Answer:
I think the answer 1
Explanation:
im probably wrong too i dont know
Answer:
ΔU = 5.21 × 10^(10) J
Explanation:
We are given;
Mass of object; m = 1040 kg
To solve this, we will use the formula for potential energy which is;
U = -GMm/r
But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.
Thus;
ΔU = -GMm((1/r_f) - (1/r_i))
Where;
M is mass of earth = 5.98 × 10^(24) kg
r_f is final radius
r_i is initial radius
G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²
Since, it's moving to altitude four times the Earth's radius, it means that;
r_i = R_e
r_f = R_e + 4R_e = 5R_e
Where R_e is radius of earth = 6371 × 10³ m
Thus;
ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)
× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))
ΔU = 5.21 × 10^(10) J
Answer: 16N
Explanation:
Given that:
mass of box M= 2 kg
Initial speed V1 = 4 m/s
Final speed V2 = 8 m/s
Time taken T= 0.5 s
Average strength of this force F = ?
Now, recall that Force is the rate of change of momentum per unit time
i.e Force = momentum / time
Hence, F = M x (V2 - V1)/T
F = 2kg x (8 m/s - 4 m/s) / 0.5s
F = 2kg x (4 m/s / 0.5s)
F = 2kg x 8 m/s/s)
F = 16N
Thus, the average strength of this
force is 16 newton.
Answer:
PE=0.29J
Explanation:
According to the description, there is a angle and in point swung upward of 70°
So,
![Y=13*10^{-2}*cos(70) \\Y=0.0444m](https://tex.z-dn.net/?f=Y%3D13%2A10%5E%7B-2%7D%2Acos%2870%29%20%5C%5CY%3D0.0444m)
Appling the equation of Potential Energy we have,
![PE=mgh\\PE=(0.35)(9.8)(0.13-0.0444)\\PE=0.29J](https://tex.z-dn.net/?f=PE%3Dmgh%5C%5CPE%3D%280.35%29%289.8%29%280.13-0.0444%29%5C%5CPE%3D0.29J)