Answer with Explanation:
We are given that
Magnetic field,B=
Length of wire,l=15 m
Current,I=19 A
a.We have to find the magnitude of magnetic force and direction of magnetic force.
Magnetic force,F=
Using the formula
Direction=
15 degree above the horizontal in the northward direction.
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
Answer:
Explanation:
Given that,
Spring constant k=200N/m
Compression x = 15cm = 0.15m
Attached mass m =2kg
Coefficient of kinetic friction uk= 0.2
The energy in the spring is given as
U =½kx²
U = ½ × 200 × 0.15²
U = 2.25J
Force in the spring is given by Hooke's law
F = ke
F = 200×0.15
F = 30N
The weight of body which is equal to the normal is give as
W = mg
W = 2 × 9.81
W = 19.62N
W = N = 19.62 Newton's 2nd Law
From law of friction,
Fr = uk•N
Fr = 0.2 × 19.62
Fr = 3.924
Using newton second law again
Fnet = F - Fr
Fnet = 30 - 3.924
Fnet = 26.076
Work done by net force is given as
W = Fnet × d
W = 26.076d
Then, the work done by this net force is equal to the energy in the spring
W = U
26.076d = 2.25
d = 2.25/26.076
d = 0.0863m
Which is 8.63cm
So the box will slide 8.63cm before stopping
