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lesantik [10]
2 years ago
14

PLEASE HELP ME I DONT UNDERSTAND THIS AND THIS IS TIMED

Physics
1 answer:
mixer [17]2 years ago
5 0

(C)

Explanation:

The circle has a radius r = 0.5 m, which means that its circumference C is

C = 2\pi r = 2\pi(0.5\:\text{m}) = 3.14\:\text{m}

One revolution means that the stopper travels a distance equal to the circumference of the circle so the velocity of the stopper is

v = \dfrac{C}{t} =\dfrac{3.14\:\text{m}}{0.2\:\text{s}} = 15.7\:\text{m/s} \approx 16\:\text{m/s}

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A cyclist travels 20km in 4hrs.What speed did the cyclist cycle at?​
Tanya [424]

Answer: 5 km/hr

Explanation:

speed= distance divided by time

20/4

= 5 km/hr

3 0
3 years ago
Read 2 more answers
Las carreras de velocidad pura en el atletismo son:
zheka24 [161]
La respuesta es la letra b
8 0
3 years ago
If the ac peak voltage across a 100-ohm resistor is 120 V, then the average power dissipated by the resistor is ________
Kazeer [188]

Answer:

The average power dissipated is 72 W.

Explanation:

Given;

peak voltage of the AC circuit, V₀ = 120 V

resistance of the resistor, R = 100 -ohm

The average power dissipated by the resistor is given by;

P_{avg} = \frac{1}{2} I_oV_o= I_{rms}V_{rms} = \frac{V_{rms}^2}{R}

where;

V_{rms} is the root-mean-square-voltage

V_{rms} = \frac{V_o}{\sqrt{2}} \\\\V_{rms} = \frac{120}{\sqrt{2}}\\\\V_{rms} = 84.853 \ V

The average power dissipated by the resistor is calculated as;

P_{avg} = \frac{V_{rms}^2}{R}\\\\P_{avg} = \frac{84.853^2}{100}\\\\P_{avg} = 72 \ W

Therefore, the average power dissipated is 72 W.

5 0
3 years ago
gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm an
Nastasia [14]

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

<u>Explanation:</u>

H₂ + Br₂ ⇒ 2HBr

PH₂ = 0.782atm

PBr₂ = 0.493atm

Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

At equilibrium:

Let 2x = pressure of HBr

PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

PHBr = 2x = 2.4 X 10⁻¹¹ atm

Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

3 0
3 years ago
?
Karo-lina-s [1.5K]

Answer:

Check Explanation.

The two statements given aren't true.

Explanation:

Although the question seems incomplete, I will address the concept of energy transfer during a wave's propagation.

The particles involved in wave's propagation move back and forth perpendicularly to the way the wave is moving, but do not move (at least, no significant movement is noticeable) in the direction of the wave. The particles ‘participate’ in the wave propagation by bumping into one another and transferring energy. This is exactly why energy can be transferred, although the average position of the particles doesn’t change.

So, the particles of the medium do not absorb energy from the atmosphere and do not significantly move from one location to another.

Hope this Helps!!!

4 0
4 years ago
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