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3 years ago

Explain why the roller coaster’s potential energy is greater at point 1 than at point 4.

2 answers:

5 0

Point 1 is PROBABLY higher up than point 4.

Another possibility is that on the way from Point 1 to point 4,
a lot of people are falling out of the coaster.

We have no way to tell, because you neglected to show us
the picture that goes with this question. It's important.

ValentinkaMS [17]3 years ago

5 0

Answer:

Potential energy is directly related to the height of an object. Point 1 is higher than point 4, so the potential energy is greater at point 1.

Explanation:

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Which system of equations and solution can be used to represent the radius if the mass of the cylinder is 11,000 grams

drek231 [11]

Explanation:

The mass of a cylinder made of barium with a height of 2 inches depends on the radius of the cylinder as defined by the

function m(r) = 7.18872.

which system of equations and solution can be used to represent the radius if the mass of the cylinder is 11,000 grams?

round to the nearest hundredth of an inch.

7 0

2 years ago

During a rockslide, a 670 kg rock slides from rest down a hillside that is 740 m along the slope and 240 m high. The coefficient

ElenaW [278]

a) $1.58\cdot 10^6 J$

b) $1.15\cdot 10^6 J$

c) $0.43\cdot 10^6 J$

d) 35.8 m/s

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its location with respect to the ground.

It is given by:

$U=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object, relative to a reference level

Here, the reference level is taken at the bottom of the hill (where the potential energy is zero).

So, we have:

m = 670 kg is the mass of the rock

$g=9.8 m/s^2$

h = 240 m is the initial height of the rock

So, the potential energy of the rock just before the slide is

$U=(670)(9.8)(240)=1.58\cdot 10^6 J$

The energy transferred to thermal energy during the slide is equal to the work done by friction, which is:

$W=F_f d$

where

$F_f$ is the force of friction

d = 740 m is the displacement of the rock along the ramp

The force of friction is given by:

$F_f=-\mu mg cos \theta$

where

$\mu=0.25$ is the coefficient of friction

m = 670 kg is the mass of the rock

$\theta$ is the angle of the ramp

Since we know the lenght of the ramp (d = 740 m) and the height (h = 240 m), we can find the angle:

$\theta=sin^{-1}(\frac{h}{d})=sin^{-1}(\frac{240}{740})=18.9^{\circ}$

Therefore, the work done by friction is:

$W=-\mu m g cos \theta d =-(0.25)(670)(9.8)(cos 18.9^{\circ})(740)=-1.15\cdot 10^6 J$

So, the energy transferred to thermal energy is $1.15\cdot 10^6 J$ .

According to the law of conservation of energy, the kinetic energy of the rock as it reaches the bottom of the hill will be equal to the initial potential energy (at the top) minus the energy transformed into thermal energy.

Therefore, we have:

$K_f = U_i -E_t$

where here we have:

$U_i=1.58\cdot 10^6 J$ is the potential energy of the rock at the top of the hill

$E_t=1.15\cdot 10^6 J$ is the energy converted into thermal energy

Substituting, we find

$K_f=1.58\cdot 10^6-1.15\cdot 10^6=0.43\cdot 10^6 J$

So, this is the kinetic energy of the rock at the bottom of the hill.

The kinetic energy of the rock at the bottom of the hill can be rewritten as

$K_f=\frac{1}{2}mv^2$

where

m is the mass of the rock

v is its final speed

In this problem, we have:

$K_f=0.43\cdot 10^6 J$ is the final kinetic energy of the hill

m = 670 kg is the mass of the rock

Therefore, the final speed of the rock is:

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.43\cdot 10^6)}{670}}=35.8 m/s$

7 0

3 years ago

The car's motion can be divided into three different stages: its motion before the driver realizes he's late, its motion after t

jolli1 [7]

Answer:

B, C, D are valid assumptions

Explanation:

In each stage the acceleration of the car varies because the velocity of car varies in each stage. At first, the acceleration of the car is moving slowly.

Then, driver realizes that he is late so the acceleration of the car increases he hits the gas.

Finally, the acceleration decreases when he sees the police car.

In the first state when driver do not know, that he is late, he will drive with a constant velocity as any one does.

When the driver hits the gas and he does not know about the police vehicle, even then he may drive with constant velocity or he may accelerate the car due to fear of being caught.

In the last part of motion when driver see the police car, even than he may accelerate the car, but acceleration will be constant throughout the motion is not possible, or even than he may continue with constant velocity.

Hence, there are only three assumptions are valid.

B. During each of the three different stages of its motion, the car is moving with constant velocity.

C . The highway is straight.

D. The highway is level.

Note that the rate of change of speed is equal to the acceleration. The acceleration is constant if change in velocity of particle is equal in equal interval of time.

8 0

3 years ago

Which of the following best represents an isolated system?

saul85 [17]

Answer:

D.Outer space is answer

7 0

3 years ago

A ball on an extremely low-friction, tilted surface will slide downhill without rotating. If the surface is rough, however, the

spayn [35]

It will go left or right

6 0

3 years ago