<u>Answer:</u>
The velocity is 30.279 m/s
<u>Explanation</u>:
Consider the initial speed of the semi-trailer be v
Then, initial kinetic energy = 
According to question, the semi-trailer coast along a ramp, which is inclined at an angle of 170, and to a distance of 160m to stop
Change in vertical position =
= 46.779m
Final potential energy of semitrailer = mgh
Applying principle of conservation of energy,
= mgh
Solving for v, we get 
= 2gh = 2*9.8*46.779 = 916.8684
= 916.8684
v = 30.279 m/s
Therefore, the velocity is 30.279 m/s
(1) The linear acceleration of the yoyo is 3.21 m/s².
(2) The angular acceleration of the yoyo is 80.25 rad/s²
(3) The weight of the yoyo is 1.47 N
(4) The tension in the rope is 1.47 N.
(5) The angular speed of the yoyo is 71.385 rad/s.
<h3> Linear acceleration of the yoyo</h3>
The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.
∑τ = Iα
rT - Rf = Iα
where;
- I is moment of inertia
- α is angular acceleration
- T is tension in the rope
- r is inner radius
- R is outer radius
- f is frictional force
rT - Rf = Iα ----- (1)
T - f = Ma -------- (2)
a = Rα
where;
- a is the linear acceleration of the yoyo
Torque equation for frictional force;
solve (1) and (2)
since the yoyo is pulled in vertical direction, T = mg 
<h3>Angular acceleration of the yoyo</h3>
α = a/R
α = 3.21/0.04
α = 80.25 rad/s²
<h3>Weight of the yoyo</h3>
W = mg
W = 0.15 x 9.8 = 1.47 N
<h3>Tension in the rope </h3>
T = mg = 1.47 N
<h3>Angular speed of the yoyo </h3>
v² = u² + 2as
v² = 0 + 2(3.21)(1.27)
v² = 8.1534
v = √8.1534
v = 2.855 m/s
ω = v/R
ω = 2.855/0.04
ω = 71.385 rad/s
Learn more about angular speed here: brainly.com/question/6860269
#SPJ1
Explanation:
potential energy =360800J
mass(m)=?
height (h)=25m
g=9.8m/s²
we have
potential energy =360800J
mgh=360800J
m×9.8×25=360800
m=360800/(9.8×25)=1472.653061kg
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Question 7:
</h2>
The graph of
• The I-V for Ohmic Metal wire conductor at constant temperature always shows a straight line between the Current(I) plotted at Y axis and Voltage(V) plotted at X axis. Picture 1
• The I-V graph for Diode shows that first the current is zero but as we increase the potential difference(voltage), it results in the increase in the current. Picture 2
<h2>_____________________________________
</h2><h2>Question 8:
</h2>
A diode is a device that allows current to flow in only one direction.
Forward Bias, When a diode is forward bias (a voltage in the "forward" direction) then the P-side of the diode is attached to the positive terminal and N-side is fixed to the negative side of the battery which is connected, current flows freely through the device. The forward bias decreases the thickness of potential barrier(The potential barrier barrier in which the charge requires additional force for crossing the region)
Reverse Bias, When a diode is Reverse bias(a voltage in the "backward direction) then the P-side of the diode is connected to the negative terminal and N-side is connected to the positive terminal of the battery which is connected. The reverse bias increases the thickness of the potential barrier resulting in the flow of no current.
The Forward bias decreases the resistance of the diode whereas the reversed bias increases the resistance of the diode. As in forward biasing the current is easily flowing through the circuit whereas reverse bias does not allow the current to flow through it.
<h2>_____________________________________
</h2><h2>Best Regards,
</h2><h2>'Borz'
</h2>
C..............................
