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3 years ago

If 12V battery maintains a 4.5A current through a resistor what is the resistance of the resistor?

Physics

1 answer:

Dimas [21]3 years ago

5 0

Answer:

2.7ohms

Explanation:

Given parameters:

Voltage of the battery = 12V

Current = 4.5A

Unknown:

Resistance of the resistor = ?

Solution:

From Ohm's law, we know that;

V = IR

V is the voltage

I is the current

R is the resistance

So;

R = $\frac{V}{I}$ = $\frac{12}{4.5}$ = 2.7ohms

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ASAP

pochemuha

Answer:

Explanation:

The fish falls from vertical rest in a time of

t = √(2h/g) = √(2(2.27)/9.81) = 0.68 s

v = d/t = 5.6 / 0.68 = 8.2 m/s

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3 years ago

What is this Regeneration

Mila [183]

Answer:

Regeneration means that an organism regrows a lost part, so that the original function is restored.

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3 years ago

Read 2 more answers

You have an incident ray from a medium with a n1=1, through a medium with a n2 =3.325. If the incident angle is equal to 0.478 r

sesenic [268]

Answer:

0.139 rad

Explanation:

We use Snell's law $n_1sin\theta_1=n_2sin\theta_2$ , where if $n_1$ is the refractive index of the medium containing the incident ray, $\theta_1$ would be the incident angle, and if $n_2$ is the refractive index of the medium containing the refracted ray, $\theta_2$ would be the refraction angle, which we want, so we do:

$sin\theta_2=\frac{n_1}{n_2}sin\theta_1$

And finally:

$\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)$

We then insert our values:

$\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)=Arcsin(\frac{1}{3.325}sin(0.478rad))=arcsin(0.13834714686
)=0.139 rad$

6 0

3 years ago

Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0

3 years ago

Angular velocity in the z direction of a flywheel is w(t)=A + Bt2 The numerical values of the constants are A=2.75 and B=1.50. W

Ivanshal [37]

Answer:

α(0) = 0 rad/s²

α(5) = 15 rad/s²

Explanation:

The angular velocity of the flywheel is given as follows:

w(t) = A + B t²

where, A and B are constants.

Now, for the angular acceleration, we must take derivative of angular velocity with respect to time:

Angular Acceleration = α (t) = dw/dt

α(t) = (d/dt)(A + B t²)

α(t) = 2 B t

where,

B = 1.5

AT t = 0 s

α(0) = 2(1.5)(0)

α(0) = 0 rad/s²

AT t = 5 s

α(5) = 2(1.5)(5)

α(5) = 15 rad/s²

6 0

3 years ago