Answer
= 60
Hope it helps:)
Answer:
9.4 m/s
Explanation:
According to the work-energy theorem, the work done by external forces on a system is equal to the change in kinetic energy of the system.
Therefore we can write:

where in this case:
W = -36,733 J is the work done by the parachute (negative because it is opposite to the motion)
is the initial kinetic energy of the car
is the final kinetic energy
Solving,

The final kinetic energy of the car can be written as

where
m = 661 kg is its mass
v is its final speed
Solving for v,

Hi there!
According to Newton's second law:
∑F = m · a, where:
∑F = net force (N = kgm/s²)
m = mass (kg)
a = acceleration (m/s²)
Rearrange to solve for acceleration:
F/m = a
20N / 4.0kg = 5 m/s²
Answer:
F = 1.24*10^4 N
Explanation:
Given
Depth of the ship, h = 25 m
Density of water, ρ = 1.03*10^3 kg/m³
Diameter of the hatch, d = 0.25 m
Pressure of air, P(air) = 1 atm
Pressure of water =
P(w) = ρgh
P(w) = 1.03*10^3 * 9.8 * 25
P(w) = 2.52*10^5 N/m²
P(net) = P(w) + P(air) - P(air)
P(net) = P(w)
P(net) = 2.52*10^5 N/m²
Remember,
Pressure = Force / Area, so
Force = Area * Pressure
Area = πr² = πd²/4
Area = 3.142 * 0.25²/4
Area = 3.142 * 0.015625
Area = 0.0491 m²
Force = 0.0491 * 2.52*10^5
F = 12373 N
F = 1.24*10^4 N
Answer:
Hope this helps you find the answer
Explanation:
The proteins, lipids, and polysaccharides that make up most of the food we eat must be broken down into smaller molecules before our cells can use them—either as a source of energy or as building blocks for other molecules.