consider east-west direction along X-axis and north-south direction along Y-axis
= velocity of migrating robin relative to air = 12 j m/s
(where "j" is unit vector in Y-direction)
= velocity of air relative to ground = 6.3 i m/s
(where "i" is unit vector in X-direction)
= velocity of migrating robin relative to ground = ?
using the equation
=
+
= 12 j + 6.3 i
= 6.3 i + 12 j
magnitude : sqrt((6.3)² + (12)²) = 13.6 m/s
direction : tan⁻¹(12/6.3) = 62.3 deg north of east
Answer:
The torque about his shoulder is 34.3Nm.
The solution approach assumes that the weight of the boy's arm acts at the center of the boy's arm length 35cm from the shoulder.
Explanation:
The solution to the problem can be found in the attachment below.
Answer:
The duration of the movie is longer than 2 hrs.
Explanation:
Given:
The duration of the movie observed by the crew on the spacecraft is 2 hrs.
According to time-dilation formula:
Here, is the required time,
is the original time,
is the velocity of the spacecraft and
is the velocity of light.
Since , so
.
So the time required will be large.
Explanation:
The given data is as follows.
F = 3.2 N, m = 18.2 kg,
t = 0.82 sec
(a) Formula for impulse is as follows.
I = Ft =
Ft =
or,
Putting the given values into the above formula as follows.
=
= 0.144 m/s
Therefore, final velocity of the mass if it is initially at rest is 0.144 m/s.
(b) When velocity is 1.85 m/s to the left then, final velocity of the mass will be calculated as follows.
Ft =
or,
=
= -1.705 m/s
Hence, we can conclude that the final velocity of the mass if it is initially moving along the x-axis with a velocity of 1.85 m/s to the left is 1.705 m/s towards the left.