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Anna35 [415]
3 years ago
13

BEST ANSWER GETS BRAINLIEST!

Physics
1 answer:
wel3 years ago
8 0

Answer:

Distance = 16.9 m

Explanation:

We are given;

Power; P = 70 W

Intensity; I = 0.0195 W/m²

Now, for a spherical sound wave, the intensity in the radial direction is expressed as a function of distance r from the center of the sphere and is given by the expression;

I = Power/Unit area = P/(4πr²)

where;

P is the sound power

r is the distance.

Thus;

Making r the subject, we have;

r² = P/4πI

r = √(P/4πI)

r = √(70/(4π*0.0195))

r = √285.6627

r = 16.9 m

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How much heat is necessary to warm 500g of water from 20°c to 65°c​
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An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,
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Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity      

       v=\sqrt{2.14^2+0.34^2}=2.17m/s

Direction,  

       \theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0
3 years ago
A 0.900-V potential difference is maintained across a 1.5m length of 2
Daniel [21]

Answer:

I = 6.42 A

Explanation:

Given that,

Potential difference, V = 0.9 V

Length of the wire, l = 1.5 m

Area of cross section, A=0.6\ mm^2=6\times 10^{-7}\ m^2

We need to find the current in the wire. Let I is current. We can find it using Ohm's law as follows :

V = IR

Where R is the resistance of the wire

I=\dfrac{V}{R}\\\\I=\dfrac{V}{\rho \dfrac{l}{A}}\\\\I=\dfrac{0.9}{5.6\times 10^{-8}\times \dfrac{1.5}{6\times 10^{-7}}}\\\\I=6.42\ A

So, the current in the wire is 6.42 A.

3 0
3 years ago
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