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2 years ago

What type of wave cannot travel in a vacuum

Physics

2 answers:

AlexFokin [52]2 years ago

7 0

The Answer is Sound Waves because in the vacuum of space, there is no medium to transmit of these mechanical waves.

lions [1.4K]2 years ago

5 0

A sound wave. Because in a vacuum there is no medium in a vacuum. And the only wave that requires a medium to travel through is a sound wave.

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Determine the magnitude and sense (direction) of the current in the 500-latex: \omega ω resistor when i = 30 ma.

VARVARA [1.3K]

Complete Question:

Check the circuit in the file attached to this solution

Answer:

Total current = 0.056 A(From left to right)

Explanation:

Let the current in loop 1 be I₁ and the current in loop 2 be I₂

Applying KVL to loop 1

30 - (I₁ - I₂)500 + I₂R + 15 = 0

45 - 500I₁ - 500I₂ + RI₂ = 0

I₁ = 30mA = 0.03 A

45 - 500(0.03) - 500I₂ + RI₂ = 0

30 -500I₂ + RI₂ = 0...............(1)

Applying kvl to loop 2

-RI₂ - 15 + 10 - 400I₁ = 0

-RI₂ = 5 + 400*0.03

RI₂ = -17 ................(2)

Put equation (2) into (1)

30 -500I₂ -17 = 0

-500I₂ = 13

I₂ = -13/500

I₂ = -0.026 A

The total current in the 500 ohms resistor = I₁ - I₂ = 0.03+0.026

Total current = 0.056 A

The current will flow from left to right

5 0

2 years ago

You toss a rock of mass m vertically upward. Air resistance can be neglected. The rock reaches a maximum height h above your han

VladimirAG [237]

Answer with Explanation:

We are given that

Mass of rock=m

Maximum height=h

a.At maximum height, velocity,v=0

We know that

$v^2=u^2-2gh$

$0+2gh=u^2$

$u^2=2gh$

Height,h=h/4

Again, $v'^2=u^2-2g\times \frac{h}{4}$

$v'^2=2gh-\frac{gh}{2}=\frac{4gh-gh}{2}=\frac{3gh}{2}$

$v'=\sqrt{\frac{3gh}{2}}=\sqrt{\frac{3\times 9.8 h}{2}}=3.83\sqrt h$

Where $g=9.8 m/s^2$

b.When height,h=3h/4

$v'^2=u^2-2gh$

$v'^2=2gh-2g\times \frac{3h}{4}=2gh-\frac{3gh}{2}=\frac{4gh-3gh}{2}=\frac{gh}{2}$

$v'=\sqrt{\frac{9.8h}{2}}=2.2\sqrt h$

$v'=2.2\sqrt h$

4 0

2 years ago

Read 2 more answers

A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f

sasho [114]

We use the kinematic equations,

$v=u+at$ (A)

$S= ut + \frac{1}{2} at^2$ (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given, $u=0$ , $a=0.21 \ m/s^2$ and $s= 280 m$ .

Substituting these values in equation (B), we get

$280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s$ .

Therefore from equation (A),

$v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s$

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s

8 0

2 years ago

where again p is the phonon momentum, E is the photon energy and c is the speed of light. When you divide the photon energy foun

creativ13 [48]

Answer:

Yes

Explanation:

p = momentum of photon

E = energy of photon

c = velocity of light

Units of p = kg m /s

Units of E = kg m^2 / s^2

Units of E / p = {kg m^2 / s^2} / {kg m /s} = m/s

It is the unit of speed, so by the division of energy to the momentum, we get the speed. yes it is correct.

8 0

2 years ago

The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from

Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

$L_T=L_S+10log(n)$

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

$L_T=60+10 log(5)$

= 66.98 db

7 0

1 year ago