Answer:
R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi
If you do it in steps
R = 9880 yd * 3 ft/yd = 29640 ft
R = 29640 ft / 5280 ft/mi = 5.61 mi
Answer:
Acceleration = 2.35 m/
Speed = 8.67 m/s
Explanation:
The coefficient of friction , u =0.3
The angle of incline = 30°
The two forces acting on block are weight and friction.
weight along the incline = mg cos60° = = 0.5 mg
Friction along incline = umg cos30° = mg
Friction along incline = 0.26 mg
Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg
Acceleration = = 0.24 g = 2.35 m/
The height of incline = 8 m
Length of the inclined edge = 16 m
v= 8.67 m/s
The electrostatic force between two charges is given by Coulomb's law:
where
ke is the Coulomb's constant
q1 is the first charge
q2 is the second charge
r is the separation between the two charges
By substituting the data of the problem into the equation, we can find the magnitude of the force between the two charges:
I think it is c density and temperature
<span>1.0x10^3 Joules
The kinetic energy a body has is expressed as the equation
E = 0.5 M V^2
where
E = Energy
M = Mass
V = Velocity
Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion
E = 0.5 * 7.2 kg * (17 m/s)^2
E = 3.6 kg * 289 m^2/s^2
E = 1040.4 kg*m^2/s^2
E = 1040.4 J
So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>