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3 years ago

5

the rock of 10 kg is falling near the Earth's surface.assume that g =10N/kg and the is no air resistance. what is the accelerati

on of the rock

1 answer:

Ivanshal [37]3 years ago

5 0

Answer:

kya faltu sawal h repetitive g=10N/kg

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14. Convert 22 degrees celsius to fahrenheit.<br> Please show your work

forsale [732]

(22°C × 9/5) + 32 = 71.6°F

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3 years ago

Read 2 more answers

ivann1987 [24]

Answer:

Required energy Q = 231 J

Explanation:

Given:

Specific heat of copper C = 0.385 J/g°C

Mass m = 20 g

ΔT = (50 - 20)°C = 30 °C

Find:

Required energy

Computation:

Q = mCΔT

Q = 20(0.385)(30)

Required energy Q = 231 J

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2 years ago

How much work is required to move

Nikitich [7]

The work is path independent since we have a conservative force.

Thus $W=d\cdot\frac{q\cdot U}{d^2}=\frac{3.0\cdot9.0}{0.010}=\boxed{2700 J}$

Answer (1)

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3 years ago

What is the resistance of : A) A 1.70 m long copper wire that is 0.700 mm in diameter? B) A 20.0 cm long piece of carbon with a

astra-53 [7]

Answer:

(I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

Explanation:

Given that,

Length of copper wire = 1.70 m

Diameter = 0.700 mm

Length of carbon piece = 20.0 cm

Cross section area $A = (2.00\times10^{-3})^2\ m$

(I). We need to calculate the area of copper wire

Using formula of area

$A=\pi r^2$

$A=3.14\times(\dfrac{0.700\times10^{-3}}{2})^2$

We need to calculate the resistance

Using formula of resistance

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.68\times10^{-8}\times1.70}{3.14\times(\dfrac{0.700\times10^{-3}}{2})^2}$

$R=0.0742\ \Omega$

(II). We need to calculate the resistance

Using formula of resistance

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{3.5\times10^{-5}\times20\times10^{-2}}{(2.00\times10^{-3})^2}$

$R=1.75\ \Omega$

Hence, (I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

8 0

3 years ago

Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 349 m/s,

SSSSS [86.1K]

Answer:

Time period between the successive beats will be 0.1703 sec

Explanation:

We have given speed of the sound v = 349 m/sec

Wavelength of piano $A\lambda _A=0.766m$

Wavelength of piano $B\lambda _B=0.776m$

So frequency of piano A $f_1=\frac{v}{\lambda _1}=\frac{349}{0.766}=455.61Hz$

Frequency of piano B $f_2=\frac{v}{\lambda _1}=\frac{349}{0.776}=449.74Hz$

So beat frequency f = 455.61 - 449.74 = 5.87 Hz

So time period $T=\frac{1}{f}=\frac{1}{5.87}=0.1703sec$

So time period between the successive beats will be 0.1703 sec

4 0

3 years ago