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blsea [12.9K]
3 years ago
9

HOW DO WE HELP OTHERS NOT BE SAD IF THEY DON'T LIKE HUGS?

Physics
2 answers:
svetlana [45]3 years ago
6 0
You need to hug them how much they need to like.
Rzqust [24]3 years ago
3 0

Explanation:

hug is not the only solution to help someone you can take them to Peaceful place and sit with them saying nothing.

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The average power dissipated in a 47 ω resistor is 2.0 w. what is the peak value i 0 of the ac current in the resistor?
pishuonlain [190]
The average dissipated power in a resistor in a ac circuit is:
P=I_{rms}^2 R
where R is the resistance, and I_{rms} is the root mean square current, defined as
I_{rms} =  \frac{I_0}{\sqrt{2}}
where I_0 is the peak value of the current. Substituting the second formula into the first one, we find
P=( \frac{I_0}{\sqrt{2} } )^2 R =  \frac{1}{2} I_0^2 R
and if we re-arrange this formula and use the data of the problem, we can find the value of the peak current I0:
I_0 =  \sqrt{ \frac{2 P}{R} }=  \sqrt{ \frac{2 \cdot 2.0 W}{47 \Omega} }=0.29 A
4 0
3 years ago
HOW FAR DOES A UNICYCLE TRAVEL AT A SPEED OF 20 M/S FOR 15 SECONDS?​
astra-53 [7]

Given:-

  • Speed of the unicycle = 20 m/s
  • Time taken = 15 s

To Find: Distance travelled by the unicycle.

We know,

s = vt

where,

  • s = Distance travelled,
  • v = Speed &
  • t = Time taken.

Therefore,

s = (20 m/s)(15 s)

→ s = (20 m)(15)

→ s = 300 m (Ans.)

8 0
3 years ago
Bob and John are pulling in different directions. If Bob is pulling to the right with a force of 10N, and John is pulling to the
Oksi-84 [34.3K]

-- The net force on the box is 2N to the left.

-- The box will move to the left and accelerate to the left.

-- F=ma . a=F/m . a=(2N)/(4kg).

a = 0.5 m/s^2 to the left.

5 0
3 years ago
A pendulum has 576 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o
Natalka [10]
Ideally, 576 J  because energy is conserved.
In the real world, a tiny tiny tiny tiny bit less than 576 J ,
because we live in a world with friction and air resistance.
3 0
3 years ago
A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when
Marianna [84]

Answer:

A) P=13.92\ J.s^{-1}

B) v=3730.9912\ m.s^{-1}

C) v=74.44\ mm.s^{-1}

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

  • Distance form the sound source, s=50\ m
  • sound intensity level at the given location, \beta=114\ dB
  • diameter of the eardrum membrane in humans, d=8.4 \times 10^{-3}\ m
  • We have the minimum detectable intensity to the human ears, I_0=10^{-12}\ W.m^{-2}

(A)

<u>Now the intensity of the sound at the given location is related mathematically as:</u>

\beta=10\ log(\frac{I}{I_0} ) ..........................................(1)

114=10\ log\ (\frac{I}{10^{-12}} )

11.4=log\ I+12\ log\ 10

I=0.2512\ W.m^{-2}

<em>As we know :</em>

I=\frac{P}{A}

0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }

P=13.92\ J.s^{-1} is the energy transferred to the  eardrums per second.

(B)

mass of mosquito, m=2\times 10^{-6}\ kg

<u>Now the velocity of mosquito for the same kinetic energy:</u>

KE=\frac{1}{2} m.v^2

13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2

v=3730.9912\ m.s^{-1}

(C)

Given:

  • Sound intensity, \beta = 20\ dB

<u>Using eq. (1)</u>

20=10\ log\ (\frac{I}{10^{-12}} )

2=log\ I+12\ log\ 10

I=10^{-10}\ W.m^{-2}

Now, power:

P=I.A

P=10^{-10}\times \pi\times \frac{8.4^2}{4}

P=5.54\times 10^{-9}\ J.s^{-1}

Hence:

KE=\frac{1}{2} m.v^2

5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2

v=0.07444\ m.s^{-1}

v=74.44\ mm.s^{-1}

(D)

mosquitoes speed in part B is very much larger than that of part C.

7 0
3 years ago
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