Which quantity is the vector? <br> a) weight<br> b) power <br> c) kinetiic energy <br> d) speed

Bogdan [553]

Because it’s hard snow that can kill you

7 0

3 years ago

if a diffraction grating produces a third-order bright spot for red light (of wavelength 650 nm ) at 68.0 ∘ from the central max

Mashutka [201]

Answer:

The angle is 25.34°.

Explanation:

Given that,

Wave length = 650 nm

Angle = 68.0°

We need to calculate the distance

For a diffraction grating

$d\sin\theta=m\lambda$

$d=\dfrac{2\times650\times10^{-9}}{\sin68.0}$

$d=2.10\times10^{-6}\ m$

We need to calculate the angle

Using formula for angle

$d\sin\theta=m\lambda$

$\sin\theta=\dfrac{m\times\lambda}{d}$

$\sin\theta=\dfrac{2\times450\times10^{-9}}{2.10\times10^{-6}}$

$\sin\theta=25.34^{\circ}$

Hence, The angle is 25.34°.

8 0

3 years ago

An electric motor rotating a workshop grinding wheel at a rate of 1.31 ✕ 102 rev/min is switched off. Assume the wheel has a con

antoniya [11.8K]

(a) 4.03 s

The initial angular velocity of the wheel is

$\omega_i = 1.31 \cdot 10^2 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=13.7 rad/s$

The angular acceleration of the wheel is

$\alpha = -3.40 rad/s^2$

negative since it is a deceleration.

The angular acceleration can be also written as

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 0$ is the final angular velocity (the wheel comes to a stop)

t is the time it takes for the wheel to stop

Solving for t, we find

$t=\frac{\omega_f - \omega_i }{\alpha}=\frac{0-13.7 rad/s}{-3.40 rad/s^2}=4.03 s$

(b) 27.6 rad

The angular displacement of the wheel in angular accelerated motion is given by

$\theta= \omega_i t + \frac{1}{2}\alpha t^2$

where we have

$\omega_i=13.7 rad/s$ is the initial angular velocity

$\alpha = -3.40 rad/s^2$ is the angular acceleration

t = 4.03 s is the total time of the motion

Substituting numbers, we find

$\theta= (13.7 rad/s)(4.03 s) + \frac{1}{2}(-3.40 rad/s^2)(4.03 s)^2=27.6 rad$

6 0

3 years ago

A sample contains 20 kg of radioactive material. The decay constant of the material is 0.179 per second. If the amount of time t

Nataly [62]

Answer:

Explanation:

Given

N0 = 20kg (original substance)

decay constant λ = 0.179/sec

time t = 300s

We are to find N(t)

Using the formula;

n(t) = N0e^-λt

Substitute the given values

N(t) = 20e^-(0.179)(300)

N(t) = 20e^(-53.7)

N(t) = 20(4.7885)

N(t) =143.055

To know how much of the original material that is active, we will find N(t)/N0 = 143.055/20 = 7.152

About 7 times the original material is still radioactive

4 0

3 years ago

The n = 2 to n = 6 transition in the bohr hydrogen atom corresponds to the ________ of a photon with a wavelength of ________ nm

Colt1911 [192]

The energy levels of the hydrogen atom are quantized and their energy is given by the approximated formula
$E=- \frac{13.6}{n^2} [eV]$
where n is the number of the level.

In the transition from n=2 to n=6, the variation of energy is
$\Delta E=E(n=6)-E(n=2)=-13.6 ( \frac{1}{6^2}- \frac{1}{2^2} )[eV]=3.02 eV$
Since this variation is positive, it means that the system has gained energy, so it must have absorbed a photon.

The energy of photon absorbed is equal to this $\Delta E$ . Converting it into Joule,
$\Delta E=3.02 eV=4.84 \cdot 10^{-19}J$
The energy of the photon is
$E=hf$
where h is the Planck constant while f is its frequency. Writing $\Delta E=hf$ , we can write the frequency f of the photon:
$f= \frac{\Delta E}{h}= \frac{4.84 \cdot 10^{-19}J}{6.63 \cdot 10^{-34}m^2 kg/s}=7.29 \cdot 10^{14}Hz$

The photon travels at the speed of light, $c=3 \cdot 10^8 m/s$ , so its wavelength is
$\lambda = \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{7.29 \cdot 10^{14}Hz}=4.11 \cdot 10^{-7}m=411 nm$

So, the initial sentence can be completed as:
The n = 2 to n = 6 transition in the bohr hydrogen atom corresponds to the "absorption" of a photon with a wavelength of "411" nm.

4 0

4 years ago

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Which quantity is the vector? a) weight b) power c) kinetiic energy d) speed