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3 years ago

10

A 24-V battery is powering a light bulb with a resistance of 3.0 ohms. What is the current flowing through the bulb? A) 7.20 A B

) 8.0 A C) 12.0 A D) 72.0 A

1 answer:

Usimov [2.4K]3 years ago

3 0

According to Ohm's law for a portion of the circuit we have:

U=RI=>I=U/R=24/3=8 A

The correct answer is B

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Is this right? plzz anwser soon

-Dominant- [34]

Answer:

yes

Explanation:

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3 years ago

Read 2 more answers

A wire along the z axis carries a current of 6.8 A in the z direction Find the magnitude and direction of the force exerted on a

Alexeev081 [22]

Answer:

Force is 14.93N along positive y axis.

Explanation:

We know that force 'F' on a current carrying conductor placed in a magnetic field of intensity B is given by

$\overrightarrow{F}=\overrightarrow{Il}\times \overrightarrow{B}$

where L is the length of the conductor

Applying values in the equation we have force F =

$\overrightarrow{F}=6.8\times 6.1\widehat{k}\times 0.36\widehat{i}\\\\\overrightarrow{F}=41.48\widehat{k}\times 0.36\widehat{i}\\\\\therefore \overrightarrow{F}=14.93N\widehat{j}$

Thus force is 14.93N along positive y axis.

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3 years ago

Two forces that are not equal in size are

Bumek [7]

They are unbalanced forces ..... Hope this helps :3

4 0

3 years ago

Vector vector b has x, y, and z components of 4.00, 4.00, and 2.00 units, respectively. calculate the magnitude of vector

Sav [38]

Good morning.

We see that $\mathsf{\overset{\to}{b}} = \mathsf{(4.00, \ 4.00, \ 2.00)}$

The magnitude(norm, to be precise) can be calculated the following way:

$\star \ \boxed{\mathsf{\overset{\to}{a}=(x, y,z)\Rightarrow ||\overset{\to}{a}|| = \sqrt{x^2+y^2+z^2}}}$

Now the calculus is trivial:

$\mathsf{\|\overset{\to}{b} \| =\sqrt{4^2+4^2+2^2} =\sqrt{16+16+4}}\\ \\ \mathsf{\|\overset{\to}{b}\|=\sqrt{36}}\\ \\ \boxed{\mathsf{\|\overset{\to}{b}\| = 6.00 \ u}}$

7 0

3 years ago

Ml(d^2θ/dt^2) =-mgθ

Nata [24]

The equation of motion of a pendulum is:

$\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,$

where $\ell$ it its length and $g$ is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for <em>small</em> angles ( $\theta \ll 1$ ), we can use:

$\sin\theta \simeq \theta.$

Additionally, let us define:

$\omega^2\equiv\dfrac{g}{\ell}.$

We can now write:

$\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.$

The solution to this differential equation is:

$\theta(t) = A\sin(\omega t + \phi),$

where $A$ and $\phi$ are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

$T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.$

This justifies that the period depends only on the pendulum's length.

4 0

3 years ago