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dem82 [27]
3 years ago
8

What is the wavelength of microwaves with a frequency of 3x10^10 Hz?​

Physics
1 answer:
RUDIKE [14]3 years ago
5 0

Answer:

0.01 m

Explanation:

Since the speed of light is 3.0×10^8 m/s

Use the equation,

Wavelength = speed ÷ frequency

Wavelength = 3.0×10^8 ÷ 3×10^10

Wavelength = 0.01m

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Radiation with a wavelength of 238 nm shines on a metal surface and ejects electrons that have a maximum speedof 3.75 X 105 m/s.
Ulleksa [173]
The Answer is e) - Gold.
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3 years ago
Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proport
netineya [11]

Answer:

[F]=[MLT^{-2}]

Explanation:

Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. The mathematical expression for the second law of motion is given by :

F = m × a

F is the applied force

m is the mass of the object

a is the acceleration due to gravity

We need to find the dimensions of force. The dimension of force m and a are as follows :

[m]=[M]

[a]=[LT^{-2}]

So, the dimension of force F is, [F]=[MLT^{-2}]. Hence, this is the required solution.

5 0
3 years ago
A tennis ball of mass 0.060 kg travels horizontally at a speed of 25m/s. The ball hits a tennis
stepladder [879]
It would be A because a is perfect
7 0
3 years ago
Read 2 more answers
Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m
aev [14]

Answer:

Explanation:

Given

charge of first body q_1=2.5\ mu C

charge of second body q_2=-3.5\ mu C

Particle 1 is at origin and particle 2 is at x=0.6\ m

third Particle which charge +q must be placed left of 2.5\mu C because it will repel the q charge while -3.5\mu C will attract it

suppose it is placed at a distance of x m

F_{1q}=\frac{kq(2.5)}{x^2}

F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}

F_{1q}+F_{2q}=0

\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0

\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}

\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}

0.6+x=1.1832x

x=3.27\ m

5 0
3 years ago
A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp
raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination \theta =11.1^{\circ}

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction \mu _k=0.39

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using v^2-u^2=2as

Final velocity v=0

0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s

s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}

s=0.68 m

5 0
3 years ago
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