The mass of a rollercoaster car moving at a velocity of 30 meters/second and has a momentum of 2.5 × 104 kilogram meters/second is 8.3 × 10²kg.
<h3>How to calculate mass?</h3>
The mass of the roller coaster car can be calculated using the following formula:
P = m × v
Where;
- P = momentum
- m = mass
- v = velocity
m = 2.5 × 10⁴ ÷ 30
m = 8.3 × 10²kg
Therefore, the mass of a rollercoaster car moving at a velocity of 30 meters/second and has a momentum of 2.5 × 104 kilogram meters/second is 8.3 × 10²kg.
Learn more about mass at: brainly.com/question/19694949
#SPJ1
Answer:
5.23km/s
Explanation:
Given
Radius of Earth = 6.37 * 10^6 m
Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m
Gravity Acceleration on Satellite Altitude = 1.87965m/s²
For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.
Centripetal Acceleration = V²/R
So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R
Make V the subject of formula
A = V²/R
V² = AR
V = √AR
Where R = (radius of earth) + (altitude of satellite)
R = 6.37 * 10^6 + 8.2 * 10^6
R = 14.57 * 10^6m
A = 1.87965m/s²
V = √(1.87965 * 14.57x10^6)
V = √27386500.5
V = 5233.211299001789
V = 5233.2113 m/s ------- Approximated
V = 5.23km/s
Steel paper clip because it can be moved by the magnet and it is lighter than the iron nail
SORRY i forgot but i think its A because the sun has a really strong gravitational pull
sorry and hope it helps
Answer:
Explanation:
Give that,
Spring constant (k)=40N/m
Force applied =75N
Since the force is applied to the right, we don't know if it is compressing or stretching the spring
So let assume it compress
Using hooke's law
F=-ke
e=-F/k
Then, e=-75/40
e=-1.875m
The deformation is 1.875m.
Let assume it stretch
Using hooke's law
-F=-ke
e=F/k
Then, e=75/40
e=1.875m
The elongation is 1.875m
