Answer:
<em>2.753*10^-11N</em>
Explanation:
According to Newton's law of gravitation, the force between the masses is expressed as;
F = GMm/d²
M and m are the distances
d is the distance between the masses
Given
M = 3.71 x 10 kg
m = 1.88 x 10^4 kg
d = 1300m
G = 6.67 x 10-11 Nm²/kg
Substitute into the formula
F = 6.67 x 10-11* (3.71 x 10)*(1.88 x 10^4)/1300²
F = 46.52*10^(-6)/1.69 * 10^6
F = 27.53 * 10^{-6-6}
F = 27.53*10^{-12}
F = 2.753*10^-11
<em>Hence the gravitational force between the asteroid is 2.753*10^-11N</em>
<em></em>
you can find it using the equation: potential energy=mass*gravitational acceleration*height.
energy=50kg*9.8N/kg*40m=19600Nm=19600J or 19.6kJ
Sometimes they use 10 instead of 9.8 for the g constant.
Rember to make me Brainliest!!!
Answer:
v=9.6 km/s
Explanation:
Given that
The mass of the car = m
The mass of the truck = 4 m
The velocity of the truck ,u= 12 km/s
The final velocity when they stick = v
If there is no any external force on the system then the total linear momentum of the system will be conserve.
Pi = Pf
m x 0 + 4 m x 12 = (m + 4 m) x v
0 + 48 m = 5 m v
5 v = 48
v=9.6 km/s
Therefore the final velocity will be 9.6 km/s.
According to Newton's 3rd law, for every force applied, there's an equal and opposite force that will occur. So when you are walking , you are pushed forward but the plain is "pushed back" from the force
Explanation:
- A force is a push or a pull that acts upon an object as a results of its interaction with another object. Forces result from interactions.
- The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.
- According to Newton, whenever objects A and B interact with each other, they exert forces upon each other.
- A variety of action-reaction force pairs are evident in nature.
Answer:
The proportion of flanges that exceeds 0.99 millimeters is 0.6
Explanation:
Given;
integral range of [0.95, 1.05]
Let X be a variable with uniform distribution over the given range.
1 - 0.95 = 0.05, 1.05 - 1 = 0.05
interval = 10 - 0.5 = 9.5
F(x) = 10x + 9.5
When, X exceeds 0.99 millimeters, then the proportion of flanges will be;
P (X > x) = 1 - F(x)
P( X > 0.99 ) = 1 - 10(0.99) + 9.5
P( X > 0.99 ) = 0.6
Therefore, the proportion of flanges that exceeds 0.99 millimeters is 0.6
