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3 years ago

A 0.22-caliber handgun fires a 29-g bullet at a velocity of 770. m/s. calculate the de broglie wavelength of the bullet.

Physics

1 answer:

True [87]3 years ago

3 0

solution:

$\lambda =\frac{h}{mv}\\
h is planck's constant=6.626\times10^-34j.sec\\
m=mass=29g\\
=28\times10^-3kg\\
v=770m/sec\\
\lambda=\frac{6.626\times10^-34j.sec}{29\times10^-3\times770}\\$

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3 years ago

Mutations can be transmitted to the next generation only if they are present in

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The germline (the gametes)

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3 years ago

Metal sphere 1 has a positive charge of 7.00 nc . metal sphere 2, which is twice the diameter of sphere 1, is initially uncharge

MariettaO [177]

Answer:

2.33 nC, 4.67 nC

Explanation:

when the two spheres are connected through the wire, the total charge (Q=7.00 nC) re-distribute to the two sphere in such a way that the two spheres are at same potential:

$V_1 = V_2$ (1)

Keeping in mind the relationship between charge, voltage and capacitance:

$C=\frac{Q}{V}$

we can re-write (1) as

$\frac{Q_1}{C_1}=\frac{Q_2}{C_2}$ (2)

where:

Q1, Q2 are the charges on the two spheres

C1, C2 are the capacitances of the two spheres

The capacitance of a sphere is given by

$C=4 \pi \epsilon_0 R$

where R is the radius of the sphere. Substituting this into (2), we find

$\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 R_2}$ (3)

we also know that sphere 2 has twice the diameter of sphere 1, so the radius of sphere 2 is twice the radius of sphere 1:

$R_2 = 2R_1$

So the eq.(3) becomes

$\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 2R_1}$

And re-arranging it we find:

$Q_2 = 2Q_1$

And since we know that the total charge is

$Q_1 + Q_2 = 7.00 nC$

we find

$Q_1 = 2.33 nC\\Q_2 = 4.67 nC$

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4 years ago

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lesantik [10]

I’d assume a tape measure?

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4 years ago

Read 2 more answers

Before a collision, a 50.0-kg object is moving at +5.0 m/s. Find the impulse that acted on the object if, after the collision, i

Virty [35]

Answer:

Explanation:

<u>Impulse And Momentum </u>

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