1.47x10^5 Joules
The gravitational potential energy will be the mass of the object, multiplied by the height upon which it can drop, multiplied by the local gravitational acceleration. And since it started at the top of a 60.0 meter hill, halfway will be at 30.0 meters. So
500 kg * 30.0 m * 9.8 m/s^2 = 147000 kg*m^2/s^ = 147000 Joules.
Using scientific notation and 3 significant figures gives 1.47x10^5 Joules.
Answer:
unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]
Explanation:
Given:
v = (-23.2, -104.4, 46.4) m/s
Above expression describes spacecraft's velocity vector v.
Find:
Find unit vector in the direction of spacecraft velocity v.
Solution:
Step 1: Compute magnitude of velocity vector.
mag (v) = sqrt ( 23.2^2 + 104.4^2 + 46.4^2)
mag (v) = 116.58 m/s
Step 2: Compute unit vector unit (v)
unit (v) = vec (v) / mag (v)
unit (v) = [ -23.2 i -104.4 j + 46.4 k ] / 116.58
unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]
Answer:
a) force between them is attraction, b) F = 1.125 10⁻² N
Explanation:
In this case the electric force is given by Coulomb's law
F =
In the exercise they give us the values of the loads
q1 = - 10 mC = -10 10⁻³ C
q2 = 5 mC = 5 10⁻³ C
d = 20 cm = 0.20 m
let's calculate
F = 9 10⁹ 
F = 1.125 10⁻² N
To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction
Answer:
The expected dynamic error is 0.019
The phase shift is -23.10°C
Explanation:
The explanation is shown on the first uploaded image
Answer:
If the line is curved, the slope is changing, which also means the velocity is changing. In a distance-time graph, the gradient of the line is equal to the speed of the object. The more the gradient (and the steeper the line) the faster the object is moving.
