Answer:
28.5 m/s
18.22 m/s
Explanation:
h = 20 m, R = 20 m, theta = 53 degree
Let the speed of throwing is u and the speed with which it strikes the ground is v.
Horizontal distance, R = horizontal velocity x time
Let t be the time taken
20 = u Cos 53 x t
u t = 20/0.6 = 33.33 ..... (1)
Now use second equation of motion in vertical direction
h = u Sin 53 t - 1/2 g t^2
20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)
t = 1.17 s
Put in equation (1)
u = 33.33 / 1.17 = 28.5 m/s
Let v be the velocity just before striking the ground
vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s
vy = uSin 53 - 9.8 x 1.17
vy = 28.5 x 0.8 - 16.66
vy = 6.14 m/s
v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2
v = 18.22 m/s
Answer:
that best describes the process is C
Explanation:
This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.
Heat absorbed by the smallest container
Q_c = m ce (
-T₀)
Heat released by the largest container is
Q_a = M ce (T_{i}-T_{f})
how
Q_c = Q_a
m (T_{f}-T₀) = M (T_{i} - T_{f})
Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.
Of the final statements, the one that best describes the process is C
since it talks about the thermal energy and the heat that is transferred in the process
Complete question is;
A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m/s² rightward. After 3s, what will be the velocity of the rocket ship?
Answer:
v = 12 m/s
Explanation:
We are given;
Initial velocity; u = 0 m/s (because ship starts from rest)
Acceleration; a = 4 m/s²
Time; t = 3 s
To find velocity after 3 s, we will use Newton's first equation of motion;
v = u + at
v = 0 + (4 × 3)
v = 12 m/s
