Answer:
The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N
Explanation:
For this exercise we will use Newton's second law, let's set a reference system where the x axis is parallel to the plane, in the adjoint we can see the forces in the system.
sin 35 = Wₓ / W
cos 35 = W_y / W
Wₓ = W sin 35
W_y = W cos 35
Wₓ = 2500 9.8 sin 35
Wₓ = 14052.6 N
let's write the equations for each axis
and
Y axis
N-W_y = 0
N = W_y
X axis
F -Wₓ = m a
F = Wₓ + m a = mg sin 35 + m a
F = m (a + g sin 35)
let's calculate
F = 2500 (5 + 9.8 sin 35)
F = 26552.6 N
The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N
Answer:
Check below for the explanation
Explanation:
Since it is stated that the ring is dropped from a height, h, through a non uniform magnetic field, two kinds of force will act on the ring, namely:
- A magnetic force (that is non uniform since the field is non uniform)
- Gravitational force
A certain amount of torque is provided by the non uniform magnetic force on the ring while the force gravity pulls it down. Due to the downward pull by the force of gravity on the ring and the torque acting on it as a result of the non uniform magnetic force, the ring begins to rotate.
Answer:
t = 103.45 n m
Explanation:
given,
refractive index of cornea = 1.38
refractive index of eye drop = 1.45
wavelength of refractive index = 600 nm
refractive index of eye drop is greater than refractive index of cornea and the air.
Formula used in this case
for constructive interference
At m = 0 for the minimum thickness, so
t = 103.45 n m
the minimum thickness of the film of eyedrops t = 103.45 n m
Both positive work and negative work have meaning: Positive work follows when the force has a component parallel to the displacement. Positive work adds energy to a system. Negative work follows when the force has a component opposite or against the displacement.
Answer:
The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.
Explanation:
Given that
q₁ = 5 μ C
q₂ = - 4 μ C
The distance between charges = 50 cm
d= 50 cm
Lets take at distance x from the charge μ C ,the electrical field is zero.
That is why the distance from the charge - 4 μ C = 50 - x cm
We know that ,electric field is given as
Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.
