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Jlenok [28]
3 years ago
13

5. For the speaker in this circuit, the voltage across it is always proportional to the current through it. Find the maximum amo

unt of power that the circuit can deliver to the speaker.
Physics
1 answer:
rosijanka [135]3 years ago
4 0

Answer:

speaker64

--------

34x

Explanation:

64-34

x

speaker

4

2

4

788

- circuit

voltage

100000

x.34

Sorry but you have no picture shown

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Let's say you have two tuning forks which are supposed to produce the same frequency, 512 Hz. One is of good quality, but the ot
solong [7]

Answer:

= 2 beats per seconds

Explanation:

  • From |f -f'| = modulus of the difference between the frequency given.
  • f = 510Hz and f' = 512Hz
  • Difference between the frequency will give us the number of beat per seconds.
  • i.e 2 beats per seconds

These also shows how to get the period of the tuning forks.

7 0
3 years ago
Suppose an event is measured to be at a = (0,-2, 3, 5) in one reference frame. Find the components of this event in another refe
LenaWriter [7]

Answer:

The components of the moving frame is (8.07c, -2, 3, 9.493)

Solution:

As per the question:

Velocity of moving frame w.r.t original frame v_{m} 0.85c

Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane

a = (0, - 2, 3, 5)

Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):

New coordinates are given by:

X = \frac{x - v_{m}t}{\sqrt{1 - \frac{v_{m}^{2}}{c^{2}}}}

X = \frac{0 - 0.85c\times 5}{\sqrt{1 - \frac{(0.85c)^{2}}{c^{2}}}}

X = 8.07 c

Now,

Y = y = - 2

Z = z = 3

Now,

t' = \frac{t - \frac{vx}{c}^{2}}{\sqrt{1 - (\frac{v}{c})^{2}}}

t' = \frac{5 - 0}{\sqrt{1 - (\frac{0.85c}{c})^{2}}} = 9.493 s

4 0
4 years ago
Plz do all plz i will give brainest and thanks to best answer do it right
lara31 [8.8K]

Answer:

I'm not really sure but I think it's choice a

5 0
3 years ago
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PLEASE HELLLLLLLLLLLLPPPPPPPPPP!!!!!! I will give brainliest
alexandr1967 [171]

Answer:

its C. The north pole of one magnet attracts the south pole of another

Explanation:

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4 0
3 years ago
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A sonar pulse sent out by a boat arrives back after 4 seconds. If the speed of sound in water is 1600m/s, how deep is the water?
Varvara68 [4.7K]

Answer:

the boat would be deeped by 3200 m

Explanation:

Given that

The boat arrives back after 4 seconds

And, the speed of the sound in water is 1,600 m/s

We need to find out how much deep is the water

So,

As we know that

Distance = ( speed × time) ÷ 2

Here we divided by 2 because the boat arrives back

= (1600 × 4) ÷ 2

= 3200 m

Therefore the boat would be deeped by 3200 m

7 0
3 years ago
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