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USPshnik [31]
2 years ago
15

State joole's law ofheating and verifyexperimentally​

Physics
1 answer:
diamong [38]2 years ago
6 0

Answer:

Joule's law is verified using Joule's calorimeter. .

Explanation:

Two thirds of the volume of the calorimeter is filled with water. The calorimeter is enclosed in a wooden box to minimise loss of heat. A battery (Bt), a key (K), a rheostat (Rh) and an ammeter (A) are connected in series with the calorimeter

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A hollow sphere of inner radius 8.82 cm and outer radius 9.91 cm floats half-submerged in a liquid of density 948.00 kg/m^3. (a)
kari74 [83]

Answer:

a) 0.568 kg

b) 474 kg/m³

Explanation:

Given:

Inner radius = 8.82 cm = 0.0882 m

Outer radius = 9.91 cm = 0.0991 m

Density of the liquid = 948.00 Kg/m³

a) The volume of the sphere = \frac{4\pi}{3}\times(0.0991^2-0.0882^2)

or

volume of sphere = 0.0012 m³

also, volume of half sphere = \frac{\textup{Total volume}}{\textup{2}}

or

volume of half sphere = \frac{\textup{0.0012}}{\textup{2}}

or

Volume of half sphere =0.0006 m³

Now, from the Archimedes principle

Mass of the sphere = Weight of the volume of object submerged

or

Mass of the sphere = 0.0006× 948.00 = 0.568 kg

b) Now, density =  \frac{\textup{Mass}}{\textup{Volume}}

or

Density = \frac{\textup{0.568}}{\textup{0.0012}}

or

Density = 474 kg/m³

8 0
3 years ago
IN WHAT CONDITION DO SOUND ECHO
DerKrebs [107]

Answer:

The conditions necessary for hearing the echo. The distance between the sound source and the reflecting surface must not be less than 17 metres where the time period between hearing the original sound and its echo should not be less than 0.1 of a second.

6 0
2 years ago
A manager at a local bank analyzed the relationship between monthly salary and three independent variables: length of service (m
Readme [11.4K]

Answer:

a

Explanation:

8 0
3 years ago
A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun
nalin [4]

Answer:

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

Explanation:

Given data

Source Frequency fs=600Hz

Length r=1.0m

RPM=100 rpm

The speed of the generator is calculated as:

v_{s}=rw\\v_{s}=r(2\pi f)

Substitute the given values

v_{s}=(1.0m)2\pi (\frac{100}{60}rev/s )\\v_{s}=10.47m/s

For approaching generator the frequency is calculated as:

f_{+}=\frac{f_{s}}{1-\frac{v_{s}}{v} }\\f_{+}=\frac{600Hz}{1-\frac{10.47m/s}{343m/s} } \\f_{+}=620Hz

On the other hand,for the receding generator,Doppler's effect is expressed as:

 f_{-}=\frac{f_{s}}{1+\frac{v_{s}}{v} }\\f_{-}=\frac{600Hz}{1+\frac{10.47m/s}{343m/s} } \\f_{-}=580Hz

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

8 0
3 years ago
the angular speed of an automobile engine is increased at a constant rate from 1300rev/min to 2000rev/min in 3s (a) what is its
GalinKa [24]

Answer:

please find attached pdf

Explanation:

Download pdf
8 0
2 years ago
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