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4 years ago

Can anyone help me I will give you 30 points

Physics

1 answer:

Aleonysh [2.5K]4 years ago

3 0

Answer:

1) ELECTRO MAGNET

2) B

3) -AMOUNT OF ELECTRICITY OF THE POWER SOURCE

- AMOUNT OF COIL WRAPPED UP TO THE NAIK

-CONDUCTIVITY OF THE MEDIUM

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HELP ASAP How are the bars in barred spiral galaxies thought to have formed?

sergey [27]

Answer:

D. slow evolution

Explanation:

8 0

3 years ago

Two identical traveling waves, moving in the same direction, are out of phase by π/5.0 rad. What is the amplitude of the resulta

andreev551 [17]

Answer:

Therefore the amplitude of the resultant wave is $=0.95 y_m$

Explanation:

The equation of wave:

y=A sin (kx-ωt)

For wave 1:

y₁=A sin (kx-ωt) = $y_{m}$ sin (kx-ωt)

For wave 2:

y₂=A sin (kx-ωt+Φ) = $y_{m}$ sin (kx-ωt+Φ)

Where A= amplitude= $y_m$

The angular frequency $\omega=\frac{2\pi}{T}$

$k=\frac{2\pi}{\lambda}$ , $\lambda$ = wave length.

t= time

T= Time period

$\phi$ = phase difference = $\frac{\pi}{5}$

The resultant wave will be

y = y₁ + y₂

= $y_m$ sin (kx-ωt) + $y_m$ sin (kx-ωt+Φ)

$=y_m$ {sin (kx-ωt) + sin (kx-ωt+Φ)}

$=y_m\ sin(\frac{kx-\omega t +\phi + kx-\omega t }2)\ cos(\frac{kx-\omega t +\phi -kx+\omega t}2)$

$=y_m\ sin({kx-\omega t +\frac\phi 2)\ cos(\frac{\phi }2)$

$=y_m\ cos(\frac{\phi }2) sin({kx-\omega t +\frac\phi 2)$

Therefore the amplitude of the resultant wave is

$=y_m\ cos(\frac{\phi }2)$

$=y_m\ cos(\frac{\pi }{10})$

$=0.95 y_m$

6 0

4 years ago

May you help me answer this

Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$