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Nookie1986 [14]
2 years ago
7

A baking dish is removed from a hot oven and placed on a cooling rack. As the dish cools down to 35 C from 175 C, its net radian

t power decreases to 12.0 W. What was the net radiant power of the baking dish when it was first removed from the oven? Assume that the temperature in the kitchen remains at 22 C as the dish cools down.

Physics
1 answer:
pogonyaev2 years ago
3 0

Answer:

net radiant power is 275.28 W

Explanation:

Given data

net radiant power P(F) = 12 W

initial temperature Ta = 175 °C = 175 + 273 = 448 K

final temperature Tb = 35 °C = 35 + 273 = 308 K

temperature in the kitchen T(f) = 22 °C = 22 + 273 = 295 K

to find out  

net radiant power

solution

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Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fish
Sunny_sXe [5.5K]

Answer:

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

Explanation:

This exercise can be analyzed with the law of refraction that establishes that a ray of light when passing from one medium to another with a different index makes it deviate from its path,

      n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the refractive indices of the incident and refracted means and the angles are also for these two means.

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

1 sin θ₁ = 1.33 sin θ₂

        θ₂ = sin⁻¹ ( 1/1.33 sin θ₁)

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

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2 years ago
Can a small child play with fat child on the seesaw?Explain how?
RoseWind [281]
Yes a small child can play with fat child in the seesaw because if the the load distance is decreased the effort will increase. That's means if the distance between the fat boy and the fulcrum is decreased the small child needs less effort.so,he can play
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Determine the centripetal force upon a 40-kg child who makes 10 revolution around the cliffhanger in 29.3 seconds.the radius of
zysi [14]

Answer:

The centripetal force acting on the child is 39400.56 N.

Explanation:

Given:

Mass of the child is, m=40\ kg

Radius of the barrel is, R=2.90\ m

Number of revolutions are, n =10

Time taken for 10 revolutions is, t=29.3\ s

Therefore, the time period of the child is given as:

T=\frac{n}{t}=\frac{10}{29.3}=0.341\ s

Now, angular velocity is related to time period as:

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.341}=18.43\ rad/s

Now, centripetal force acting on the child is given as:

F_{c}=m\omega^2 R\\F_{c}=40\times (18.43)^2\times 2.90\\F_{c}=40\times 339.66\times 2.90\\F_{c}=39400.56\ N

Therefore, the centripetal force acting on the child is 39400.56 N.

8 0
2 years ago
Two wires are stretched between two fixed supports and have the same length. One wire A there is a second-harmonic standing wave
lina2011 [118]

(a) Greater

The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, f_1:

f_n = n f_1

So we have:

- On wire A, the second-harmonic has frequency of f_2 = 660 Hz, so the fundamental frequency is:

f_1 = \frac{f_2}{2}=\frac{660 Hz}{2}=330 Hz

- On wire B, the third-harmonic has frequency of f_3 = 660 Hz, so the fundamental frequency is

f_1 = \frac{f_3}{3}=\frac{660 Hz}{3}=220 Hz

So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

(b) f_1 = \frac{v}{2L}

For standing waves on a string, the fundamental frequency is given by the formula:

f_1 = \frac{v}{2L}

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

v=2Lf_1

We know that the two wires have same length L. For wire A, f_1 = 330 Hz, while for wave B, f_B = 220 Hz, so we can write the ratio between the speeds of the waves in the two wires:

\frac{v_A}{v_B}=\frac{2L(330 Hz)}{2L(220 Hz)}=\frac{3}{2}

So, the waves travel faster on wire A.

7 0
2 years ago
A length of 20-gauge copper wire (of diameter 0.8118 mm) is formed into a circular loop with a radius of 30.0 cm. A magnetic fie
IrinaK [193]

Answer: 3 x 10^-24 watt

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r= 30 cm.

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r1= 0.8118/2 * 10^-3 m

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I=v/R

v=E,

I = 3.95 / 3.24 x 10^-8 = 1.22 x 10^8

P=I^2 x R.

= 3 x 10^-24 watt

7 0
2 years ago
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