Answer:
Explanation:
given that
mass = 10kg
distance = 4m
force = 50N
to calculate the workdone when the force is applied in the same direction of displacement
mathematically,
workdone = force × distance
Workdone = 50 × 4
workdone = 200 joules
2) to calculate the workdone at an angle of 30° with the displacement we apply the formula
workdone = force × distance × cos Ф
workdone = 50 × 4 × cos 30°
workdone = 200 × 0.866
workdone = 173 . 2 joules
Answer:
Explanation:
The given function is
Now h = the height from the surface of the Earth
Here the building is 458 m tall
So,
Answer:
The man moves across the ice with a speed of 0.345m/s.
Explanation:
From the conservation of linear momentum, we have that the total linear momentum before the book throw is equal to the total linear momentum just after it. Since the initial velocity of the system is zero (so the initial momentum is zero), we have that:
Where 
is the mass of the man, 
is the mass of the book, and 
and 
are their velocities. Plugging in the given values, we can compute the speed of the man (ignoring the negative sign, because we care about the magnitude, not the direction):
In words, the resulting speed of the man is 0.345m/s.
Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s
Answer:
The answer is X
Explanation:
Cause the highest points will most likely have the most potential energy
