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2 years ago

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If a heat engine pulls 4390.0 J of heat from the hot reservoir and exhausts 3582.2 J of heat to the cold reservoir, what was the

work done by the engine? What is the efficiency of this engine in percent? (do not enter the percent sign)

1 answer:

Oksanka [162]2 years ago

8 0

Answer:

18.4

Explanation:

Q1 = 4390 J

Q2 = 3582.2 J

The efficiency of heat engine is given by

n = 1 - Q2 / Q1

n = 1 - 3582.2 / 4390

n = 0.184

n = 18.4 %

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A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t

Morgarella [4.7K]

Answer:

(A). The work done is $1.50\times10^{-6}\ J$ .

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy $K.E=1.50\times10^{-6}\ J$

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

$W = \Delta K.E$

$W=K.E_{f}-K.E_{i}$

Put the value into the formula

$W= 1.50\times10^{-6}-0$

$W=1.50\times10^{-6}\ J$

The work done is $1.50\times10^{-6}\ J$ .

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

$\Delta P.E=\Delta K.E$

So, $U = 1.50\times10^{-6}$

Using formula of potential

$V=\dfrac{U}{q}$

Put the value into the formula

$V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}$

$V=357.14\ V$

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

$W=F\times r$ ....(I)

Using formula of force

$F=qE$

Put the value in the equation (I)

$W=qE\times r$

$E=\dfrac{W}{q\times r}$

Put the value into the formula

$E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}$

$E=5952.38\ N/C$

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

7 0

2 years ago

If the average speed of an orbiting space shuttle is 27 800 km/h, determine the time required for it to circle Earth. Assume tha

balu736 [363]

Answer:

t = 1.51 hours

Explanation:

given,

Speed of space shuttle. v = 27800 Km/h

Radius of earth, R = 6380 Km

height of shuttle above earth, h = 320 Km

Total radius of the shuttle orbit

r' = R + h

r' = 6380 + 320

r' = 6700 Km

distance, d = 2 π r

d = 2 π x 6700

$time = \dfrac{distance}{speed}$

$time = \dfrac{2\pi\times 6700}{27800}$

t = 1.51 hours

Time require by the shuttle to circle the earth is equal to 1.51 hr.

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2 years ago

The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational

s344n2d4d5 [400]

Answer:

v = √2G $M_{earth}$ / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

Eo = K + U = ½ m1 v² - G m1 m2 / r1

Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

Eo = Ef

½ m1v² - G m1 $M_{earth}$ / R = - G m1 $M_{earth}$ / R

v² = 2G $M_{earth}$ (1 / R - 1 / Rinf)

If we do Rinf = infinity 1 / Rinf = 0

v = √2G $M_{earth}$ / R

Ef = = - G m1 m2 / R

The mechanical energy is conserved

Em = -G m1 $M_{earth}$ / R

Em = - G m1 $M_{earth}$ / R

R = int ⇒ Em = 0

6 0

2 years ago

An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far r from the center

Diano4ka-milaya [45]

Answer:

2023857702.507m

Explanation:

$f=\frac{GMm}{r^{2} }$

recall from newton's law of gravitation

G=gravitational constant

mshew=50g

melephant=5*10^3kg

rearth=radius of the earth 6400km or 6400000m

mearth= masss of the earth

Gm(shrew)m(earth)/r(earth)^2 = Gm(elephant)m(earth)/r^2

strike out the left hand side and right hand side variables

m(shrew)/r(earth)^2 = m(elephant)/r^2

r^2 = m(elephant).r(earth)^2 / m(shrew) .........make r^2 the subject of the equation

r^2= $(5*10^{3} *(6400000)^{2} )/.05$

r^2=40960000000000

r=2023857702.507m

4 0

2 years ago

A box of mass 3kg is lifted 1.5m onto a shelf. Calculate the change in its gravitational potential energy. The gravitational fie

Vedmedyk [2.9K]

Answer:

The change in gravitational potential energy is 45 J.

Explanation:

Given that,

Mass = 3 kg

Distance = 1.5 m

Gravitational field strength = 10 N/kg

We need to calculate the change in gravitational potential energy

Using formula of gravitational potential energy

$Change\ in\ gravitational\ potential\ energy =gravitational\ field\ strength\times mass\times distance$

Put the value into the formula

$Change\ in\ gravitational\ potential\ energy =10\times3\times1.5$

$Change\ in\ gravitational\ potential\ energy =45\ J$

Hence, The change in gravitational potential energy is 45 J.

5 0

2 years ago