The lightbulb, the fire, and the Sun all create light from some energy source.
The Moon doesn't create light. The parts of it that the Sun shines on
are visible, the parts of it that the Sun doesn't shine on are not visible.
Just like a rock on the ground at night, the Moon must be illuminated
by a flashlight in order to be seen. The Sun is the flashlight.
The correct answer to the question is : A) The velocity of the cart after it hits the wall.
EXPLANATION:
Before answering this question, first we have to understand impulse.
Impulse of a body is defined as the change in momentum or the product of force with time.
Mathematically impulse = m ( v- u ).
Here, v is the final momentum and u is the initial momentum.
Hence, we need the velocity of the cart after it hits the wall in order to calculate the impulse of the lab cart.
V=IR can be changed to V/R=I so 10V/2 ohms = 5amps so 5 amps is your answer boss
Answer:
6400 m
Explanation:
You need to use the bulk modulus, K:
K = ρ dP/dρ
where ρ is density and P is pressure
Since ρ is changing by very little, we can say:
K ≈ ρ ΔP/Δρ
Therefore, solving for ΔP:
ΔP = K Δρ / ρ
We can calculate K from Young's modulus (E) and Poisson's ratio (ν):
K = E / (3 (1 - 2ν))
Substituting:
ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)
Before compression:
ρ = m / V
After compression:
ρ+Δρ = m / (V - 0.001 V)
ρ+Δρ = m / (0.999 V)
ρ+Δρ = ρ / 0.999
1 + (Δρ/ρ) = 1 / 0.999
Δρ/ρ = (1 / 0.999) - 1
Δρ/ρ = 0.001 / 0.999
Given:
E = 69 GPa = 69×10⁹ Pa
ν = 0.32
ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)
ΔP = 64.0×10⁶ Pa
If we assume seawater density is constant at 1027 kg/m³, then:
ρgh = P
(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa
h = 6350 m
Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.
