Explanation:
Below is an attachment containing the solution.
Answer:
a) 16.32 m/s
b) 640 N
Explanation:
A) mass of rocket m_r = 1000 g = 1 kg
initial speed of rocket u_r = 15 m/sec
initial speed of ball is u_b = 18 m/sec
final speed of ball is v_b = 40 m/sec
Let m_b be the mass of the ball= 60 g and v_r be the final velocity of the rocket
from law of conservation of momentum
momentum of the system remains zero
m_r×(u_r-v_r)+m_b(16-42) = 0
1×(15-v_r) = -0.060(18-40)
15-v_r = -1.32
v_r = 15+1.32 = 16.32 m/sec.
B) Average force that the rocket exert's on the ball is F_avg can be calculated as
contact time t=7.00 ms
F_avg = m(v-u)/t = 0.06×(40+18)/0.007 = 640 N
Answer:
The actual elevation angle is 12.87 degrees
Explanation:
In the attachment you can clearly see the situation. The angle of elevation as seen for the scuba diver is shown in magenta, we conclude that .
Using Snell's Law we can write:
,
Let's approximate the index of refraction of the air (medium 1 in the picture) to 1.
We thus have:
. Calling the actual angle of elevation, we get from the picture that
Answer:
D. An area of low pressure might move more quickly on Tuesday and Wednesday than expected.
Explanation:
I took the test and got it wrong :/