Diameter = 0.170 meter
Circumference = 0.170 π meters
530 rpm = 530 circumferences / minute
= (530 x 0.170 π meters) / minute
= 283.06 meter.minute
= 4.72 meters/second
(B)Option b is your answer..........
Gravity is a force because it pulls down on objects.
Answer:
the ship's energy is greater than this and the crew member does not meet the requirement
Explanation:
In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship
W =∫ F dx = ΔK
Let's replace
∫ (α x³ + β) dx = ΔK
α x⁴ / 4 + β x = ΔK
Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J
x (α x³ + β) =
- K₀
= K₀ + x (α x³ + β)
Assuming that the low limit is x = 0, measured from the cargo hangar
Let's calculate
= 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)
Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)
Kf = 2.7 10¹¹ - 1.1475 10¹¹
Kf = 1.55 10¹¹ J
In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement
We evaluate the kinetic energy if the System is well calibrated
W = x F₀ =
–K₀
= K₀ + x F₀
We calculate
= 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶
= (2.7 -2.625) 10¹¹
= 7.5 10⁹ J
Answer:
D)Not enough information
Explanation:
According to Pascal's principle, the pressure exerted on the two pistons is equal:

Pressure is given by the ratio between force F and area A, so we can write

The force exerted on each piston is just equal to the weight of the corresponding mass:
, where m is the mass and g is the gravitational acceleration. So the equation becomes

Now we can rewrite the mass as the product of volume, V, times density, d:

We also know that 
So we can further re-arrange the equation (and simplify g as well):


We are also told that block B has bigger volume than block A:
. However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.