Answer:
9.44 eV
Explanation:
q = Amount of charge = 86.3 μC = 86.3 x 10⁻⁶ C (Since 1 μC = 10⁻⁶ C)
V = Electric potential of the charge = 17.5 Volts
U = Amount of electric potential energy carried by charge
Amount of electric potential energy carried by charge is given as
U = q V
Inserting the values
U = (86.3 x 10⁻⁶) (17.5)
U = 0.00151 J
U = 
U = 9.44 eV
KE = Kinetic energy the charge could have
Using conservation of energy
KE = U
KE = 9.44 eV
Because it records speed of the car at a certain time, the independent variable should be time and dependent would be speed or velocity. Since it's taken every second, it would be considered instantaneous velocity, which is D.
The principle that
describes the force on a charge in a magnetic field was named for Hendrik Lorentz, and the principle
is named as Lorentz force.
Hendrik Antoon Lorentz<span> <span>(18 July 1853 – 4 February 1928) was a Dutch </span></span>physicist<span> <span>who shared the 1902 </span></span>Nobel Prize in Physics<span> <span>with </span></span>Pieter Zeeman<span> <span>for the discovery and theoretical explanation
of the </span></span>Zeeman effect<span>.<span> </span></span>
Answer:
The speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.
Explanation:
Given;
mass of block, m = 4 kg
coefficient of kinetic friction, μk = 0.25
angle of inclination, θ = 30°
initial speed of the block, u = 5 m/s
From Newton's second law of motion;
F = ma
a = F/m
Net horizontal force;
∑F = mgsinθ + μkmgcosθ

At the top of the ramp, energy is conserved;
Kinetic energy = potential energy
¹/₂mv² = mgh
¹/₂ v² = gh
¹/₂ x 5² = 9.8h
12.5 = 9.8h
h = 12.5/9.8
h = 1.28 m
Height of the ramp is 1.28 m
Now, calculate the speed of the block (in m/s) when it has returned to the bottom of the ramp;
v² = u² + 2ah
v² = 5² + 2 x 7.022 x 1.28
v² = 25 + 17.976
v² = 42.976
v = √42.976
v = 6.56 m/s
Therefore, the speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.