Answer:
m = 3 kg
The mass m is 3 kg
Explanation:
From the equations of motion;
s = 0.5(u+v)t
Making t thr subject of formula;
t = 2s/(u+v)
t = time taken
s = distance travelled during deceleration = 62.5 m
u = initial speed = 25 m/s
v = final velocity = 0
Substituting the given values;
t = (2×62.5)/(25+0)
t = 5
Since, t = 5 the acceleration during this period is;
acceleration a = ∆v/t = (v-u)/t
a = (25)/5
a = 5 m/s^2
Force F = mass × acceleration
F = ma
Making m the subject of formula;
m = F/a
net force F = 15.0N
Substituting the values
m = 15/5
m = 3 kg
The mass m is 3 kg
Answer:
This is known as a Galilean transformation where
V' = V - U
Where the primed frame is the Earth frame and the unprimed frame is the frame moving with respect to the moving frame
V - speed of object in the unprimed frame
U - speed of primed frame with respect to the unprimed frame
Here we have:
V = -15 m/s speed of ball in the moving frame (the truck)
U = -20 m/s speed of primed (rest) frame with respect to moving frame
So V' = -15 - (-20) = 5 m/s
It may help if you draw a vector representing the moving frame and then add
a vector representing the speed of the ball in the moving frame.
Answer:
500000000
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Answer:
θ = 13.16 °
Explanation:
Lets take mass of child = m
Initial velocity ,u= 1.1 m/s
Final velocity ,v=3.7 m/s
d= 22.5 m
The force due to gravity along the incline plane = m g sinθ
The friction force = (m g)/5
Now from work power energy
We know that
work done by all forces = change in kinetic energy
( m g sinθ - (m g)/5 ) d = 1/2 m v² - 1/2 m u²
(2 g sinθ - ( 2 g)/5 ) d = v² - u²
take g = 10 m/s²
(20 sinθ - ( 20)/5 ) 22.5 = 3.7² - 1.1²
20 sinθ - 4 =12.48/22.5
θ = 13.16 °
Answer:
(a) I_A=1/12ML²
(b) I_B=1/3ML²
Explanation:
We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².
(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².
(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

Finally, using the Parallel Axis Theorem, we calculate I_B:
