Answer:
B. Gravity and air resistance
Explanation:
Experimentation has proven that :
When the parachute opens, air resistance becomes much greater than gravity. The forces are no longer balanced and this changes the
speed
Answer:
d) 12 V
Explanation:
Due to the symmetry of the problem, the potential (relative to infinity) at the midpoint of the square, is the same for all charges, provided they be of the same magnitude and sign, and be located at one of the corners of the square.
We can apply the superposition principle (as the potential is linear with the charge) and calculating the total potential due to the 4 charges, just adding the potential due to any of them:
V = V(Q₁) + V(Q₂) +V(Q₃) + V(Q₄) = 4* 3.0 V = 12. 0 V
Answer:
J = 0.7 N-s away from wall
Explanation:
Given that,
Mass of the ball, m = 0.2 kg
Initial speed of the ball, u = 2 m/s
Final speed of the ball, v = -1.5 m/s (as it rebounds)
Impulse is equal to the change in momentum. So,

So, the impulse given to the ball is 0.7 N-m and it is away from the wall. Hence, the correct option is (c).
L = length of the incline = 75 m
θ = angle of incline = 22 deg
h = height of skier at the top of incline = L Sinθ = (75) Sin22 = 28.1 m
μ = Coefficient of friction = 0.090
N = normal force by the surface of incline
mg Cosθ = Component of weight of skier normal to the surface of incline opposite to normal force N
normal force "N" balances the component of weight opposite to it hence we get
N = mg Cosθ
frictional force acting on the skier is given as
f = μN
f = μmg Cosθ
v = speed of skier at the bottom of incline
Using conservation of energy
potential energy at the top of incline = kinetic energy at the bottom + work done by frictional force
mgh = f L + (0.5) m v²
mgh = μmg Cosθ L + (0.5) m v²
gh = μg Cosθ L + (0.5) v²
(9.8 x 28.1) = (0.09 x 9.8 x 75) Cos22 + (0.5) v²
v = 20.7 m/s
Answer:
The magnetic force will be 0.256 N in +y direction.
Explanation:
It is given that, a wire along the z axis carries a current of 6.4 A in the z direction. Length of the wire is 8 cm. It is placed in uniform magnetic field with magnitude 0.50 T in the x direction.
The magnetic force in terms of length of wire is given by :

For direction,

So, the magnetic force will be 0.256 N in +y direction.