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Umnica [9.8K]
3 years ago
12

you and a friend sit still on a spinning merry-go-round. The merry-go-round spins 5 times every second, and to a stationary obse

rver, you have a speed of 9m/s. From your frame of reference, what is the speed of your friend?
Physics
2 answers:
iris [78.8K]3 years ago
6 0

Answer:

the other person is right, the answer is zero.

Explanation:

dedylja [7]3 years ago
4 0

Answer:

  • <u> From your frame of reference, the speed of your friend is zero.</u>

Explanation:

The<em> speed</em> from a <em>frame of reference</em> is the magnitude of the change of the position per unit of time in that frame of reference.

When it is said that, to a <em>stationary observer</em> you have a <em>speed of 9m/s</em>, it means that the observer "measures"  tha your position, in a moment, changes 9 m per second.

Since both you and your friend are sit still on the same spinning merry-go-round, neither of you will change positions with respect to the other, meaning that you will perceive that your friend will not move, from your frame of reference. That is, if you are isolated from the rest of the universe, and the merry-go-round continues spining uniformly, you cannot detect any movement of your friend, which means that from your frame of reference, the speed of your friend is zero.

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A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls
nekit [7.7K]

Given:

density of air at inlet, \rho_{a} = 1.20 kg/m_{3}

density of air at inlet, \rho_{b} = 1.05 kg/m_{3}

Solution:

Now,

\dot{m} = \dot{m_{a}} = \dot{m_{b}}

\rho_{a} A v_{a} = \rho _{b} Av_{b}                        (1)

where

A = Area of cross section

v_{a} = velocity of air at inlet

v_{b} = velocity of air at outlet

Now, using eqn (1), we get:

\frac{v_{b}}{v_{a}} = \frac{\rho_{a}}{\rho_{b}}

\frac{v_{b}}{v_{a}} = \frac{1.20}{1.05} = 1.14

% increase in velocity = 1.14\times 100 =114%

which is 14% more

Therefore % increase in velocity is 14%

5 0
3 years ago
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Answer:

counter question if you get out the shower clean then how does your towel get dirty?

3 0
3 years ago
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The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 ​5 ​​ m/s then ho
Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10​⁵​​ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

F = \frac{MV}{t}

solving this two equations together;

\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9}  \ s \ = 1.639 \ ns

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0
3 years ago
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balandron [24]
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7 0
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Allisa [31]

Answer:

The SI unit of intensity is the watt per square meter/metre (W/m^2.)

Explanation:

Intensity is equal to the power transferred per unit area. Since power is measured in watts (W) and 1 W = 1 J/s, then intensity can be viewed as how fast energy goes through a certain area.

In physics, intensity is often used when studying light, sound, or other phenomena that involve waves or energy transfer. (With waves, the power value is taken as the average power transfer over the wave's period.)

8 0
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