A quarterback throws a football with an initial velocity v at an angle θ above horizontal. Assume the ball leaves the quarterbac
1 answer:
You might be interested in
Answer: (A) velocity = 2.8m/s
(B) Average force = 1.9536 Newtons
Explanation:
Find detailed explanation in the attached picture
Answer:
a) Fₓ = 23.5 N
b) Net force = Fₓ
Explanation:
An image of the question as described is attached to this solution.
From the image attached, the forces acting on the box include the weight of the box, the normal reaction of the surface on the box, the applied force on the box and the Frictional force opposing the motion of the box (which is negligible and equal to 0)
a) From the diagram, the horizontal component of the force is
Fₓ = 25 cos 20° = 23.49 N = 25 N
b) Again, from the diagram attached, doing a force balance on the box, in the horizontal direction, we obtain
Net force = Fₓ - Frictional force
But frictional force is 0 N
Net force = Fₓ
Hope this Helps!!!
Answer:
The magnitudes of the net magnetic fields at points A and B is 2.66 x T
Explanation:
Given information :
The current of each wires, I = 4.7 A
dH = 0.19 m
dV = 0.41 m
The magnetic of straight-current wire :
B= μI/2πr
where
B = magnetic field (T)
μ = 1.26 x
(N/
)
I = Current (A)
r = radius (m)
the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,
BH = μI/2πr
= (1.26 x )(4.7)/(2π)(0.19)
= 4.96 x T
BV = μI/2πr
= (1.26 x )(4.7)/(2π)(0.41)
= 2.3 x T
hence,
the net magnetic field = BH - BV
= 4.96 x - 2.3 x
= 2.66 x T