Answer:
The first part can be solved via conservation of energy.
For the second part,
the free body diagram of the car should be as follows:
- weight in the downwards direction
- normal force of the track to the car in the downwards direction
The total force should be equal to the centripetal force by Newton's Second Law.
where 
because we are looking for the case where the car loses contact.
Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.
Explanation:
The point that might confuse you in this question is the direction of the normal force at the top of the loop.
We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.
The label means that the bulb will consume 75 joules of electrical energy in 1 sec
Answer:
≅3666.67 N
Explanation:
Use Newton's 2nd law, F = ma where F=force applied, m = mass of the object,
a = acceleration acquired by the object.
a= (v-u)/t where v = final velocity, u = initial velocity and t = time taken
calculate a = (30-0)/9 ≅ 3.33 m/s2
Then F = 1100×a = 3666.67 N
Answer:
Time, t = 0.57 hours
Explanation:
It is given that,
Speed of the school bus, v = 35 mi/hr
Distance covered by the bus, d = 20 miles
We need to find the time taken by the bus to get to school. Time taken by the bus is given by :
t = 0.57 hours
So, the time taken by the bus to reach school is 0.57 hours. Hence, this is the required solution.
Answer:
a) 210 Ω
b) 21 V
c) 0.07 A, 0.03 V
d) VAE = 7 V
VEB = 14 V
VAF = 9 V
VFB = 12 V
Explanation:
a) 100 Ω and 200 Ω are in series, equivalent = 100 + 200 = 300 Ω,
300 Ω and 400 Ω are in series, equivalent = 300 + 400 = 700 Ω
300 Ω and 700Ω are in parallel, so the equivalent resistance between A and B is R where
1/300 + 1/700 = 1/R, get R = 210 Ω
b) Given I = 0.1 A
so VAB = I * R = 1 * 210 = 21.0 V
c) I₁ + I₂ = I = 0.1, so I₂ = 0.1 - I₁
I₁ * 300 = I₂ * 700
I₁ * 300 = (0.1 - I₁) * 700
get I₁ = 0.07 A, I₂ = 0.1 - I₁ = 0.03 A
d) VAE = I₁ * 100 = 7 V
VEB = I₁ * 200 = 14 V
VAF = I₂ * 300 = 9 V
VFB = I₂ * 400 = 12 V
