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ICE Princess25 [194]
2 years ago
11

A 120-kg roller coaster cart is being tested on a new track, and a crash-test dummy is loaded into itThe roller coaster starts f

rom rest at point A which is 35 m above the ground. The cart is observed to roll down the track to point Bwhich is 10 m above the ground . In your own words, what physics ideas would you be able to use to predict how fast the cart is moving at point B? What calculations would you use, what things would you be measuring , and what physics principles or laws would you use to make this calculation ?
Physics
1 answer:
Black_prince [1.1K]2 years ago
7 0

Answer:

a) variation of the energy is equal to the work of the friction force

b) W = Em_{f} -Em₀ ,  c) he conservation of mechanical energy

Explanation:

a) In an analysis of this problem we can use the energy law, where at the moment the mechanical energy is started it is totally potential, and at the lowest point it is totally kinetic, we can suppose two possibilities, that the friction is zero and therefore by equalizing the energy we set the velocity at the lowest point.

 Another case is if the friction is different from zero and in this case the variation of the energy is equal to the work of the friction force, in value it will be lower than in the calculations.

b) the calluses that he would use are to hinder the worker's friction force and energy

          W = Em_{f} -Em₀

          N d = ½ m v² - m g (y₂-y₁)

          y₂-y₁ = 35 -10 = 25m

c) if there is no friction, the physical principle is the conservation of mechanical energy

 If there is friction, the principle is that the non-conservative work is equal to the variation of the energy

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a car travels from station A to B at 30kmph during its return trip A it travels 30 km pH for the first half distance and it 70 k
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Explanation:

If you cannot visualize it, just assume that the distance from station A to B is 420km. Each half is 210km.

When the car travels from A to B, it takes 420/30 = 14 hours.

When the car travels from B to the halfway point, it takes 210/30 = 7 hours.

When the car travels from the halfway point to A, it takes 210/70 = 3 hours.

Total time taken = 14 + 7 + 3 = 24 hours.

Total distance = 420km * 2 = 840km.

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3 0
3 years ago
a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B su
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Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

U=(M+m)gh       (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m

The potential energy is:

U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

Mv_1+mv_2=(M+m)v    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

M(0)+mv_2=(M+m)v    

v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N    

The force on the cord after the impact is 2.59N

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Answer

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RT - 30Ω

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