Answer:
(D) 42.4N
Explanation:
Since the frame is at rest, the net force acting on it must be 0. There are three forces acting on it: the gravity and the opposing forces of the two cables.
Since the gravity is a vertical force, we are only interested in the vertical components of the remaining forces. The net force equation is
F_net = 0 = F_g -2 * F_y
The vertical force of one cable (using the information in the drawing) is:
F_y = 30N * sin 45 deg = 21.21N
Now the weight can be determined:
0 = F_g - 2 * F_y
F_g= 2 * F_y = 2 * 21.21N = 42.4N
The weight of the frame is about 42.4N.
Answer:
<h2>760 m/s</h2>
Explanation:
The jet's speed or final velocity can be found by using the formula
<h3>v = u + at</h3>
where
v is the final velocity
u is the initial velocity
t is the time taken
a is the acceleration
From the question we have
v = 600 + 20(8)
= 600 + 160
We have the final answer as
<h3>760 m/s</h3>
Hope this helps you
Answer
given,
constant speed of cart on right side = 2 m/s
diameter of nozzle = 50 mm = 0.05 m
discharge flow through nozzle = 0.04 m³
One-fourth of the discharge flows down the incline
three-fourths flows up the incline
Power = ?
Normal force i.e. Fn acting on the cart

v is the velocity of jet
Q = A V


v = 20.37 m/s
u be the speed of cart assuming it to be u = 2 m/s
angle angle of inclination be 60°
now,

F n = 2295 N
now force along x direction



Power of the cart
P = F x v
P = 1987.52 x 20.37
P = 40485 watt
P = 40.5 kW
(a) The frequency of the motion after the collision is 0.71 Hz.
(b) The maximum angular displacement of the motion after the collision is 16.3⁰.
<h3>Speed of the 2.2 kg ball when it collides with 2.7 kg ball</h3>
The speed of the 2.2 kg ball which was initially 10 cm higher that 2.7 kg ball is calculated as follows;
K.E = P.E
¹/₂mv² = mgh
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 0.1)
v = 1.4 m/s
<h3>Final speed of both balls after collision</h3>
The final speed of both balls after the collision is determined from the principle of conservation of linear momentum.
Pi = Pf
m₁v₁ + m₂v₂ = vf(m₁ + m₂)
2.2(1.4) + 2.7(0) = vf(2.2 + 2.7)
3.08 = 4.9vf
vf = 3.08/4.9
vf = 0.63 m/s
<h3>Maximum displacement of the balls after the collision</h3>
P.E = K.E

<h3>Maximum angular displacement</h3>
The maximum angular displacement of the balls after the collision is calculated as follows;

<h3>Frequency of the motion</h3>

Learn more about maximum angular displacement here: brainly.com/question/13665036