Answer:
a) V = -0.227 mV
b) V = -0.5169 mV
Explanation:
a)
Inside a sphere with a uniformly distributed charge density, electric field is radial and has a magnitude
E = (qr) / (4πε₀R³)
As we know that
V = -
By solving above equation, we get
V = (-qr²) / (8πε₀R³)
When
R = 1.81 cm
r = 1.2 cm
q = +2.80 fC
ε₀ = 8.85 × 10⁻¹²
V = (-2.80 × 10⁻¹⁵ × (1.2 × 10⁻²)²) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²)³)
V = -2.27 × 10⁻⁴ V
V = -0.227 mV
b)
When
r = R
R = 1.81 cm
q = +2.80 fC
ε₀ = 8.85 × 10⁻¹²
V = (-qR²) / (8πε₀R³)
V = (-q) / (8πε₀R)
V = (-2.80 × 10⁻¹⁵) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²))
V = -5.169 × 10⁻⁴ V
V = -0.5169 mV
Answer:
The base runner
Explanation:
To know the correct answer to the question, we shall determine the time taken for the baseball and the base runner to get to the home plate. This is illustrated below:
For the baseball:
Horizontal velocity (u) = 30 m/s
Horizontal distance (s) = 60 m
Time (t) =?
s = ut
60 = 30 × t
Divide both side by 30
t = 60 / 30
t = 2 s
Thus, it will take the baseball 2 s to get to the home plate.
For the base runner:
Horizontal velocity (u) = 5 m/s
Horizontal distance (s) = 5 m
Time (t) =?
s = ut
5 = 5 × t
Divide both side by 5
t = 5 / 5
t = 1 s
Thus, it will take the base runner 1 s to get to the home plate.
SUMMARY:
Time taken for the baseball to get to the home plate = 2 s
Time taken for the base runner to get to the home plate = 1 s
From the calculations made above, we can conclude that the base runner will arrive at the home plate first because it took him 1 s to get to the home plate whereas the baseball took 2 s to get there.
For this case, in the next item we have gravitational potential energy:
An apple in a tree.
Suppose we define our reference system at the floor level.
Suppose the apple is at a height h from the floor and has mass m.
The gravitational potential energy of the apple is given by:
U = mgh
Where,
m: apple mass
h: height of the apple with respect to the floor
g: acceleration due to gravity
Answer:
C) an apple on a tree
This is a classic example of conservation of energy. Assuming that there are no losses due to friction with air we'll proceed by saying that the total energy mus be conserved.
Now having information on the speed at the lowest point we can say that the energy of the system at this point is purely kinetic:
Where m is the mass of the pendulum. Because of conservation of energy, the total energy at maximum height won't change, but at this point the energy will be purely potential energy instead.
This is the part where we exploit the Energy's conservation, I'm really insisting on this fact right here but it's very very important, The totam energy Em was
It hasn't changed! So inserting this into the equation relating the total energy at the highest point we'll have:
Solving for h gives us:
It doesn't depend on mass!
When two stars are bound together gravitationally and orbit a common mass, theyre known as B. BINARY STARS.
