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3 years ago

Which of the following statements concerning revolution is true?

Physics

2 answers:

Scorpion4ik [409]3 years ago

8 0

The Earth rotates around its axis, this rotating movement is called rotation.

The Earth also revolves around the sun, this movement it calls revolution.

The answer is: A. The earth revolves around the sun, and B. The earth revolves around its axis.

irakobra [83]3 years ago

4 0

Only 'A' is true. ..

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Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of

max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

$Q =\frac{KA(\delta T)}{L}$

$Q =\frac{25.87*1*20}{1}$

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0

3 years ago

Read 2 more answers

If 20 beats are produced within one second, which of the following frequencies could possibly be held by two sound waves traveli

NeTakaya

Answer:

D. 22 Hz and 42 Hz

Explanation:

When two waves with different frequency travelling in the same medium meet each other, they produce an interference pattern called beat.
The frequency of the beat produced is equivalent to the difference between the individual frequencies of the two waves involved.
Therefore; in this case since the frequency of the beat is 20 Hz, that is from 20 beats per second.
We need to find a pair from the choices whose frequency difference is 20 Hz.
This happens to be choice D. 22 Hz and 42 Hz, that is 42 Hz - 22 Hz = 20 Hz

8 0

3 years ago

Read 2 more answers

Hitungkan pecutan bagi blok di bawah: / Cal (a) m= 2 kg F= 8.0 N

ioda

Answer:

Acceleration = 4 m/s²

Explanation:

Given the following data;

Force = 8 N

Mass = 2 kg

To find the acceleration of the block;

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

$Acceleration = \frac {Net \; force}{mass}$

Substituting into the formula, we have;

$Acceleration = \frac {8}{2}$

Acceleration = 4 m/s²

4 0

3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$