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2 years ago

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Russell bradley carried 207 kg of bricks 3.65 m up a ladder. If the amount of work required to perform that task is used to comp

ress a gas at a constant pressure of 1.8 X 10^6 Pa, what is the change in volume of the gas?

2 answers:

MrRa [10]2 years ago

8 0

Answer:

Explanation:

Work done in carrying bricks

mgh

= 207 x 9.8 x 3.65

-= 7404.4 J

Work done in compressing gas

PΔV

Pressure x change in volume

1.8 x 10⁶ ΔV = 7404.4

ΔV = 7404.4 / 1.8 x 10⁶m³

= 4113.33 x 10⁻⁶ m³

= 4113.33 cc

vodka [1.7K]2 years ago

6 0

Answer:

$dV=4.1136\times 10^{-4}\ m^3$

Explanation:

Given:

Mass of the bricks carried, $m=207\ kg$

height of displacement, $h=3.65\ m$

Constant pressure of the gas, $P=1.8\times 10^6\ Pa$

<u>Now the work done to displace the brick along the length of the ladder:</u>

$W=(m.g)\times h$

$W=(207\times 9.8)\times 3.65$

$W=740.439\ J$

<u>As we know that the work done in compressing the gas at constant pressure is given as:</u>

$W=P.dV$

where:

$dV=$ change in volume of the gas

$740.439=1.8\times 10^6\times dV$

$dV=4.1136\times 10^{-4}\ m^3$

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Lina20 [59]

Explanation:

Given that,

Force with which a child hits a ball is 350 N

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We need to find the impulse received by the ball. The impulse delivered is given by :

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Which one of the following statements is true concerning an object executing simple harmonic motion?

timurjin [86]

Answer:

D) The objects velocity is zero when its acceleration is a maximum

Explanation:

In a simple harmonic motion, the total energy is constant (if we neglect air resistance and friction), and it is equal to the sum of the elastic potential energy U and the kinetic energy K:

$E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2$ (1)

where

m is the mass

v is the velocity

k is the spring constant

x is the displacement

As a consequence, since E must remain constant, when K increases U decreases, and vice-versa.

Also, in a simple harmonic motion the acceleration of the system is proportional to the negative of the displacement:

$a\propto - x$ (2)

So, combining (1) with (2), we have the following situations:

- When the displacement is zero (x=0), the acceleration is also zero (a=0), and so the velocity is maximum, because the kinetic energy is maximum

- When the displacement is maximum (x=max), the acceleration is also maximum, while the velocity is zero because the kinetic energy is zero

So, the correct statement is

D) The objects velocity is zero when its acceleration is a maximum

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2 years ago

550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c

Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

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Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work

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2 years ago

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Answer:

Honda won by 0.14 s

Explanation:

We are given that

Distance =S=200 m

Initial velocity of Honda=u=0m/s

Initial velocity of Porsche=u'=0m/s

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Acceleration of Porsche's= $3.5m/s^2$

Time taken by Honda to start=1 s

$s=ut+\frac{1}{2}at^2$

Substitute the values

$200=0(t)+\frac{1}{2}(3)t^2$

$200=\frac{3}{2}t^2$

$t^2=\frac{200\times 2}{3}=\frac{400}{3}$

$t=\sqrt{\frac{400}{3}}=11.55s$

Time taken by Honda=11.55 s

Now, time taken by Porsche

$200=\frac{1}{2}(3.5)t^2$

$t^2=\frac{200\times 2}{3.5}$

$t=\sqrt{\frac{400}{3.5}}=10.69 s$

Total time taken by Porsche=10.69+1=11.69 s

Because it start 1 s late

Time taken by Honda is less than Porsche .Therefore, Honda won and

Time =11.69-11.55=0.14 s

Honda won by 0.14 s

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mash [69]

Answer:

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Vf = 0 m/s

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a = - 1 m/s^2

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Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.

5 0

2 years ago