Answer:
The car's displacement during this time is 25.65 meters.
Explanation:
Given that,
Final velocity of the car, v = 4.5 m/s
Deceleration of the car, 
Let u is the initial speed of the car. It is given by :
u = 12.6 m/s
Let d is the car's displacement during this time. It can be calculated using second equation of motion as :
d = 25.65 meters
So, the car's displacement during this time is 25.65 meters. Hence, this is the required solution.
The first and third choices could both do it, but the first choice makes a much better, clear demonstration.
Answer:
28.5 m/s
18.22 m/s
Explanation:
h = 20 m, R = 20 m, theta = 53 degree
Let the speed of throwing is u and the speed with which it strikes the ground is v.
Horizontal distance, R = horizontal velocity x time
Let t be the time taken
20 = u Cos 53 x t
u t = 20/0.6 = 33.33 ..... (1)
Now use second equation of motion in vertical direction
h = u Sin 53 t - 1/2 g t^2
20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)
t = 1.17 s
Put in equation (1)
u = 33.33 / 1.17 = 28.5 m/s
Let v be the velocity just before striking the ground
vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s
vy = uSin 53 - 9.8 x 1.17
vy = 28.5 x 0.8 - 16.66
vy = 6.14 m/s
v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2
v = 18.22 m/s
See coulomb's law. Force is inversely proportional to the distance squared. So if you multiply r by 2, the force is multiplied by (½)² = ¼.
a. F/4
Answer:
B)
The magnitude of induced emf in the conducting loop is 0.99 mV.
Explanation:
Rate of increase in magnetic field per unit time = 0.090 T/s
Area of the conducting loop = 110 cm^2 = 0.0110 m^2
Electromagnetic induction is the production of an emf or voltage in a coil of wire due to a changing magnetic field through the coil.
Induced e.m.f is given as:
EMF = (-N*change in magnetic field/time)*Area
EMF = rate of change of magnetic field per unit time * Area
EMF = 0.090 * 0.0110
EMF = 0.00099 V
EMF = 0.99 mV
