Answer:
the distance traveled by the sports car is 90 m
Explanation:
Given;
mass of the sports car, m = 500 kg
initial velocity of the sports car, u = 0
final velocity of the sport car, v = 30 m/s
time of motion of the car, t = 6 s
The distance traveled by the sports car is calculated as;
Therefore, the distance traveled by the sports car is 90 m
Answer:
1.31×10¯⁶ N
Explanation:
From the question given above, the following data were obtained:
Mass of John (M₁) = 81 Kg
Mass of Mike (M₂) = 93 Kg
Distance apart (r) = 0.620 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force (F) =?
The gravitational force between the two students, John and Mike, can be obtained as follow:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 81 × 93 / 0.62²
F = 6.67×10¯¹¹ × 7533 / 0.3844
F = 1.31×10¯⁶ N
Therefore, the gravitational force between the two students, John and Mike, is 1.31×10¯⁶ N
Answer:
Weight (mass) = 16.5 kg
velocity = 0 m/a
acceleration =2.6 m/s^2
displacement = 13.2m
now,
acceleration = velocity/ time
2.6 = 0 / t
t = o / 2.6
t = o
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1) The average velocity is
2) The instantaneous velocity is
Explanation:
1)
The average velocity of an object is given by
where
d is the displacement
t is the time elapsed
In this problem, the position of the particle is given by the function
where t is the time.
The position of the particle at time t = 6 sec is
While the position at time t = 12 sec is
So, the displacement is
And therefore the average velocity is
2)
The instantaneous velocity of a particle is given by the derivative of the position vector.
The position vector is
By differentiating with respect to t, we find the velocity vector:
Therefore, the instantaaneous velocity at any time t can be found by substituting the value of t in this expression.
Learn more about velocity:
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