Answer:
1.4m/s
Explanation:
Average velocity is the total distance covered divided by the total time taken.
Average velocity =
Total time taken = 5s + 6s = 11s
The first distance covered = velocity x time = 1.4 x 5 = 7m
second distance covered = velocity x time = 1.4 x 6 = 8.4m
So;
Average velocity =
= 1.4m/s
Answer:
Explanation:
Given that,
Mass of star M(star) = 1.99×10^30kg
Gravitational constant G
G = 6.67×10^−11 N⋅m²/kg²
Diameter d = 25km
d = 25,000m
R = d/2 = 25,000/2
R = 12,500m
Weight w = 690N
Then, the person mass which is constant can be determined using
W =mg
m = W/g
m = 690/9.81
m = 70.34kg
The acceleration due to gravity on the surface of the neutron star is can be determined using
g(star) = GM(star)/R²
g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²
g (star) = 8.49 × 10¹¹ m/s²
Then, the person weight on neutron star is
W = mg
Mass is constant, m = 70.34kg
W = 70.34 × 8.49 × 10¹¹
W = 5.98 × 10¹³ N
The weight of the person on neutron star is 5.98 × 10¹³ N
E=mgh. 196=5kg*9.81m/s^2*h. So h=196/(5*9.81)=4m
Answer:
A) coil A
Explanation:
According to Faraday, Induced emf is given as;
E.M.F = ΔФ/t
ΔФ = BACosθ
where;
ΔФ is change in magnetic flux
θ is the angle between the magnetic field, B, and the normal to the loop of area A
A is the area of the loop
B is the magnetic field
From the equation above, induced emf depends on the strength of the magnetic field.
Both coils have the same area and are oriented at right angles to the field.
Coil A has a magnetic field strength of 10-T which is greater than 1 T of coil B, thus, coil A will have a greater emf induced in it.
Answer:
3.25 × 10^7 m/s
Explanation:
Assuming the electrons start from rest, their final kinetic energy is equal to the electric potential energy lost while moving through the potential difference (ΔV)
Ek = 1/2 mv2 = qΔV .................. 1
Given that V is the electron speed in m/s
Charge of electron = 1.60217662 × 10-19 coulombs
Mass of electron = 9.109×10−31 kilograms
ΔV = 3.0kV = 3000V
Make V the subject of the formula in eqaution 1
V = sqr root 2qΔV/m
V = 2 × 1.60217662 × 10-19 × 3000 / 9.109×10−31
V = 3.25 × 10^7 m/s