Answer:
find the diagram in the attachment.
Explanation:
Let vi = 12 m/s be the intial velocy when the ball is thrown, Δy be the displacement of the ball to a point where it starts returning down, g = 9.8 m/s^2 be the balls acceleration due to gravity.
considering the motion when the ball thrown straight up, we know that the ball will come to a stop and return downwards, so:
(vf)^2 = (vi)^2 + 2×g×Δy
vf = 0 m/s, at the highest point in the upward motion, then:
0 = (vi)^2 + 2×g×Δy
-(vi)^2 = 2×g×Δy
Δy = [-(vi)^2]/2×g
Δy = [-(-12)^2]/(2×9.8)
Δy = - 7.35 m
then from the highest point in the straight up motion, the ball will go back down and attain the speed of 12 m/s at the same level as it was first thrown
Frictional forces are exerting force against the basketball, causing it to stop.
Answer:
7 orbitals are allowed in a sub shell if the angular momentum quantum number for electrons in that sub shell is 3.
Explanation:
We that different values of m for a given l provide the total number of ways in which a given s, p,d and f sub shells in presence of magnetic field can be arranged in space along x, y ,z- axis or total number of orbitals into which a given subshell can be divided.
Range for given l lies between -l to +l .
The possible values of m are -3 , -2 , -1 , 0 , 1 ,2 , 3 .
Total number of orbitals are 7.
Answer:
(a) 0.204 Weber
(b) 0.22 Volt
Explanation:
N = 100, radius, r = 10 cm = 0.1 m, B = 0.0650 T, angle is 90 degree with the plane of coil, so theta = 0 degree with the normal of coil.
(a) Magnetic flux, Ф = N x B x A
Ф = 100 x 0.0650 x 3.14 x 0.1 0.1
Ф = 0.204 Weber
(b) B1 = 0.0650 T, B2 = 0.1 T, dt = 0.5 s
dB / dt = (B2 - B1) / dt = (0.1 - 0.0650) / 0.5 = 0.07 T / s
induced emf, e = N dФ/dt
e = N x A x dB/dt
e = 100 x 3.14 x 0.1 x 0.1 x 0.07 = 0.22 V
Answer:
the relative speed should be 40mph
Explanation: