*g
where:
W = weight [N]
m = mass [kg]
g = gravity acceleration [m/s²]
Now we have
m = 50 [kg]
ge = Earth gravity acceleration = 10 [m/s²]
gp = Distant planet gravity acceleration = 4 [m/s²]
We = ge*m
We = 10*50 = 500 [N]
Wp =gp*m
Wp = 4*50 = 200 [N]
Therefore the answer is D
For the third image, The mass is always going to be preserved, regardless of where the body or object is in space, its weight is the only one that changes since the gravitational force is modified. That is, the mass on the moon and on Earth will always be the same.
m = 70 [kg]
First, we must calculate the acceleration, by means of the following equation of kinematics.
where:
Vf = final velocity = 20 [m/s]
Vo = initial velocity = 0 (because stars from the rest)
a = acceleration [m/s²]
t = time = 4 [s]
20 = 0 + a*4
20 = 4*a
a = 5 [m/s²]
Now using Newton's second law which tells us that the total force acting on a body is equal to the product of mass by acceleration.
F = m*a
where:
F = force [N] (units of Newtons)
m = mass = 2 [kg]
a = acceleration = 5 [m/s²]
F = 2*5
F = 10 [N]
The body of Figure D, since a total force of 25 [N] to the left acts on it, in the rest of cases the force is zero or much less than 25 [N]
50 + 40 - 35 - 30 = F
F = 25 [N]
Answer:
a) 23.2 e V
b) energy of the original photon is 36.8 eV
Explanation:
given,
energy at ground level = -13.6 e V
energy at first exited state = - 3.4 e V
A photon of energy ionized from ground state and electron of energy K is released.
h ν₁ - 13.6 = K
K combine with photon in first exited state giving out photon of energy
= 26.6 e V
h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°
K + ( 3.4 ) = 26.6 e V
a) energy of free electron
K = 26.6 - 3.4 = 23.2 e V
b) energy of the original photon
h ν₁ - 13.6 = K
h ν₁ = 23.2 + 13.6
= 36.8 e V
energy of the original photon is 36.8 eV