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3 years ago

7

A 720 kg roller-coaster starts off from Location A. Assuming friction does not impede the car's motion, what will be the change

in its potential energy by moving from Location A to Location B

2 answers:

Alekssandra [29.7K]3 years ago

8 0

The correct answer is: 6.0 x 10^5 J

vaieri [72.5K]3 years ago

7 0

We know, Potential Energy = m * g * h
Here, mass & gravity would be same, but their height will change so it will be:

ΔU = U₂ - U₁
ΔU = mgh₂ - mgh₁
ΔU = mg (h₂ - h₁)

Hope this helps!

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A ball is thrown with a speed of 20 m/s at an angle of 40o above the horizontal from the top of a 22-m tall building.

MrRissso [65]

Answer:

(a). The distance is 49.79 m.

(b). The speed of the ball is 24.39 m/s.

Explanation:

Given that,

Speed = 20 m/s

Angle = 40°

Height = 22 m

Time = 3.25 sec

(a). We need to calculate the distance

Using formula of distance

$d=u\cos\theta\times t$

Put the value into the formula

$d=20\cos40\times3.25$

$d=49.79\ m$

(b). We need to calculate the horizontal velocity

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=20\times\cos40$

$v_{x}=15.3\ m/s$

We need to calculate the vertical velocity

Using equation of motion

$v_{y}=u\sin\theta-gt$

Put the value into the formula

$v_{y}=20\sin40-9.8\times3.25$

$v_{y}=-19\ m/s$

Negative sign shows the opposite direction.

We need to calculate the speed of ball

Using formula of speed

$v=\sqrt{v_{x}^2+v_{y}^2}$

$v=\sqrt{(15.3)^2+(19)^2}$

$v=24.39\ m/s$

Hence, (a). The distance is 49.79 m.

(b). The speed of the ball is 24.39 m/s.

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3 years ago

Explain the correspondence that lets us easily translate between linear motion and rotational motion. What are the linear analog

RideAnS [48]

Explanation:

The linear analog of angle is angle itself.

The linear analog of angular velocity is linear velocity.

ω is angular velocity, therefore linear velocity is given by v

∴ for linear velocity, $v^{2} = u^{2}+2.a.S$

for angular velocity, $\omega_{f}^{2} = \omega _{i}^{2}+2.a.S$

The linear analog of angular acceleration is acceleration.

α is angular acceleration whereas as a is linear acceleration.

∴ for linear acceleration, v = u + a.t

for angular acceleration, $\omega_{f}= \omega _{i}+\alpha .t$

The linear analog of moment of inertia is mass.

I is moment of inertia and m is mass,

∴ for linear analog, F = m.a

for angular analog, τ - I.α

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3 years ago

A 80-kg person sits on a 2.7kg chair. Each leg of the chair makes contact with the floor on a circle that is 1.3 cm in diameter.

Rashid [163]

Answer:

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Explanation:

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6 0

3 years ago

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At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravit

Over [174]

Answer:

v=32.9m/s

Explanation:

The acceleration needed to mantain a circular motion of radius r and speed v is given by the equation $a=v^2/r$

This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).

We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:

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3 years ago

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Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

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We need to calculate the acceleration of sled

Using formula of acceleration

$a = \dfrac{F}{m}$

Where, F = frictional force

m = mass

Put the value into the formula

$a=\dfrac{\mu mg}{m}$

$a=\mu g$

$a=0.229\times9.8$

$a=2.244\ m/s^2$

We need to calculate the speed of the sled

Using equation of motion

$v^2=u^2-2as$

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

$v ^2=(5.49)^2-2\times2.244\times3.89$

$v=\sqrt{(5.49)^2-2\times2.244\times3.89}$

$v=3.56\ m/s$

Hence, The speed of the sled is 3.56 m/s.

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3 years ago