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jeka94
3 years ago
8

Question is in the picture below

Physics
1 answer:
Sati [7]3 years ago
4 0

may vary depending on the organization.

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A perfect bouncy ball would continue bouncing the same height forever. But in reality, the ball stops bouncing because energy is
galina1969 [7]

Answer:

D. all answers are true

6 0
3 years ago
A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.0 km, the je
jenyasd209 [6]
<h2>The answer got is reasonable.</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 300 m/s  

Acceleration, a = ?

Final velocity, v = 400 m/s  

Displacement,s = 4 km = 4000 m

Substituting  

v² = u² + 2as

400² = 300² + 2 x a x 4000

a = 8.75 m/s² = 8.8 m/s²

The acceleration is 8.8 m/s²

The answer got is reasonable.

7 0
3 years ago
Find the wavelength in meters for a transverse mechanical wave with an amplitude of 10 cm and a radian frequency of 20π rad/s if
nydimaria [60]

Answer:

The wavelength of the wave is 20 m.

Explanation:

Given that,

Amplitude = 10 cm

Radial frequency \omega = 20\pi\ rad/s

Bulk modulus = 40 MPa

Density = 1000 kg/m³

We need to calculate the velocity of the wave in the medium

Using formula of velocity

v=\sqrt{\dfrac{k}{\rho}}

Put the value into the formula

v=\sqrt{\dfrac{40\times10^{6}}{10^3}}

v=200\ m/s

We need to calculate the wavelength

Using formula of wavelength

\lambda =\dfrac{v}{f}

\lambda=\dfrac{v\times2\pi}{\omega}

Put the value into the formula

\lambda=\dfrac{200\times2\pi}{20\pi}

\lambda=20\ m

Hence, The wavelength of the wave is 20 m.

5 0
3 years ago
n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How f
pashok25 [27]

Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

v = \sqrt{\frac{2K.E}{m} } \\\\v =  \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.

8 0
3 years ago
How long does it take a microwave of power 0.2kW to sue 10000 J of energy
olasank [31]

Answer:

50s .

Explanation:

\frak{\pink{Given}}\begin{cases}\textsf{ The power of microvave is 0.2kW .}\\\textsf{ Amount of energy is 10000 J .}\end{cases}

Here the power of the microwave is 0.2kW . And as we know that ,

  • Rate of doing work is called power .

So from the definition , we have ;

\sf\longrightarrow Power =\dfrac{Work}{time}

  • Here the work done is equal to the energy consumed by the microwave i.e. 10000 J .So we can write it as ,

\sf\longrightarrow Power =\dfrac{Energy}{time}

\sf\longrightarrow 0.2kW = \dfrac{10^4 J }{t} \\

\sf\longrightarrow 0.2 * 1000 W =  \dfrac{10^4 J }{t}

Cross multiply ,

\sf\longrightarrow t = \dfrac{ 10^4 }{ 0.2 * 10^3}s=\dfrac{10^4}{2*10^2} s

Simplify ,

\sf\longrightarrow \boxed{\bf t = 50s}

<h3>Hence the time taken is 50s .</h3>
5 0
2 years ago
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