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iren2701 [21]
2 years ago
13

A car engine has power than a horse because a car engine does the same amount of work in time. Yasmin and Raj each had 10 boxes

of equal weight to stack next to each other on the same shelf, at the same height and in the same arrangement. Yasmin completed the task in 2 minutes, while Raj took 3 minutes to stack his boxes. Raj applied power than Yasmin because his stacking took time to do the same amount of work.
Physics
2 answers:
vfiekz [6]2 years ago
6 0

Answer:

more less less more

Explanation:

labwork [276]2 years ago
4 0

Answer:

A car engine has more power than a horse because a car engine does the same amount of work in time. Yasmin and Raj each had 10 boxes of equal weight to stack next to each other on the same shelf, at the same height and in the same arrangement. Yasmin completed the task in 2 minutes, while Raj took 3 minutes to stack his boxes. Raj applied less power than Yasmin because his stacking took more time to do the same amount of work.

Explanation:

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"1.0 kg mass is attached to the end of a spring. The mass has an amplitude of 0.10 m and vibrates 2.0 times per second. Find its
prisoha [69]

Answer:

1.3m/s

Explanation:

Data given,

Mass,m=1.0kg,

Amplitude,A=0.10m,

Frequency,f=2.0Hz.

From the equation of a simple harmonic motion, the displacement of the object at a given time is define as

x=Acos\alpha \\

we can express the velocity by the derivative of the displacement,

Hence

V=-Awsin\alpha \\

at equilibrium, the velocity becomes

V=wA\\w=2\pi f

Hence if we substitute values we arrive at

V=2\pi fA\\V=2\pi *2*0.1\\V=1.3m/s

4 0
3 years ago
Read 2 more answers
Pls ans 10 no. From laws of motion
Vlad1618 [11]

Answer:

40N

Explanation:

Since both weights are connected to one string, you can say that the tensions above each are equal to each other.

If you do the sum of forces for the 4kg mass, then the tension comes out to 40N (if we take gravity to be 10m/s²). But that seemed too good to be true, so I decided to do the work for the 7kg mass as well [which included finding the normal force (N) and plugging it into the sum of forces for the 7kg mass] to find that it also gives 40N as the answer.

If I were to put my process into steps:

  1. Write out the sum of Forces for both masses
  2. Set them equal to each other to find normal force (because this is the only unknown)
  3. Calculate and compare the two tensions to see if they are equal

*This all seems to line up perfectly, but do let me know if my answer doesn't match up with what you might find to he the answer later on.

4 0
3 years ago
(5, 3) and (7, 3) are two coordinate points for a single object on a position-versus-time graph. Assume time is measured in seco
Maru [420]
Since the y axis stayed consistent, we can assume it did not move at all.
(So your answer would be A)
6 0
3 years ago
Read 2 more answers
A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts
prohojiy [21]

Answer:

Explanation:

The relation between activity and number of radioactive atom in the sample is as follows

dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms

For the beginning period

dN₀ / dt = λ N₀

58.2 = λ N₀

similarly

41 = λ N

dividing

58.2 / 41 = N₀ / N

N = N₀ x .70446

formula of radioactive decay

N=N_0e^{-\lambda t }

.70446 =e^{-\lambda t }

- λ t = ln .70446 =   - .35

t = .35 / λ

λ = .693 / half life

= .693 / 5715

= .00012126

t = .35 / .00012126

= 2886.36

= 2900 years ( rounding it in two significant figures )

7 0
3 years ago
A heat engine operates between two reservoirs at 300K. During each cycle, it absorbs 200 calories from the high temperature rese
yanalaym [24]

(a) Zero

The maximum efficiency (Carnot efficiency) of a heat engine is given by

\eta=1-\frac{T_C}{T_H}

where

T_C is the low-temperature reservoir

T_H is the high-temperature reservoir

For the heat engine in the problem, we have:

T_C = 300K

T_H = 300K

Therefore, the maximum efficiency is

\eta=1-\frac{300}{300}=0

(b) Zero

The efficiency of a heat engine can also be rewritten as

\eta = \frac{W}{Q_H}

where

W is the work performed by the engine

Q_H is the heat absorbed from the high-temperature reservoir

In this problem, we know

\eta=0

Therefore, since the term Q_H cannot be equal to infinity, the numerator of the fraction must be zero as well, which means

W = 0

So the engine cannot perform any work.

5 0
2 years ago
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