The final velocity is 5.87 m/s
<u>Explanation:</u>
Given-
mass,
= 72 kg
speed,
= 5.8 m/s
,
= 45 kg
,
= 12 m/s
Θ = 60°
Final velocity, v = ?
Applying the conservation of momentum:
X
+
X
= (
+
) v
72 X 5.8 + 45 X 12 X cos 60° = (72 + 45) v
v = 417.6 + 540 X 
v = 417.6 + 
v = 5.87 m/s
The final velocity is 5.87 m/s
Answer:
2 kg
Explanation:
Acceleration = 5 m/s^2
Force = 10 N
Force = mass * acceleration
mass = force / acceleration
mass = 10 / 5
mass = 2 kg
Answer:
Time taken, 
Explanation:
It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.
From the figure,
The sum of forces in y direction is :


Sum of forces in x direction,

.............(1)
Also, 
Equation (1) becomes :

...............(2)
Let t is the time taken for the ball to rotate once around the axis. It is given by :

Put the value of T from equation (2) to the above expression:


On solving above equation :

Hence, this is the required solution.
Answer:
diverging light rays of the bulb are collected by the reflector.
Explanation:
<h2>
<em><u>A</u></em><em><u>N</u></em><em><u>S</u></em><em><u>W</u></em><em><u>E</u></em><em><u>R</u></em><em><u>S</u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em></h2>
<em>1) The relationship in between the electrical energy carriesd by the transmission wires and the amount of the heat loss in it is due to the reason that when the electricity is flown through the wires there are some resistance found in these wires which creates a disturbance in the efficient flow of electricty.Also we know that current have an heating effect when it is in motion as due to if a large amount or magnitude of electricity is flown through the transmission wires it will carry a larger heat effected and also due to the resistance is provided by the wires and so the process of heat loss takes place.</em>
<em>2)It is important to minimize current in transmission wires due to minimize the heat loss and resistance on flowing electric current to make the system more efficient </em>
<em><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u></em><em> 3)Given Resistance = 250 ohms </em>
<em>Electric potential = 150 volts </em>
<em>so we know Power = </em>
<em>volt^2/Resistance = </em>
<em>=</em><em>(150^2/250)(ohms/volts)</em>
<em>=</em><em>(22500/250)watt = 9</em><em>0</em><em> </em><em>w</em><em>a</em><em>t</em><em>t</em><em> </em>
<em>4)Heat energy (H) = Power(P)×Time(t)</em>
<em>4)Heat energy (H) = Power(P)×Time(t)= (90×2)joules = 180 joul</em><em>e</em><em>s</em>
<em>H</em><em>o</em><em>p</em><em>e</em><em> </em><em>i</em><em>t</em><em> </em><em>h</em><em>e</em><em>l</em><em>p</em><em>s</em>