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Allisa [31]
3 years ago
11

What temperature increase is necessary to increase the power radiated from an object by a factor of 8?

Physics
1 answer:
Aleks04 [339]3 years ago
7 0

To solve this problem it is necessary to apply the concepts related to the Power defined from the Stefan-Boltzmann equations.

The power can be determined as:

P = \sigma T^4

Making the relationship for two states we have to

\frac{P_1}{P_2} = \frac{T_1^4}{T_2^4}

Since the final power is 8 times the initial power then

P_2 = 8P_1

Substituting,

\frac{1}{8} = \frac{T_1^4}{T_2^4}

T_2 = T_1 8*(\frac{1}{4})

T_2 = 1,68T_1

The temperature increase would then be subject to

\Delta T = T_2-T_1

\Delta T = 1.68T_1 -T_1

\Delta T = 0.68T_1

The correct option is D, about 68%

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A car travels a distance of 540km in 6hours.What speed did he travel at?​
Sloan [31]

Answer: 90 km/hr

Explanation:

Speed= distance divided by time

540/6

= 90km/hr

5 0
3 years ago
2. A common physics experiment involves lowering an open tube into a cylinder of water and moving the tube up and down to adjust
notsponge [240]

Answer:

Explanation:

This question pertains to resonance in air column.  It is the case of closed air column in which fundamental note is formed at a length which is as follows

l = λ / 4 where l is length of tube and λ is wave length.

here l = .26 m

λ = .26 x 4 = 1.04 m

frequency of sound = 330 Hz

velocity of sound = frequency x wave length

= 330 x 1.04

= 343.2 m /s

b )

Next overtone will be produced at 3 times the length

so next length of air column = 3 x 26

= 78 cm

c )

If frequency of sound = 256 Hz

wavelength = velocity / frequency

= 343.2 / 256

= 1.34 m

= 134 cm

length of air column for resonance

= wavelength / 4

134/4

= 33.5 cm

7 0
3 years ago
3 resistors are connected in parallel, each with resistance between 1 and 10 Ω. What is NOT a possible value for the overall equ
Leno4ka [110]

Answer:

Therefore,

1/3 ≤ Re ≤ 10/3 ohms

The equivalent resistance CANNOT be any value outside the boundary above

That is

Re CANNOT be greater than 10/3 ohms and CANNOT be less than 1/3 ohms, given that R1,R2,R3 are all between 1 and 10ohms.

Explanation:

For a parallel resistor arrangement.

The equivalent resistance (Re) of the three resistors is given by;

1/Re = 1/R1 + 1/R2 + 1/R3

1/Re = (R2R3 + R1R3 + R1R2)/R1R2R3

Re = R1R2R3/(R2R3 + R1R3 + R1R2)

Therefore, if R1,R2,R3 are between 1-10ohms

We need to calculate the range of values of Re.

Taking the lower bound 1

R1= R2=R3= 1ohms

Re = 1/(1+1+1)

Re = 1/3 ohms

Taking the upper bound 10 ohms

R1= R2=R3= 10 ohms

Re = 1000/(100+100+100)

Re = 1000/300

Re = 10/3 ohms

Therefore,

1/3 ≤ Re ≤ 10/3 ohms

The equivalent resistance CANNOT be any value outside the boundary above

That is

Re cannot be greater than 10/3 ohms and cannot be less than 1/3 ohms, given that R1,R2,R3 are all between 1 and 10ohms.

7 0
4 years ago
A string of 18 identical holiday tree lights is connected inseries to a 190V source. The string dissipates 62.0 W. One of thebul
PtichkaEL [24]

Answer:

549.9 ohms, 65.65 watts, the power went up

Explanation:  power P = V²/R

v= 190volts , P =62watts

from P =V²/R

62=190²/R

making R the subject

R×62= 190²

62R =36100

R=36100/62

R=586.25 ohms

Resistance of each lamp = 586.26/18 =32.3ohms

a) resistance of the light string now = 17×32.3 = 549.9 ohms

b) P=V²/R

where R=549.9ohms , P=?

P=190²/549.9

P=36100/549.9

P=65.65watts

Power dissipated has increased( went up )

7 0
3 years ago
It has been suggested that rotating cylinders about 17.0 mi long and 4.99 mi in diameter be placed in space and used as colonies
Sladkaya [172]

Answer:

\omega = 49.86*10^{-3}rad/s

Explanation:

We start converting to SI units,

A mile = 1609m

L=17mi=27000m

D=4.99mi=7884m

We know that the expression, which can relate linear acceleration and angular velocity is given by,

a_c = r\omega^2

Where \omega is the angular velocity

r=radius

a_c = linear acceleration,

Re-arrange for \omega,

\omega = \sqrt{\frac{a_c}{r}}

Our acceleration is equal to the gravity force, so replacing,

\omega = \sqrt{\frac{9.8}{(7884/2)}}

\omega = 49.86*10^{-3}rad/s

8 0
3 years ago
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