Answer:
answer is DE I hope it will help you please follow me
Answer:
The radius of the curve that Car 2 travels on is 380 meters.
Explanation:
Speed of car 1, 
Radius of the circular arc, 
Car 2 has twice the speed of Car 1, 
We need to find the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration. We know that the centripetal acceleration is given by :

According to given condition,


On solving we get :

So, the radius of the curve that Car 2 travels on is 380 meters. Hence, this is the required solution.
Answer:
Chemical potential energy is the energy stored in the chemical bonds of a substance. The various chemicals that make up gasoline contain a large amount of chemical potential energy that is released when the gasoline is burned in a controlled way in the engine of the car.
Explanation:
The value of normal force as the slider passes point B is
The value of h when the normal force is zero
<h3>How to solve for the normal force</h3>
The normal force is calculated using the work energy principle which is applied as below
K₁ + U₁ = K₂
k represents kinetic energy
U represents potential energy
the subscripts 1,2 , and 3 = a, b, and c
for 1 to 2
K₁ + W₁ = K₂
0 + mg(h + R) = 0.5mv²₂
g(h + R) = 0.5v²₂
v²₂ = 2g(1.5R + R)
v²₂ = 2g(2.5R)
v²₂ = 5gR
Using summation of forces at B
Normal force, N = ma + mg
N = m(a + g)
N = m(v²₂/R + g)
N = m(5gR/R + g)
N = 6mg
for 1 to 3
K₁ + W₁ = K₃ + W₃
0 + mgh = 0.5mv²₃ + mgR
gh = 0.5v²₃ + gR
0.5v²₃ = gh - gR
v²₃ = 2g(h - R)
at C
for normal force to be zero
ma = mg
v²₃/R = g
v²₃ = gR
and v²₃ = 2g(h - R)
gR = 2gh - 2gR
gR + 2gR = 2gh
3gR = 2gh
3R/2 = h
Learn more about normal force at:
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Answer:
B. 7.07 m/s
Explanation:
The velocity of the stone when it leaves the circular path is its tangential velocity,
, which is given by

where
is the angular speed and
is the radius of the circular path.
is given by

where
is the frequency of revolution.
Thus

Using values from the question,

<em>Note the conversion of 75 cm to 0.75 m</em>
