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2 years ago

A rifle is aimed horizontally at a target 47 m away. The bullet hits the target 2.3 cm below the aim point.

Physics

1 answer:

inna [77]2 years ago

3 0

Answer:

Is your question asking for the muzzle velocity of the bullet?

Explanation:

I will assume it does

The bullet travels horizontally to the target in the same amount of time it falls 2.3 cm from vertical rest

s = ½at²

t = √(2s/g) = √(2(0.023) / 9.8) = 0.0685118...s

v = d/t = 47/0.0685118 = 686.01242...

v = 690 m/s

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In an electricity demonstration at the Deutsches Museum in Munich, Germany, a person sits inside a metal sphere of radius 0.90 m

vladimir2022 [97]

Answer:

E_interior = 0

Explanation:

As the sphere is metallic, the electrical charges are distributed on its surface, as far away as possible from each other.

If we apply Gauss's law, as the charge is on the surface, when drawing a spherical Gaussian surface, we see that there is no charge inside, therefore there is no electric field inside the metallic sphere.

E_interior = 0

6 0

3 years ago

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24

Oksi-84 [34.3K]

Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}$
Now At xp we have:
$\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0$
$\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0$
Which is a second order equation, using the quadratic formula to solve for xp would give us:
$xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
or
$xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

6 0

3 years ago

In a velocity selector having electric field E and magnetic field B, the velocity selected for positively charged particles is v

maw [93]

Answer:

True or False

Explanation:

Because.....

easy 50% chance you are right

6 0

3 years ago

Help me with this please!

alexgriva [62]

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3 0

3 years ago

A series circuit consists of a 100-ω resistor, a 10.0-μf capacitor, and a 0.350-h inductor. the circuit is connected to a 120-v

Tpy6a [65]

Current will be $I=\dfrac{V_{rms}}{Z}=\dfrac{V_{rms}}{\sqrt{ R^{2}+(X_{C}-X_{L})^{2}}}\\where~X_{c}=\dfrac{1}{j.\omega .C}~and~X_{L}=j.\omega.L~where~\omega=2.\pi f~and~f=60Hz$
now just pluf in the values and Voila..

7 0

3 years ago