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Dafna11 [192]
3 years ago
15

What is the best description of the acceleration of an object?

Physics
1 answer:
marin [14]3 years ago
8 0
The rate of change of velocity
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What amount of heat is required to raise the temperature of 200 g of water by 15°C (the specific heat of water is 1 cal/g°C)
Sergeeva-Olga [200]

Answer:

Heat energy required (Q) = 3,000 J

Explanation:

Find:

Mass of water (M) = 200 g

Change in temperature (ΔT) = 15°C

Specific heat of water (C) = 1 cal/g°C

Find:

Heat energy required (Q) = ?

Computation:

Q = M × ΔT × C

Heat energy required (Q) = Mass of water (M) × Change in temperature (ΔT) × Specific heat of water (C)

Heat energy required (Q) = 200 g × 15°C × 1 cal/g°C

Heat energy required (Q) = 3,000 J

4 0
3 years ago
Monochromatic light of wavelength 385 nm is incident on a narrow slit. On a screen 3.00 m away, the distance between the second
LiRa [457]

To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.

For this purpose we have that the constructive interference in waves can be expressed under the function

a sin\theta = m\lambda

Where

a = Width of the slit

d = Distance of slit to screen

m = Number of order which represent the number of repetition of the spectrum

\theta = Angle between incident rays and scatter planes

At the same time the distance on the screen from the central point, would be

sin\theta = \frac{y}{d}

Where y = Represents the distance on the screen from the central point

PART A ) From the previous equation if we arrange to find the angle we have that

\theta = sin^{-1}(\frac{y}{d})

\theta = sin^{-1}(\frac{1.4*10^{-2}}{3})

\theta = 0.2673\°

PART B) Equation both equations we have

a sin\theta = m\lambda

a \frac{y}{d} = m\lambda

Re-arrange to find a,

a = \frac{(2)(385*10^{-9})(3)}{(1.4*10^{-2})}

a = 1.65*10^{-4}m

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3 years ago
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Rickshaw ke pahiye laga do ge tho bhai india ke roado ki haal waali feeling aaygi
8 0
3 years ago
Can anyone please help me with this one
Vinvika [58]
Ur in k12 ALVA too???
6 0
4 years ago
A gyroscope slows from an initial rate of 49.3 rad/s at a rate of 0.731 rad/s^2. (a) How long does it take (in s) to come to res
olasank [31]

Answer:

(a) 67.44 second

(b) 529.5 revolutions

Explanation:

ωo = 49.3 rad/s

α = 0.731 rad/s^2

(a) Let it takes t time to come to rest.

ω = 0

Use first equation of motion for rotational motion

ω = ωo + α t

0 = 49.3 - 0.731 x t

t = 67.44 second

(b) Let it turns an angle θ rad before coming to rest. use third equation of motion for rotational motion.

ω^2 = ωo^2 + 2 α θ

0 = 49.3 x 49.3 -  2 x 0.731 x θ

2430.49 = 1.462 θ

θ = 1662.44 rad

Number of revolutions = θ / 2π = 1662.44 / 3.14 = 529.5 revolutions

6 0
3 years ago
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