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harina [27]
3 years ago
6

A radio changes what energy to what energy

Physics
1 answer:
koban [17]3 years ago
3 0
Electrical energy to sound energy
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Two pans of a balance are 24.1 cm apart. The fulcrum of the balance has been shifted 1.33 cm away from the center by a dishonest
Citrus2011 [14]

As we know that fulcrum is shifted to one side by 1.33 cm

so here let say left side mass is m1 while on other side mass is m2

so we will have

m_1g (24.1 - 1.33) = m_2 g(24.1 + 1.33)

m_2 = 0.895 m_1

now in order to find the percentage we have

percentage = \frac{m_1 - m_2}{m_1}\times 100

percentage = (1 - 0.895)\times 100

percentage = 10.5%

so it is 10.5 % less mass

8 0
3 years ago
For the circuit shown, which of the following is true?
liraira [26]
You can't tell the relative values of I-1 and I-2 from the information given,
but that's not for any super-technical reason. 

The reason is because the drawing doesn't show WHERE in the circuit
I-1 and I-2 are measured. 

It makes a BIG difference if they're the currents in the resistors, or
the currents in and out of the battery.

-- The currents in and out of the battery are equal.

-- The current through either resistor is less than
the current in or out of the battery.

-- The sum of the currents through both resistors
is equal to the current in or out of the battery.

-- The currents through the two resistors may be equal
or unequal, and either one may be greater or less than
the other.  It all depends on the values of the resistors.
6 0
3 years ago
Read 2 more answers
Marx argued that what happens when a worker is separated
kicyunya [14]

Answer:

well its simple the worker wouldn't actually be working

3 0
3 years ago
A particular 12 V car battery can send a total charge of 110 A·h (ampere-hours) through a circuit, from one terminal to the othe
DiKsa [7]
<h2>Answer:</h2>

(a) 3.96 x 10⁵C

(b) 4.752 x 10⁶ J

<h2>Explanation:</h2>

(a) The given charge (Q) is 110 A·h (ampere hour)

Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;

=> Q = 110A·h

=> Q = 110 x 1A x 1h          [1 hour = 3600 seconds]

=> Q = 110 x A x 3600s

=> Q = 396000A·s

=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C

Therefore, the number of coulombs of charge is 3.96 x 10⁵C

(b) The energy (E) involved in the process is given by;

E = Q x V           -----------------(i)

Where;

Q = magnitude of the charge = 3.96 x 10⁵C

V = electric potential = 12V

Substitute these values into equation (i) as follows;

E = 3.96 x 10⁵ x 12

E = 47.52 x 10⁵ J

E = 4.752 x 10⁶ J

Therefore, the amount of energy involved is 4.752 x 10⁶ J

8 0
3 years ago
A 0.500-kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long. The stone is moving around the
Semenov [28]

Answer:

B. 7.07 m/s

Explanation:

The velocity of the stone when it leaves the circular path is its tangential velocity, v, which is given by

v=\omega r

where \omega is the angular speed and r is the radius of the circular path.

\omega is given by

\omega = 2\pi f

where f is the frequency of revolution.

Thus

v=2\pi fr

Using values from the question,

v=2\pi\times 1.50\times0.75

<em>Note the conversion of 75 cm to 0.75 m</em>

v=2\times3.14\times 1.50\times0.75 = 9.42\times0.75 = 7.065=7.07

6 0
3 years ago
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