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Bess [88]
2 years ago
11

While working out a man preformed 200 J of work in 3 seconds. What was his power?

Physics
1 answer:
AysviL [449]2 years ago
4 0

Answer:

C

Explanation:

A watt is a joule per second. Since he is working over 3 seconds you have to divide his work over three seconds which is 200/3. This gets you 66.7 watts.

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What was the average speed in km/h of a car that travels 400.0 km in 4.5 h?
Yanka [14]
V= s/t = 400/4.5=800/9 (km/h)
6 0
3 years ago
What does a velocity measurement include that a speed measurement does not? A: Time B:direction C:distance D: acceleration​
andrew-mc [135]

Answer:

B. Direction

Explanation:

Speed is a scalar quantity and doesn't keep track of direction : Velocity is a vector quantity and is direction aware.

8 0
2 years ago
(a) If two sound waves, one in a gas medium and one in a liquid medium, are equal in intensity, what is the ratio of the pressur
GarryVolchara [31]

Answer:

(a) The ratio of the pressure amplitude of the waves is 43.21

(b) The ratio of the intensities of the waves is 0.000535

Explanation:

Given;

density of gas, \rho _g = 2.27 kg/m³

density of liquid, \rho _l = 972 kg/m³

speed of sound in gas, C_g = 376 m/s

speed of sound in liquid, C_l = 1640 m/s

The of the sound wave is given by;

I = \frac{P_o^2}{2 \rho C} \\\\P_o^2 = 2 \rho C I\\\\p_o = \sqrt{2 \rho CI}

Where;

P_o is the pressure amplitude

P_o_g= \sqrt{2 \rho _g C_gI} -------(1)\\\\P_o_l= \sqrt{2 \rho _l C_lI}---------(2)\\\\\frac{P_o_l}{P_o_g} = \frac{\sqrt{2 \rho _l C_lI}}{\sqrt{2 \rho _g C_gI}} \\\\\frac{P_o_l}{P_o_g} = \sqrt{\frac{2 \rho _l C_lI}{2 \rho _g C_gI} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ \rho _l C_l}{ \rho _g C_g} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ (972)( 1640)}{ (2.27)( 376)} }\\\\\frac{P_o_l}{P_o_g} = 43.21

(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;

I = \frac{P_o^2}{2 \rho C}\\\\I_g = \frac{P_o^2}{2 \rho _g C_g}-------(1)\\\\I_l = \frac{P_o^2}{2 \rho _l C_l}-------(2)\\\\\frac{I_l}{I_g} = (\frac{P_o^2}{2 \rho _l C_l})*(\frac{2\rho_gC_g}{P_o^2} )\\\\\frac{I_l}{I_g} = \frac{\rho _gC_g}{\rho_lC_l} \\\\\frac{I_l}{I_g} = \frac{(2.27)(376)}{(972)(1640)}\\\\ \frac{I_l}{I_g} = 0.000535

3 0
3 years ago
(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is . Find the time i
a_sh-v [17]

Answer:

a)<em> It took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves. </em>

b) <em>The minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds. </em>

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensitie. That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies:

  • Gamma rays
  • X-rays
  • Ultraviolet rays
  • Visible region
  • Infrared
  • Microwave
  • Radio waves.

Any radiation that belongs to electromagnetic spectrum has a speed in vacuum of 3x10^{8}m/s.  

<em>a) Find the time it took for his voice to reach the Earth via radio waves.</em>

To know the time that took for Neil Armstrong's voice to reach the Earth via radio waves, the following equation can be used:

c = \frac{d}{t}  (1)

Where v is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

t = \frac{d}{c}  (2)

The distance from the Earth to the Moon is 3.85x10^{8} m, therefore.

t = \frac{3.85x10^{8} m}{3x10^{8}m/s}

t = 1.28s

Hence, it took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

<em>b) Determine the minimum time that will be required for a message from Mars to reach the Earth via radio waves.</em>

The distance from the Earth to the Mars at its closest approach is 5.76x10^{10}m, therefore.

t = \frac{5.76x10^{10}m}{3x10^{8}m/s}

t = 192s

Hence, the minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

3 0
3 years ago
how many kilograms off mercury would fill a 5litre container if the density of mercury is 13.6grams per cm3
Sidana [21]

Answer:

68kg

Explanation:

1 cm^3 is the same as 1 mL and there are 5000mL in 5L

Therefore if the density is 13.6g/mL we multiply 13.6 by 5000 to get the amount of grams required = 68000g which is 68kg

7 0
3 years ago
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