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just olya [345]
3 years ago
14

I am in plato cuzz wauup what u doing

Physics
1 answer:
Irina-Kira [14]3 years ago
8 0
Yoohoo dudio!!! whatz up??!!

MARK AS THE BRAINLIEST SWEETY!!
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A 1,650 kg SUV comes uniformly to a stop. If the vehicle is accelerating at -1.3 m/s^2,
lions [1.4K]

Answer:b

Explanation:

4 0
3 years ago
The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
How can the speed of a mechanical wave energy be calculated
Slav-nsk [51]
Speed = frequency * wavelength
5 0
3 years ago
A typical nuclear fission power plant produces about 1.00 GW of electrical power. Assume the plant has an overall efficiency of
mamaluj [8]

Answer:

mass consumed by 235U each day = 2 kg

Explanation:

electrical power produced = 1 GW = 1 × 10⁹ × (6.24151 × 10¹⁸ ) eV

                                            = 6.24151× 10²¹ MeV/s

thermal energy =  0.420 * 250 = 105 MeV

\dfrac{1 GW}{150 MeV}= \dfrac{6.24151\times 10^{21}}{105}

                                      = 5.94 × 10¹⁹ fission/second

                                       =5.94 × 10¹⁹× 24 × 60 ×60)

                                      =  5.13 × 10²⁴ fission/day

mu = 235.04393 ×  1.660× 10 ⁻²⁷ = 390.1729× 10⁻²⁷ Kg

M = mu ×5.13 × 10²⁴

   = 390.1729× 10⁻²⁷ ×5.13 × 10²⁴

M   =  2 kg(approx.)

mass consumed by 235U each day = 2 kg

3 0
3 years ago
Light in vacuum is incident on the surface of a slab of transparent material. In the vacuum the beam makes an angle of 39.9° wit
wariber [46]

Answer:

n=2.053

Explanation:

We will use Snell's Law defined as:

n_{1}*Sin\theta_{1}=n_{2}*Sin\theta_{2}

Where n values are indexes of refraction and \theta values are the angles in each medium. For vacuum, the index of refraction in n=1. With this we have enough information to state:

1*Sin(39.9)=n_{2}*Sin(18.2)

Solving for n_{2} yields:

n_{2}=\frac{Sin(39.9)}{Sin(18.2)}=2.053

Remember to use degrees for trigonometric functions instead of radians!

6 0
3 years ago
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