Answer:
idk what the answer plz help!!!
Explanation:
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:
11250 N
Explanation:
From the question given above, the following data were obtained:
Normal force (R) = 15000 N
Coefficient of static friction (μ) = 0.75
Frictional force (F) =?
Friction and normal force are related by the following equation:
F = μR
Where:
F is the frictional force.
μ is the coefficient of static friction.
R is the normal force.
With the above formula, we can calculate the frictional force acting on the car as follow:
Normal force (R) = 15000 N
Coefficient of static friction (μ) = 0.75
Frictional force (F) =?
F = μR
F = 0.75 × 15000
F = 11250 N
Therefore, the frictional force acting on the car is 11250 N
Answer:
25.6 V
Explanation:
The kinetic energy of electron associated with its potential difference is given by eV which is equal to the 1/2 mv^2.
m = 9.1 x 10^-31 kg, v = 3 x 10^6 m/s, e = 1.6 x 10^-19 C
eV = 1/2 m v^2
V = mv^2 / 2 e
V = (9.1 x 10^-31) x (3 x 10^6)^2 / (2 x 1.6 x 10^-19)
V = 25.6 V
The forces acting on the book (Fn and Fg) are equal. Therefore, the netforce is zero and it is an unbalanced force.
