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4 years ago

8

If a roller coaster car had 40,000 J of gravitational potential energy when at rest on the top of a hill how much kinetic energy

does it have when it is 3/4 of the way down the hill

1 answer:

Inessa [10]4 years ago

6 0

Answer:

K.E = 30,000 J

Explanation:

Given,

The potential energy of the roller coaster car, P.E = 40000 J

The kinetic energy at height h/4, K.E = ?

According to the law of conservation of energy, the total energy of the system is conserved.

At height 'h', the total energy is,

P.E = mgh

K.E = 0

At height 'h/4', the total energy is

P.E + K.E = mgh

P.E = mgh/4

K.E = 1/2 mv²

Therefore,

mgh/4 + 1/2 mv² = mgh

gh/4 + v²/2 = gh

Hence,

v² = 3gh/2

Substituting in the K.E equation

K.E = 1/2 mv²

= 1/2 m (3gh/2)

= 3/4 mgh

= 3/4 x 40000

= 30000 J

Hence, the K.E of the roller coaster car is, K.E = 30000 J

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A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a

Elza [17]

Answer:

$\dfrac{dz}{dt}=0.65\ ft/s$

Explanation:

Given that

x= 150 ft

$\dfrac{dy}{dt}= 7\ ft/s$

y= 14 ft

From the diagram

$z^2=x^2+y^2$

When ,x= 150 ft and y= 14 ft

$z^2=150^2+14^2$

$z=\sqrt{150^2+15^2}$

z=150.74 ft

$z^2=x^2+y^2$

By differentiating with respect to time t

$2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$

$z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$

Here x is constant that is why

$\dfrac{dx}{dt}=0$

$z\dfrac{dz}{dt}= y\dfrac{dy}{dt}$

Now by putting the values in the above equation we get

$150.74\times \dfrac{dz}{dt}=14\times 7$

$\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s$

$\dfrac{dz}{dt}=0.65\ ft/s$

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

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3 years ago

Use Hooke's Law, which states that the distance a spring stretches (or compresses) from its natural, or equilibrium, length vari

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Answer:

706.68 N

Explanation:

By Hooke's law,

$F = ke$

$k=\dfrac{F}{e}$

Using the values in the question,

$k=\dfrac{265\text{ N}}{0.15 \text{ m}}=1766.7\text{ N/m}$

When e = 0.4 m,

$F = 1766.7\text{ N/m}\times0.4\text{ m}=706.68\text{ N}$

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3 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

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3 years ago

Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL o

garri49 [273]

Answer:

$P_{sol}=50.4\ mm.Hg$

Explanation:

According to given:

molecular mass of glycerin, $M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}$

molecular mass of water, $M_w=2+16=18\ g.mol^{-1}$

∵Density of water is $0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}$

∴mass of water in 316 mL, $m_w=316\times 0.992=313.5 g$

mass of glycerin, $m_g=154\ g$

pressure of mixture, $P_x=55.32\ torr= 55.32\ mm.Hg$

temperature of mixture, $T_x=40^{\circ}C$

<em>Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.</em>

<u>moles of water in the given quantity:</u>

$n_w=\frac{m_w}{M_w}$

$n_w=\frac{313.5}{18}$

$n_w=17.42 moles$

<u>moles of glycerin in the given quantity:</u>

$n_g=\frac{m_g}{M_g}$

$n_g=\frac{154}{92}$

$n_g=1.674 moles$

<u>Now the mole fraction of water:</u>

$X_w=\frac{n_w}{n_w+n_g}$

$X_w=\frac{17.42}{17.42+1.674}$

$X_w=0.912$

<em>Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.</em>

$\therefore P_{sol}=X_w\times P_x$

$\therefore P_{sol}=0.912\times 55.32$

$P_{sol}=50.4\ mm.Hg$

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3 years ago

Define acceleration what it's formula

Paha777 [63]

Answer:

The definition of acceleration is a change in the rate of motion, speed or action.

<h2>hope it helps.</h2><h2>stay safe healthy and happy..</h2>

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