The mass of the object does not change by moving it to another place. ... At the center of the earth the net gravitational force is zero, so the weight will be zero, but its masses will remain same. Hence the mass at the centre of earth will be equl to 50 kg.
Answer:
The ball takes 5s to reach the ground
Explanation:
in order to solve this problem we use the kinematics equation with gravity as acceleration:
we replace the values
We solve this quadratic equation:
t=5s
t=-6s (this solution has not physical sense)
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
Answer:
3420.39 N
Explanation:
Applying,
Fd = 1/2(mv²-mu²)................. Equation 1
Where F = force on the bumber, d = distance, m = mass of the car, v = final velocity, u = initial velocity.
make F the subject of the equation
F = (mv²-mu²)/2d............... Equation 2
From the question,
Given: m = 890 kg, v = 0 m/s (to rest), u = 1.4 m/s, d = 0.255 m
Substitute these values into equation 2
F = [(890×0²)-(890×1.4²)]/(2×0.255)
F = -1744.4/0.51
F = -3420.39 N
The negative sign denotes that the force in opposite direction to the motion of the car.
