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2 years ago

15

The speed of a car is 40m/s after 10 sec suddenly there was a crowd and the driver reduces its speed to 20m/s. What is the decel

eration of the car?

1 answer:

aliya0001 [1]2 years ago

4 0

Explanation:

the number 0.4 in P4 from east

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The height of the Empire State Building is 318.00 meters. If a stone is dropped from the top of the building, what is the stone'

Mars2501 [29]

I belive its like 1200 mile per hour ive done he math for it

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3 years ago

How to find i1, i2,i3

MrRissso [65]

to find i1, i2, and i3 we need to find the total current.

to find the total current, you need to find the total resistance

you're already given the total voltage, Vs

to find Rtotal, start from the resistors furthest from the voltage source.

R3 and R4 are in series so

Rtotal= R3+R4 = 6+3 = 9 ohms

9 ohms is now in parallel with R2 so,

Rtotal= $(\frac{1}{R3+R4}) ^{-1}\\$ + (\frac{1}{R2}) ^{-1})^-1= (1/18)^-1 +( 1/9)^-1 = 6 ohms

6 ohms is in series with R1 so

Rtotal= 4+6=10 ohms

itotal= (\frac{Vtotal}{Rtotal})

= 120 v/10 ohms = 12 A

i total = i1 because all the current flows through it

i1= 12A

so the current splits into i2 and i3 and the amount of current that flows through a branch depends on the total resistance in each branch.

we already calculated the resistance in the R3+R4 & R2 branch as 6 ohms

since r3 and r4 are in series, the same current will flow through them

r3+r4 = 9 ohms

r2= 18 ohms

so the current in r2 will be half that of R3 & R4 (V=IR)

using the current divider rule

$Ix = Itotal * \frac{Rtotal}{Rx}$

i2= 12A x (6 ohms/18 ohms)= 4 A

i3= 12A x (6 ohms/9 ohms) = 8 ohms

6 0

3 years ago

In the Bohr model of the atom, an electron in an orbit has a fixed ____.

worty [1.4K]

The answer is C. an electron in an orbit has a fixed energy.

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3 years ago

If the accepted value of a wave is 121 m/s, who has the most accurate method of measuring the speed of a wave?

qaws [65]

The answer would be erin out of all of them thank me later :)

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3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

3 years ago