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irakobra [83]
2 years ago
10

2. On January 21 in 1918, Granville, North Dakota, had a surprising change in temperature. Within 12 hours, the temperature chan

ged from 237 K to 283 K. What is this change in temperature in the Celsius and Fahrenheit scales?
Physics
1 answer:
wolverine [178]2 years ago
7 0

Answer:

The temperature change in Celsius is 46°C.

The temperature change in Fahrenheit is 82.8°F.

Explanation:

A degree of Celsius scale is equal to that of kelvin scale; therefore,

\Delta C = \Delta K  = 283K-237K\\\\ \boxed{\Delta C = 46^oC}

A degree in Fahrenheit is 1.8 times the Celsius degree; therefore

\Delta F  = 1.8(46^o)

\boxed{\Delta F = 82.8^oF}

Hence, the temperature change in Celsius is 46°C, and the temperature change in Fahrenheit is 82.8°F.

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A cylinder is 0.10 m in radius and 0.20 in length. Its rotational inertia, about the cylinder axis on which it is mounted, is 0.
Bumek [7]

Answer:5 rad/s^2

Explanation:

Given

Radius of cylinder r=0.1 m

Length L=0.2 in.

Moment of inertia I=0.020 kg-m^2

Force F=1 N

We Know Torque is given by

Torque =I\alpha =F\cdot r

where \alpha =angular\ acceleration

I\alpha =F\cdot r

0.02\cdot \alpha =1\cdot 0.1

\alpha =5 rad/s^2    

5 0
2 years ago
A 250 Hz tuning fork is struck and the intensity at the source is I1 at a distance of one meter from the source. (a) What is the
Zina [86]

Answer:

a) 0.0625 I_1

b) 3.16 m

Explanation:

<u>Concepts and Principles  </u>

The intensity at a distance r from a point source that emits waves of power P is given as:  

I=P/4π*r^2                         (1)

<u>Given Data</u>

f (frequency of the tuning fork) = 250 Hz

I_1 is the intensity at the source a distance r_1 = I m from the source.  

<u>Required Data</u>

- In part (a), we are asked to determine the intensity I_2 a distance r_2 = 4 in from the source.

- In part (b), we are asked to determine the distance from the tuning fork at which the intensity is a tenth of the intensity at the source.  

<u>solution:</u>

(a)  

According to Equation (1), the intensity a distance r is inversely proportional to the distance from the source squared:

I∝1/r^2

Set the proportionality:  

I_1/I_2=(r_2/r_1)^2                                 (2)

Solve for I_2 :  

I_2=I_1(r_2/r_1)^2  

I_2=0.0625 I_1

(b)  

Solve Equation (2) for r_2:  

r_2=(√I_1/I_2)*r_1

where I_2 = (1/10)*I_1:

r_2=(√I_1/1/10*I_1)*r_1

     =3.16 m

3 0
2 years ago
The Lyman series comprises a set of spectral lines. All of these lines involve a hydrogen atom whose electron undergoes a change
mihalych1998 [28]

Answer:

a) 1.2*10^-7 m

b) 1.0*10^-7 m

c) 9.7*10^-8 m

d) ultraviolet region

Explanation:

To find the different wavelengths you use the following formula:

\frac{1}{\lambda}=R_H(1-\frac{1}{n^2})

RH: Rydberg constant = 1.097 x 10^7 m^−1.

(a) n=2

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(2)^2})=8227500m^{-1}\\\\\lambda=1.2*10^{-7}m

(b)

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(3)^2})=9751111,1m^{-1}\\\\\lambda=1.0*10^{-7}m

(c)

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(4)^2})=10284375m^{-1}\\\\\lambda=9.7*10^{-8}m

(d) The three lines belong to the ultraviolet region.

8 0
2 years ago
the brightest , hottest, and most massive stars are the brilliant blue stars designated as spectral class O. if a class O star w
4vir4ik [10]

The speed is 0.956 m / s.

<u>Explanation</u>:

The kinetic energy is equal to the product of half of an object's mass, and the square of the velocity.

                   K.E = 1/2 \times m \times v^{2}

where K.E represents the kinetic energy,

           m represents the mass,

            v represents the velocity.

                  K.E = 1/2 \times m \times v^{2}

    1.10 \times 10^42 = 1/2 \times 3.26 \times 10^31 \times v^{2}

                     v^{2} = (1.10 \times 10^42 \times 2) / (3.26 \times 10^31)

                     v = 0.956 m / s.

6 0
2 years ago
In the equation vx^2=v0x^2+2ax(x-x0) what does the terms vx, v0x, x, and x0 stand for respectively?
tatuchka [14]

B. velocity at position x, velocity at position x=0, position x, and the original position

In the equation

v_{x}^{2} = v_{ox}^{2} +2 a x (x - x₀)

v_{x} = velocity at position "x"

v_{ox} = velocity at position "x = 0 "

x = final position

x_{o} = initial position of the object at the start of the motion

6 0
2 years ago
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