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IgorC [24]
3 years ago
5

Water has a specific heat of 4.186 J/g°C. How much heat is required to increase 10.0 g of water from 25.0°C to 30.0°C?

Physics
1 answer:
poizon [28]3 years ago
6 0

Answer:

=0.2093J

Explanation:

Heat required(Q) = mcø

= (10/1000)*4.186*(30-25)

= 0.2093J

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explain the fleming left-hand rule with the diagram and what will the direction of the induced current in the figure if the magn
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Hi sorry

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4 0
2 years ago
You are 15.0m from the source of a sound. At that distance, you hear it at a sound level of 20.0dB. How close must you move to t
Andrej [43]
The answer is 45.0cm
5 0
3 years ago
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A ray of light is incident on a flat surface of a block of polystyrene, with an index of refraction of 1.49, that is submerged i
Crazy boy [7]

Answer:

21.5°

Explanation:

Given,

Refractive index of water, n₁ = 1.33

Refractive index of polystyrene, n₂ = 1.49

Angle of reflection = ?

Angle of refraction = 19.1°

Using Snell's law

n₁ sin θ₁ = n₂ sin θ₂

1.33 x sin θ₁ = 1.49 x sin 19.1°

sin θ₁ = 0.366

θ₁  = 21.5°

According to law of reflection angle of incidence is equal to angle of reflection.

Angle of reflection =  21.5°

4 0
3 years ago
A 0.25-m string, vibrating in its sixth harmonic, excites a 0.96-m pipe that is open at both ends into its second overtone reson
Andrews [41]

Answer:

option D

Explanation:

given,

length of the pipe, L = 0.96 m

Speed of sound,v = 345 m/s

Resonating frequency when both the end is open

f = \dfrac{nv}{2L}

n is the Harmonic number

2nd overtone = 3rd harmonic

so, here n = 3

now,

f = \dfrac{3\times 345}{2\times 0.96}

f = 540 Hz

The common resonant frequency of the string and the pipe is closest to 540 Hz.

the correct answer is option D

7 0
3 years ago
49. \ A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner
faust18 [17]

Answer:

    \frac{L_1}{L_2} = \sqrt{(n^2 - 1)}

Explanation:

For this interesting problem, we use the definition of centripetal acceleration  

      a = v² / r  

angular and linear velocity are related  

     v = w r  

we substitute  

    a = w² r

the rectangular body rotates at an angular velocity w  

We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A  

the distance OB = L₂  

the distance AB = L₁

the sides of the rectangle  

It is indicated that the acceleration in in A and B are related  

      a_A = n \ a_B  

we substitute the value of the acceleration  

    w² r_A = n r_B  

the distance from the each corner is  

    r_B = L₂  

    r_A = \sqrt{L_1^2 + L_2^2}  

we substitute  

   \sqrt{L_1^2 + L_2^2} = n L₂  

    L₁² + L₂² = n² L₂²  

    L₁² = (n²-1) L₂²  

4 0
3 years ago
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