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3 years ago

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The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiato

r it has a temperature of 175 °F. Calculate the amount of heat transferred from the engine to the surroundings by one gallon of water with a specific heat of 4.184 J/g °C.

1 answer:

djverab [1.8K]3 years ago

4 0

Answer:571.09 kJ

Explanation:

Given

Temperature of cooling water from engine exit $=240^{\circ} F\approx 115.55^{\circ}C$

After Passing through the radiator its temperature decreases to $175^{\circ}F\approx 79.44^{\circ}F$

specific heat of water $=4.184 J/g^{\circ}C$

Volume of water $= 1 gallon\approx 3.78 L$

density of water $\rho =1 gm/mL$

Thus mass of water $=\rho \times V=3.78\times 1=3.78 kg$

Heat transferred to the surrounding is equal to heat absorbed by cooling water

$Q=m\cdot c\cdot \Delta T$

$Q=3.78\times 4.184\times 1000\times (115.55-79.44)$

$Q=3.78\times 4.184\times 1000\times (36.11)$

$Q=571.09 kJ$

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lutik1710 [3]

Answer:

speed of electrons = 3.25 × $10^{7}$ m/s

acceleration in term g is 3.9 × $10^{17}$ g.

radius of circular orbit is 2.76 × $10^{-4}$ m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = $\sqrt{\frac{2qV}{m}}$

v = $\sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}$

v = 3.25 × $10^{7}$ m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = $\frac{Bqv}{m}$

a = $\frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}$

a = 3.82 × $10^{18}$ m/s²

and acceleration in term g

a = $\frac{3.82\times 10^{18}}{9.81}$

a = 3.9 × $10^{17}$ g

acceleration in term g is 3.9 × $10^{17}$ g.

and

electron moving in circular orbit has centripetal force

F = $\frac{mv^2}{r}$

Bqv = $\frac{mv^2}{r}$

r = $\frac{mv}{Bq}$

r = $\frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}$

r = 2.76 × $10^{-4}$ m

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8 0

3 years ago

A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and

riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

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3 years ago

Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin

likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0 and T - mgcos57 - F = m(0) = 0

Mg - T = 0 (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

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Scientists measure the time between the arrival of an earthquake's _______ and _______ waves to help determine the distance betw

soldier1979 [14.2K]

Scientists measure the time between the arrival of an earthquake's __P____ and ___S____ waves to help determine the distance between the recording seismograph and the earthquake epicenter.

Explanation:

P- (compressional) and S- (shear) waves produced in earthquakes travel at different speeds. P waves are faster than S waves and hence will be detected first by a seismograph after an earthquake. The further away a seismograph is from the epicenter of an earthquake, the longer the time difference between the two (2) waves will be.

Using several, at least 3, seismographs located at different geoghraphical locations and detecting earthquakes, geologists can extrapolate the epicenter of an earthquake using the time differences in arrivals of the two waves in each of the seismographs, using the mathematics of triangulation.

Learn More:

For more on P- and S-- waves check out;

brainly.com/question/11915788

brainly.com/question/11334414

brainly.com/question/2530620

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