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3 years ago

WHAT WILL BE YOUR CONCLUSI0N ABOUT ELECTROMAGNETIC WAVE

Physics

1 answer:

andrezito [222]3 years ago

8 0

Answer:

It's really important waves type for fulfill human's needs

Explanation:

Electromagnetic waves are transverse waves composed by the perpendicular oscillating electric and magnetic fields.

EM waves have both Electrical and magnetic features.

they travel in the velocity of light (3*10⁸ ms⁻¹)

they does not require any media to travel. It has two perpendicular electric field and the magnetic field which are perpendicular to each other

They travel perpendicular to each of those electric and magnetic fields.

Example :

Radio Wave
Micro Wave
IR wave
Light Wave
UV rays
X rays
Gamma rays
Cosmic rays

The main importance of em waves is they allow energy to be stored within them and then can be propagated over a large distance using the dielectric and magnetic properties of materials .

This performance has been used in many fields wisely and effectively to make the things easy.

Ex : medicine , Telecommunication , energy , Engineering etc

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3 years ago

A 12.0 kg block rests on an inclined plane. The plane makes an angle of 31.0° with the horizontal, and the coefficient of fricti

allochka39001 [22]

Answer:

The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.

Explanation:

Given that,

Mass = 12.0 kg

Angle = 31.0°

Friction coefficient = 0.158

Mass of second block = 38.0 kg

Using formula of frictional force

$f_{\mu} = \mu N$ ....(I)

Where, N = normal force

$N = mg\cos\theta$

Put the value of N into the formula

$N =12\times9.8\times\cos 31^{\circ}$

$N=100.80\ N$

Put the value of N in equation (I)

$f_{mu}=0.158\times100.80$

$f_{mu}=15.9264\ N$

Now, Weight of second block

$W = mg$

$W=38.0\times9.8$

$W=372.4\ N$

The horizontal force is

$F = mg\sintheta$

$F=12\times9.8\times\sin 31^{\circ}$

$F=60.5684\ N$ ....(II)

(I). We need to calculate the acceleration

$a=m_{2}g-\dfrac{f_{\mu}+mg\sin\theta}{m_{1}+m_{2}}$

$a=\dfrac{372.4-(15.9264+60.5684)}{12+38}$

$a=5.92\ m/s^2$

(II). We need to calculate the tension in the rope

$m_{2}g-T=m_{2}a$

$-T=38\times5.92-38\times9.8$

$T=147.44\ N$

Hence, The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.

5 0

4 years ago

miniature spring-loaded, radio-controlled gun is mounted on an air puck. The gun's bullet has a mass of 5.00 g, and the gun and

Tamiku [17]

Answer:

12 m/s

Explanation:

From the question,

Applying the law of conservation of momentum,

total momentum before collision = Total momentum after collision

mu+Mu' = mv+Mv'........................... Equation 1

Where m = mass of the bullet, u = initial velocity of the bullet, M = combined mass of the gun and the puck, u' = initial velocity of the gun and the puck, v = final velocity of the bullet, v' = final velocity of the gun and the puck

make v the subeject of the equation

v = [(mu+Mu')-Mv']/m................. Equation 2

Given: m = 5.00 g = 0.005 kg, M = 120 g = 0.12 kg, u = u' = 0 m/s (at rest), v' = 0.5 m/s

Substitute these values into equation 2

v = [0-(0.12×0.5)]/0.005

v = -0.06/0.005

v = -12 m/s

The negative sign can be ignored since we are looking for the speed, which has only magnitude.

Hence the speed of the bullet is 12 m/s

5 0

3 years ago