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3 years ago

14

Students in a science class were shown three identical bulbs, each with a resistance of 50 Ω, connected to a 1.5-volt battery as

shown in the diagram. The students were told to assume the resistance of the wire is 0 Ω and make a claim as to which bulb has the most current. Which student has correctly justified a claim?

1 answer:

Rama09 [41]3 years ago

7 0

Answer:

It B

Explanation:

i just now its B because you can just read it and you will get the anwers

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What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?

serg [7]

Answer: $6.268(10)^{-16}J$

Explanation:

The kinetic energy of an electron $K_{e}$ is given by the following equation:

$K_{e}=\frac{(p_{e})^{2} }{2m_{e}}$ (1)

Where:

$K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}$

$p_{e}$ is the momentum of the electron

$m_{e}=9.11(10)^{-31}kg$ is the mass of the electron

From (1) we can find $p_{e}$ :

$p_{e}=\sqrt{2K_{e}m_{e}}$ (2)

$p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}$

$p_{e}=2.091(10)^{-24}\frac{kgm}{s}$ (3)

Now, in order to find the wavelength of the electron $\lambda_{e}$ with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

$\lambda_{e}=\frac{h}{p_{e}}$ (4)

Where:

$h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s}$ is the Planck constant

So, we will use the value of $p_{e}$ found in (3) for equation (4):

$\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}$

$\lambda_{e}=3.168(10)^{-10}m$ (5)

We are told the wavelength of the photon $\lambda_{p}$ is the same as the wavelength of the electron:

$\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m$ (6)

Therefore we will use this wavelength to find the energy of the photon $E_{p}$ using the following equation:

$E_{p}=\frac{hc}{lambda_{p}}$ (7)

Where $c=3(10)^{8}m/s$ is the spped of light in vacuum

$E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}$

Finally:

$E_{p}=6.268(10)^{-16}J$

4 0

3 years ago

After 24.0 days 2.00 milligrams of an original 128.0. Milligram sample remain what is the half life of the sample

alina1380 [7]

Answer:

4 days

either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2

n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2

Explanation:

5 0

3 years ago

Which statements about velocity are true?

OleMash [197]

<span>the speed of a direction</span>

5 0

3 years ago

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I really need help with this to be able to pass this last semester this is about (Circular Motion ) on physics

ankoles [38]

Question 1
To find centripetal acceleration, use the formula : centripetal acceleration = v^2/r
so answer would be (3.71)^2/42.85=0.32 (2d.p.)
Question 2
Force =ma
a= (9.98)^2/31.77=3.1350
Force= 3.1350 * 56.63 = 177.54 (2 d.p.)

4 0

3 years ago

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A body is projected upward at an angle of 30 degree to the horizontal at an initial speed of 200ms-.In how many seconds will it

Crazy boy [7]

Answer:

20.41 s

3534.80 m

Explanation:

<em><u>In how many seconds will it reach the ground?</u></em>

We are given the initial velocity of the body, which is 200 m/s at a 30° angle.

We know the acceleration in the vertical direction is -9.8 m/s², assuming that the upwards/right direction is positive and the downwards/left direction is negative.

Since we are using acceleration in the y-direction, let's use the vertical component of the initial velocity.

200 · sin(30) m/s

Let's use the fact that at the top of its trajectory, the body will have a final velocity of 0 m/s.

Now we have one missing variable that we are trying to solve for: time t.

Find the constant acceleration equation that contains v₀, v, a, and t.

v = v₀ + at

Substitute known values into the equation.

0 = 200 · sin(30) + (-9.8)t
-200 · sin(30) = -9.8t
t = 10.20408163

Recall that this is only half of the body's trajectory, so we need to double the time value we found to find the total time the body is in the air.

2t = 20.40816327

The body will reach the ground in 20.41 seconds.

<em><u>How far from the point of projection would it strike? </u></em>

We want to find the displacement in the x-direction for the body.

Let's find the constant acceleration equation that contains time t, that we just found, and displacement (Δx).

Δx = v₀t + 1/2at²

Substitute known values into the equation. Remember that we want to use the horizontal component of the initial velocity and that the acceleration in the x-direction is 0 m/s².

Δx = (200 · cos(30) · 20.40816327) + 1/2(0)(20.40816327)²
Δx = 3534.797567

The body will strike 3534.80 m from the point of projection.

4 0

2 years ago