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2 years ago

Which planet has the most extreme temperature difference between day and night?

Physics

1 answer:

Softa [21]2 years ago

4 0

Answer:

Explanation:

Mercury, the first planet from the Sun, endures drastic temperature changes from day to night. During the day, the planet is incredibly near to the Sun, with temperatures reaching 430°C.

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An open pipe of length 0.58 m vibrates in the third harmonic with a frequency of 939Hz. What is the speed of sound through the a

vaieri [72.5K]

363 m/s is the speed of sound through the air in the pipe.

Answer: Option B

<u>Explanation:</u>

The formula used to calculate the wavelength given as below,

$Wavelength (\lambda)=\frac{\text { wave velocity }(v)}{\text { frequency }(f)}$

$v=\lambda \times f$ --------> eq. 1

In power system, harmonics define by positive integers of the fundamental frequency. So the third order harmonic is a multiple of the third fundamental frequency. Each harmonic creates an additional node and an opposite node, as well as an additional half wave within the string.

If the number of waves in the circuit is known, the comparison between standing wavelength and circuit length can be calculated algebraically. The general expression for this given as,

$L=\frac{n \lambda}{2}$

For first harmonic, n =1

$L=\frac{\lambda}{2}$

For second harmonic, n =2

$L=\frac{2 \lambda}{2}=\lambda$

For third harmonic, n =3

$L=\frac{3 \lambda}{2}$

$\lambda=\frac{2 L}{3}$ -------> eq. 2

Here given f = 939 Hz, L = 0.58 m...And, substitute eq 2 in eq 1 and values, we get

$v=\frac{2 \times 0.58 \times 989}{3}=\frac{1089.24}{3}=363.08 \mathrm{m} / \mathrm{s}$

3 0

3 years ago

When a positively charged conductor touches a neutral conductor, the neutral conductor will:

frutty [35]

Answer:

Lose electrons

Explanation:

When a positively charged conductor touches a neutral conductor, the neutral conductor will lose electrons. Only electrons can move from one conductor to another, so if the neutral conductor ended up with a positive charge it means it lost electrons. The conductor touching and the neutral conductor both end up being charged positively.

6 0

4 years ago

A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg.

rodikova [14]

Answer:

the moment of inertia of the merry go round is 38.04 kg.m²

Explanation:

We are given;

Initial angular velocity; ω_1 = 37 rpm

Final angular velocity; ω_2 = 19 rpm

mass of child; m = 15.5 kg

distance from the centre; r = 1.55 m

Now, let the moment of inertia of the merry go round be I.

Using the principle of conservation of angular momentum, we have;

I_1 = I_2

Thus,

Iω_1 = I'ω_2

where I' is the moment of inertia of the merry go round and child which is given as I' = mr²

Thus,

I x 37 = ( I + mr²)19

37I = ( I + (15.5 x 1.55²))19

37I = 19I + 684.7125

37I - 19 I = 684.7125

18I = 684.7125

I = 684.7125/18

I = 38.04 kg.m²

Thus, the moment of inertia of the merry go round is 38.04 kg.m²

7 0

4 years ago

Read 2 more answers

The initial kinetic energy imparted to a 0.020 kg bullet is 1200 J. (a) Assuming it

Lilit [14]

Answer:

(a) Power= 207.97 kW

(b) Range= 5768.6 meter

Explanation:

Given,

Mass of bullet, $m=0.02 kg$

Kinetic energy imparted, $K=1200 J$

Length of rifle barrel, $d=1 m$

(a)

Let the speed of bullet when it leaves the barrel is $v$ .

Kinetic energy, $K=\frac{1}{2} mv^{2}$

$v=\sqrt{\frac{2K}{m} }$

$=\sqrt{\frac{2\times1200}{0.02} }$

$=346.4m/s$

Initial speed of bullet, $u=0$

The average speed in the barrel, $v_a_v_g=\frac{u+v}{2}$

$=\frac{0+346.4}{2} \\=173.2 m/s$

Time taken by bullet to cross the barrel, $t=\frac{d}{v}$

$=\frac{1}{173.2}\\ =0.00577 second$

Power, $P_a_v_g=\frac{W}{t}$

$=\frac{1200}{0.00577} \\=207.97kW$

(b)

In projectile motion,

Maximum height, $H_m=\frac{v^2\sin^2\theta}{2g} \\$

Range, $R=\frac{v^2\sin2\theta}{g}$

given that, $H_m=R$

then, $\frac{v^2\sin^2\theta}{2g}=\frac{v^2\sin2\theta}{g}\\\sin^2\theta=2\sin\theta\cos\theta\\\\\tan\theta=4\\\theta=\tan^-^14\\\theta=75.96^0\\R=\frac{v^2\sin2\theta}{g}\\=\frac{346.4^2\times\sin(2\times75.96)}{9.8}\\5768.6 meter$

5 0

3 years ago

Which statement accurately compares the heating and cooling rates for water and land?

AVprozaik [17]

Answer:

C is the answer Hope I helped

6 0

3 years ago

Read 2 more answers