Answer:
The value is
Explanation:
The radius is
The mass of the ordinary matter is
Generally the speed of the star is mathematically represented as
Here G is the gravitational constant with a value
So
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A 1500kg car traveling at 25m/s skids to a stop. The force of friction between the tires and the road is 10500N. How far does th
1 answer:
44.64m
Explanation:
Given parameters:
Mass of the car = 1500kg
Initial velocity = 25m/s
Frictional force = 10500N
Unknown:
Distance moved by the car after brake is applied = ?
Solution:
The frictional force is a force that opposes motion of a body.
To solve this problem, we need to find the acceleration of the car. After this, we apply the appropriate motion equation to solve the problem.
-Frictional force = m x a
the negative sign is because the frictional force is in the opposite direction
m is the mass of the car
a is the acceleration of the car
a = =
= -7m/s²
Now using;
V² = U² + 2as
V is the final velocity
U is the initial velocity
a is the acceleration
s is the distance moved
0² = 25² + 2 x 7 x s
0 = 625 - 14s
-625 = -14s
s = 44.64m
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Velocity problems brainly.com/question/10932946
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Answer:
The speed decreases 75%.
Explanation:
- Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
- For the first collission, only mass 1 is moving before it, so we can write the following equation:
- Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:
- From (1) and (2) we get:
- v₁ = v₀/2 (3)
- Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:
- Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:
- From (4) and (5) we get:
- v₂ = v₀/3 (6)
- Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:
- Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:
- From (7) and (8) we get:
- v₃ = v₀/4
- This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.