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lesantik [10]
3 years ago
12

The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, dete

rmine the minimum height h of the hill crest so that the car travels around both inside the loops without leaving the track. Neglect friction, the mass of the wheels, and the size of the car. What is the normal reaction on the car when the car is at B and when it is at C? Take ???????? = 7.5 m and ???????? = 5 m.
Physics
1 answer:
sweet [91]3 years ago
7 0

Answer:

h = 18.75 m

Now when it will reach at point B then its normal force is just equal to ZERO

N_B = 0

F_n = 1.72 \times 10^4

Explanation:

Since we need to cross both the loops so least speed at the bottom must be

v = \sqrt{5 R g}

also by energy conservation this is gained by initial potential energy

mgh = \frac{1}{2}mv^2

v = \sqrt{2gh}

so we will have

\sqrt{2gh} = \sqrt{5Rg}

now we have

h = \frac{5R}{2}

here we have

R = 7.5 m

so we have

h = \frac{5(7.5)}{2}

h = 18.75 m

Now when it will reach at point B then its normal force is just equal to ZERO

N_B = 0

now when it reach point C then the speed will be

mgh - mg(2R_c) = \frac{1}{2]mv_c^2

v_c^2 = 2g(h - 2R_c)

v_c = 13.1 m/s

now normal force at point C is given as

F_n = \frac{mv_c^2}{R_c} - mg

F_n = \frac{700\times 13.1^2}{5} - (700 \times 9.8)

F_n = 1.72 \times 10^4

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nataly862011 [7]

Answer:

Train accaleration = 0.70 m/s^2

Explanation:

We have a pendulum (presumably simple in nature) in an accelerating train. As the train accelerates, the pendulum is going move in the opposite direction due to inertia. The force which causes this movement has the same accaleration as that of the train. This is the basis for the problem.

Start by setting up a free body diagram of all the forces in play: The gravitational force on the pendulum (mg), the force caused by the pendulum's inertial resistance to the train(F_i), and the resulting force of tension caused by the other two forces (F_r).

Next, set up your sum of forces equations/relationships. Note that the sum of vertical forces (y-direction) balance out and equal 0. While the horizontal forces add up to the total mass of the pendulum times it's accaleration; which, again, equals the train's accaleration.

After doing this, I would isolate the resulting force in the sum of vertical forces, substitute it into the horizontal force equation, and solve for the acceleration. The problem should reduce to show that the acceleration is proportional to the gravity times the tangent of the angle it makes.

I've attached my work, comment with any questions.

Side note: If you take this end result and solve for the angle, you'll see that no matter how fast the train accelerates, the pendulum will never reach a full 90°!

8 0
3 years ago
A baseball pitcher throws the ball towards the batter at 90 mph. His bat connects with the ball for a line drive, after which th
forsale [732]

Answer:

F=-18412.9N, where the minus indicates the direction is opposite to that of the throw.

Explanation:

a)

Since MKS stands for meter-kilogram-second and we know that:

1\ hour = 3600\ seconds

1\ mile = 1600\ meters

1000g = 1kg

We can write that:

\frac{1\ hour}{3600\ seconds}=1

\frac{1600\ meters}{1\ mile}=1

\frac{1kg}{1000g}=1

These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.

So we have that:

90 mph=90 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=40m/s

110 mph=110 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=48.89m/s

145 g=145 g(\frac{1kg}{1000g})=0.145kg

b)

Newton's 2nd Law tells us that F=ma, and the definition of acceleration is a=\frac{\Delta v}{\Delta t}, so we have:

F=m\frac{\Delta v}{\Delta t}=m\frac{v_f-v_i}{t}

Taking the throw direction as the positive one, for our values we have:

F=m\frac{v_f-v_i}{t}=(0.145kg)\frac{(-48.89m/s)-(+40m/s)}{0.0007s}=-18412.9N

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Applied force: It is the external force that acts directly on a body.

Therefore, we can say that if you have an object and push it towards yourself, you are exerting an external force on the object.

This external force was not acting on the object previously, therefore, it is a force that you are applying at that moment.

Answer:

you exert an Applied Force on an object when you pull it towards you

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Answer:

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