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3 years ago

13

metal cap is the negative terminal of an electric cell true or false please send me the answer please

1 answer:

USPshnik [31]3 years ago

7 0

Answer:

false. it's positive terminal of an electric cell

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Springfield's "classic rock" radio station broadcasts at a frequency of 102.1 mhz. what is the length of the radio wave in meter

Mila [183]

The frequency of the radio wave is:
$f=102.1 MHz = 102.1 \cdot 10^6 Hz$

The wavelength of an electromagnetic wave is related to its frequency by the relationship
$\lambda= \frac{c}{f}$
where c is the speed of light and f the frequency. Plugging numbers into the equation, we find
$\lambda= \frac{3 \cdot 10^8 m/s}{102.1 \cdot 10^6 Hz}= 2.94 m$
and this is the wavelength of the radio waves in the problem.

7 0

4 years ago

How much force must a locomotive exert on a 12840-kg boxcar to make it accelerate forward at 0.490 m/s2?

ddd [48]

Data given:

m=12840kg

a=0.490m/s²

Formula:

F=ma

Solution:

F= 12840kg×0.490m/s²

F=6291.6N

6 0

3 years ago

A source charge of 5.0 µC generates an electric field of 3.93 × 105 at the location of a test charge. How far is the test charge

kap26 [50]

Answer:

B. ) 0.34 m

I definitely guessed and got the right answer so :))

6 0

4 years ago

Read 2 more answers

The type of function that describes the amplitude of damped oscillatory motion is _______. The type of function that describes t

Salsk061 [2.6K]

Answer:

exponential

Explanation:

type of function that describes the amplitude of damped oscillatory motion is exponential because as we know that here function is

y = A × $e^{\frac{-bt}{2m}}$ × cos(ωt + ∅ ) ..................................... ( 1 )

here function A × $e^{\frac{-bt}{2m}}$ is amplitude

as per equation ( 1 )it is exponential

so that we can say that amplitude of damped oscillatory motion is exponential

8 0

3 years ago

Two long, parallel wires are attracted to each other by a force per unit length of 350 µN/m. One wire carries a current of 22.5

pishuonlain [190]

Answer

given,

force per unit length = 350 µN/m

current, I = 22.5 A

y = y = 0.420 m

$\dfrac{F}{L}= \dfrac{KI_1I_2}{d}$

$I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}$

$I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}$

I₂ = 32.67 A

distance where the magnetic field is zero

$\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}$

$y_1 = 0.248\ m$

there the distance at which the magnetic field is zero in the two wire is at 0.248 m.

3 0

3 years ago