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Schach [20]
3 years ago
10

A wave has a period of 2s and a wavelength of 5m. What is the speed of the wave?

Physics
1 answer:
horrorfan [7]3 years ago
8 0

Answer:

10m per Second

Explanation:

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A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it
Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface \Phi it is possible to calculate the magnitude \frac{\Phi}{A}=B\ cos \theta.

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux \Phi is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is (m^{2}). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with \theta being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

                                                     \Phi=B\ A\ cos\theta

We are told the values of \Phi and B, then we can calculate the magnitude

                                                      \frac{\Phi}{A}=B\ cos\theta

3 0
3 years ago
A force of F = (2.00ˆi + 3.00ˆj) N is applied to an object that is pivoted about a fixed axle aligned along the z coordinate axi
Vladimir [108]

Explanation:

It is given that,

Force applied to object, F=(2i+3j)\ N

Position, r=(4i+5j)\ m  

(b) The cross product of force and position vector is used to find the net torque about the z axis. It is given by :

\tau=F\times r

\tau=(2i+3j) \times (4i+5j)

\tau=\begin{pmatrix}0&0&-2\end{pmatrix}

or

\tau=(-2k)\ N-m

The torque is acting in -z axis.

(a) The magnitude of torque is given by :

|\tau|=\sqrt{(-2)^2}

|\tau|=2\ N-m

Hence, this is the required solution.

5 0
3 years ago
2 cannons fire projectiles upwards with the same velocity. The
dybincka [34]

Answer:

Explanation:

Given

Two projectile is fired vertically upward

One has 4 times the mass of other

When Projectile is fired their trajectory is independent of mass of object. Also if they launched with same speed then both achieved same maximum height in same time and will hit the ground at the same moment.

3 0
3 years ago
A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J
Lyrx [107]

Explanation:

Given that,

Wavelength of the light, \lambda=4170\ A=4170\times 10^{-10}\ m

Work function of sodium, W_o=4.41\times 10^{-19}\ J

The kinetic energy of the ejected electron in terms of work function is given by :

KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J

The formula of kinetic energy is given by :

KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s

Hence, this is the required solution.

7 0
3 years ago
An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, KE_{t} = \frac{3}{2}k_{b}T

k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

\lambda_{e} = \frac{h}{p_{e}}

\lambda_{e} = \frac{h}{m_{e}{v_{e}}              (1)

where

h = Planck's constant = 6.626\times 10^{- 34}m^{2}kg/s

p_{e} = momentum of an electron

v_{e} = velocity of an electron

m_{e} = 9.1\times 10_{- 31} kg = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

v_{e} = 2.86\times 10^{5} m/s                    (2)

Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

where

m_{p} = 1.6726\times 10_{- 27} kg = mass of proton

v_{p} = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T

}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}

v_{p} = 6.674\times 10^{3} m/s                               (4)                    

Using eqn (4) in (3):

\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

7 0
3 years ago
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