Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.
By Newton's second law,
- the net horizontal force acting on the beam is

where
are the magnitudes of the tensions in ropes 1 and 2, respectively;
- the net vertical force acting on the beam is

where
and
.
Eliminating
, we have





Solve for
.



Answer:
47.4 m
Explanation:
When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.
In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

where
s is the vertical displacement
u = 0 is the initial velocity
t = 3.11 s is the time
is the acceleration of gravity (taking downward as positive direction)
Solving the formula, we find

Answer:
You would be watching tv for 1 hour and 40 mins
Explanation:
How much time take it take for 3:50 am to 5:30
Answer:
- 278.34 kg m/s^2
Explanation:
The rate of the change of momentum is the same as the force.
The force that an object feels when moviming in a circular motion is given by:
F = -mrω^2
Where ω is the angular speed and r is the radius of the circumference
Aditionally, the tangential velocity of the body is given as:
v = rω
The question tells us that
v = 25 m/s
r = 7m
mv = 78 kg m/s
Therefore:
m = (78 kg m/s) / (25 m/s) = 3.12 kg
ω = (25 m/s) / (7 m) = 3.57 (1/s)
Now, we can calculate the force or rate of change of momentum:
F = - (3.12 kg) (7 m)(3.57 (1/s))^2
F = - 278.34 kg m/s^2
Answer:
a)3.5s
b)28.57m/S
c)34.33m/S
d)44.66m/S
Explanation:
Hello!
we will solve this exercise numeral by numeral
a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

where
Vo = Initial speed
=0
T = time
g=gravity=9.81m/s^2
y = height=60m
solving for time

T=3.5s
b)The horizontal speed remains constant since there is no horizontal acceleration.
with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

c)
to find the final vertical velocity we use the equations for motion with constant velocity as follows
Vf=Vo+g.t
Vf=0+(9.81 )(3.5)=34.335m/S
d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares
