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polet [3.4K]
3 years ago
8

The flow of electrons in a circuit follows a path from the negative terminal of the battery to the positive terminal

Physics
1 answer:
SVEN [57.7K]3 years ago
3 0

Answer:

No it's not true electric grow from Positive terminal to negative terminal

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a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi
zubka84 [21]

Answer:

The work done on the car is, W = 75,000 J

The power delivered by the engine, P = 25,000 watts

Explanation:

Given,

The mass of the car, m = 1500 Kg

The initial velocity of the car, u = 0

The final velocity of the car, v = 10 m

The time duration of the travel, t = 3 s

Using the first equation of motion

                     v = u + at

                     a = (v - u) / t

Substituting the given values in the above equation

                    a = (10 - 0) / 3

                       = 3.33 m/s²

Using the second equations of motion

                      s = ut + 1/2 at²

                         = 0 + 0.5 x 3.33 x 3²

                         = 15 m

The force exerted by the car

                         F = m x a

                            = 1500 Kg x 3.33 m/s²

                            = 5000 N

The work done by the car,

                          W = F x S

                               = 5000 N x 15 m

                               = 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

                            P = W / t

                                = 75,000 J / 3 s

                                 = 25,000 watts

The power delivered by the engine, P = 25,000 watts

5 0
4 years ago
Changing the velocity of a moving object will have little effect on its kinetic energy.
Alinara [238K]

Answer:Faulse

Explanation:

5 0
3 years ago
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The closeness of particles of gas and their low speeds allow intermolecular forces to become important at certain pressures and
Lilit [14]
This is a limitation of kinetic-molecular energy. Right?
4 0
3 years ago
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An airplane traveling at one third the speed of sound (i.e., 114 m/s) emits a sound of frequency 3.72 kHz. At what frequency doe
nlexa [21]

Answer:

f'=5.58kHz

Explanation:

This is an example of the Doppler effect, the formula is:

f'=\frac{(v+v_o)}{(v+v_s)}f

Where f is the actual frequency, f' is the observed frequency, v is the velocity of the sound waves, v_o the velocity of the observer (which is negative if the observer is moving away from the source)  and v_s the velocity of the source  (which is negative if is moving towards the observer). For this problem:

f=3.72kHz\\v=342m/s\\v_o=0m/s\\v_s=-114m/s

f'=\frac{(342+0)}{(342-114)}3.72\times10^3\\f'=\frac{342}{228}3.72\times10^3\\f'=(1.5)3.72\times10^3\\f'=5580Hz=5.58kHz

5 0
3 years ago
One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau
suter [353]

Answer:

actual efficiency is  47.78 %

Carnot efficiency  is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency  and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy  = 7.0 × 10^9 cal  (4.184 J/1 cal)  = 29.29 × 10^9 J

actual efficiency  =  1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency  =  0.4778

actual efficiency is  47.78 %

and

Carnot efficiency  is = 1 - ( 703 / 2173 )

so Carnot efficiency  is  = 0.67648

Carnot efficiency  is 67.65 %

and

power output  = work / time

power output  =  1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

5 0
3 years ago
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