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nataly862011 [7]
3 years ago
14

A boy drops a 0.10 kg stone down a 150 m well and listens for the echo. The air temperature is 20°C. How long after the stone is

dropped will the boy hear the echo? The speed of sound through air at 20°C is v = 343 m/s.
Physics
1 answer:
S_A_V [24]3 years ago
4 0
Use formula for Echo which is Velocity=2(Distance)/time so 343=2(150)/T 343T=300..T=300/343=0.9 seconds
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Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
A 55.0-g sample of hot metal initially at 99.5oC was added to 40.0 g of water in a Styrofoam coffee cup calorimeter. The water a
Kaylis [27]

Answer:

Cp= 0.44 J/g.C

This is heat capacity of metal.

Explanation:

From energy conservation

Heat lost by metal = Heat gain by water +Heat gain by  calorimeter

Because here temperature of metal is high that is why it loose the heat.The temperature of water and  calorimeter is low that is why they gain the heat.

final temperature is T= 30.5 C

We know that sensible heat transfer given as

Q= m Cp ΔT

m=Mass

Cp=Specific heat capacity

ΔT=Temperature difference

By putting the values

55 x Cp ( 99.5 - 30.5) = 40 x 4.184 ( 30.5- 21 ) + 10 x ( 30.5 - 21)

Cp ( 99 .5- 30.5) = 30.65

Cp= 0.44 J/g.C

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4 0
4 years ago
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dimulka [17.4K]
That would be only rotational motion
5 0
3 years ago
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givi [52]
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3 0
3 years ago
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a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

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Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
2 years ago
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