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2 years ago

9

A dog walks 12 meters to the west and then 16 meters back to the east

2 answers:

kupik [55]2 years ago

5 0

4 meters east dude woot

Anna71 [15]2 years ago

4 0

Answer:

Explanation:

4 meters east

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EXPLAIN HOW ENERGY IS TRANSMITTED THROUGH A MEDIUM

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A truck pulls a block 8 meters across a level surface at a force of 216 N over the course of 12 seconds. How much power did the

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Answer:

work = 1728

Power = 134

Explaination:

by using the formula,

Work(W)= Force(F)×Distance(D)

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3 years ago

Question 5 of 10

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A. The particle model, because only high-energy frequencies of light can remove electrons .

Explanation:

Each photon of blue light has higher energy than each photon of red light has . So when each photon strikes each electron , it gets ejected . But the photon of red light has not sufficient energy to eject electron . Once the photon of red light strikes the electron , the energy is wasted off . Energy of photon can not be accumulated . Thus photon behaves like particle .

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3 years ago

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.

From the above we see that F ∝ mm' and $F\alpha \frac{1}{r^{2} }$

Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

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Hence the gravitational force $f_{2} =G\frac{m m_{2} }{r^{2} }$

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$= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }$

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3 years ago

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Consider a Carnot heat-engine cycle executed in a closed system using 0.025 kg of steam as the working fluid. It is known that t

ArbitrLikvidat [17]

Answer:

The temperature of the steam during the heat rejection process is 42.5°C

Explanation:

Given the data in the question;

the maximum temperature T $_H$ in the cycle is twice the minimum absolute temperature T $_L$ in the cycle

T $_H$ = 0.5T $_L$

now, we find the efficiency of the Carnot cycle engine

η $_{th$ = 1 - T $_L$ /T $_H$

η $_{th$ = 1 - T $_L$ /0.5T $_L$

η $_{th$ = 0.5

the efficiency of the Carnot heat engine can be expressed as;

η $_{th$ = 1 - W $_{net$ /Q $_H$

where W $_{net$ is net work done, Q $_H$ is is the heat supplied

we substitute

0.5 = 60 / Q $_H$

Q $_H$ = 60 / 0.5

Q $_H$ = 120 kJ

Now, we apply the first law of thermodynamics to the system

W $_{net$ = Q $_H$ - Q $_L$

60 = 120 - Q $_L$

Q $_L$ = 60 kJ

now, the amount of heat rejection per kg of steam is;

q $_L$ = Q $_L$ /m

we substitute

q $_L$ = 60/0.025

q $_L$ = 2400 kJ/kg

which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)

q $_L$ = h $_{fg$ = 2400 kJ/kg

now, at h $_{fg$ = 2400 kJ/kg from saturated water tables;

T $_L$ = 40 + ( 45 - 40 ) ( $\frac{2400-2406.0}{2394.0-2406.0}\\}$ )

T $_L$ = 40 + (5) × (0.5)

T $_L$ = 40 + 2.5

T $_L$ = 42.5°C

Therefore, The temperature of the steam during the heat rejection process is 42.5°C

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