Answer:
Tension in the string will increase
Explanation:
As we know that tension in the string at any angle with the vertical is given as

now we have

also we know that
angular speed of the stone is directly depending on the time period of the motion
so it is given as

since the frequency of the revolution is increased from n = 1 rev/s to 2 rev/s
so the angular speed would be doubled
So here we can say that
tension in the string will increase when we will increase the frequency of revolution.
Answer: 6250 joules
Explanation:
The work needed to lift an object of mass M by a height H is equal to:
w = M*g*H
where h = 10m/s^2
then the total work that he did is equal to the sum of the work for every stone:
W = (100kg*g*H) + (120kg*g*H) + (140kg*g*H) + (160kg*g*H) + (180kg*g*H)
= (100kg + 120kg + 140kg + 160kg + 180kg)*g*H
= (500kg)*g*H
and now we can repalce g by 10m/s^2 and H by 125cm
But you can notice that we have two different units of distance, so knowing that 100cm = 1m
we can write H = 125cm = (125/100) m = 1.25 m
Then we have:
H = 500kg*10m/s^2*1.25m = 6250 J
Answer:
The answer your looking for is option 2 - Inertia
Principles<span> of </span>arc welding<span>. </span>Arc welding<span> is a </span>welding<span> process, in which heat is generated by an </span>electric arc<span> struck between an electrode and the work piece. </span>Electric arc<span> is luminous</span>electrical<span> discharge between two electrodes through ionized gas.</span>
Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s