PHYSICS HIGH SCHOOL

What is the power of a refrigerator with voltage 110 V and
current 0.8 A?

Answers

Answer 1
Answer:

Answer: 88

Explanation:

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Which one of the following lines best illustrates personification? A.She floated graceful as a dove

B.Spring is a dream unsung

C.A narrow wind complains all day

D.The fog comes on little cat feet

Answers

C. A narrow wind complains all day.

Personification is a figure of speech wherein animals, objects, and ideas are given human traits or characteristics.

The narrow wind, which is an unseen object, was given a human trait which is complains.

The lines that best illustrates personification is (D), The fog comes on little cat feet.

What is personification?

Personification is a literary device where inanimate objects are given human qualities. In the line "The fog comes on little cat feet," the fog is given the human qualities of walking and having feet. This is a clear example of personification.

The other lines do not illustrate personification. In line A, "She floated graceful as a dove," the dove is being compared to a graceful woman, but the dove is not being given any human qualities. In line B, "Spring is a dream unsung," spring is being compared to a dream, but spring is not being given any human qualities. In line C, "A narrow wind complains all day," the wind is being compared to a person who is complaining, but the wind is not being given any human qualities.

Find out more on personification here: brainly.com/question/23487716

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What is the approximate difference in elevation between Coolage Lake and the rectangular building in the top center portion of the map? A. 27 ft
B. 5 ft
C. 35 ft
D. 25 ft

Answers

The right answer for the question that is being asked and shown above is that:  "C. 35 ft." The approximate difference in elevation between Coolage Lake and the rectangular building in the top center portion of the map is that of C. 35 ft 

What do you mean by average velocity​

Answers

Answer:

Here is the answer. Hope this helps you!

Explanation:

Average velocity is the sum of initial and final velocity divided by 2. I t is the same as total Displacement divided by total time. Average velocity is calculated when the body is in non-uniform motion (also when total displacement and time is not given). The formula is as follows:

u + v/2 = v_(av)

Average velocity = Average speed

during motion in a straight line. therefore, the above mentioned formula can be used for calculating average speed as well, when the direction is one and only the same, that is, during motion in a straight line. The S.I unit remains the same-

m/s.

Since velocity is vector, average VELOCITY is also vector. However, Average SPEED is scalar as speed is scalar. both can be equal only when the distance = displacement and when they are following the same direction of motion.

an empty propane tank dropped from a hot air balloon hits the ground with a speed of 143.8 m/s. from what height was the tank released?

Answers

What you know:
Vi=0m/s
Vf=143.8m/s
A=-9.8m/s
d=???
Use the equation Vf^2=Vi^2+2A(d)
Rearrange to isolate d: d=Vf^2/2A
d=(143.8)^2/2(-9.8)
d=20678.4/-19.6
d=-1055m
The tank was released from a height of 1055m

You're conducting a physics experiment on another planet. You drop a rock from a height of 2.3 m and it hits the ground 1.1 seconds later. What is the acceleration due to gravity on this planet?A. 4.2 m/s2
B. 3.8 m/s2
C. 2.4 m/s2
D. 9.8 m/s2

Answers

Acceleration due to gravity on this planet will be 3.802 m / s^2

What are equations of motion?

Equation of motion are defined as equations that describe the behavior of a physical system in terms of its motion as a function of time

Using equation of motion

u=0

s= 2.3 m

t = 1.1 sec

to find = g (acceleration due to gravity on this planet)

s = u t + 1/2 (a ) (t ^(2))

s = 1/2 (g) (t^2)

2.3 = 1/2 (g) (1.1^2)

g = 2 * 2.3 /(1.1)^2

g = 4.6 /1.21= 3.802

correct answer is b) g = 3.802 m / s^2

Learn more about equation of motion

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You get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (B)

An internal combustion engine uses fuel, of energy content 44.4 MJ/kg, at a rate of 5 kg/h. If the efficiency is 28%, determine the power output and the rate of heat rejection

Answers

Efficiency =  Power Output / Power Input

Power Input = Rate of Energy input = 44.4 MJ/kg * 5 kg/h

 = 222 MJ/h

But 1 hour = 3600seconds

222 MJ/h = 222 MJ/3600s =  0.061667 MW              J/s = Watts

Power input = 0.061667 MW = 61 667 W

From  Efficiency =  Power Output / Power Input

   28% =  Power Output / 61667

   Power Output = 0.28 * 61667

   Power Output = 17266.76 W
 
  Power Output = 17 267 W

 Rate of heat Rejection = Power input - Power output

                                        = 61667 - 17267 = 44400 W

Rate of heat Rejection = 44 400 W.


C- Copyright.
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