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3 years ago

BEST ANSWER GETS BRAINLIEST!!!!!!!!!!

Physics

1 answer:

denis-greek [22]3 years ago

8 0

Answer:

There are Microwaves, the type of electro magnetic radiation is a Micro-wave. We use x-rays, the type of electro magnetic radiation is a gamma wave. We also use radios, the type of electro magnetic radiation is a radio wave.

Explanation:

I remember doing this assignment too

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qwelly [4]

Answer:

Explanation:

6 0

3 years ago

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

5 0

3 years ago

A pulse of red light and a pulse of blue light enter a glass block normal to its surface at the same time. after passing through

pshichka [43]

I believe the red light would go faster than the blue.

7 0

3 years ago

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1. Strzała o masie 20 g tuż po wystrzale ma prędkość 50 m/s. Oblicz pracę wykonaną

kow [346]

Answer:

Explanation:

Question 1

An arrow weighing 20g shortly after firing has a speed of 50m / s. Calculate the work done by the athlete. What is the potential energy of the elasticity of the tensed string?

mass m = 20g = 20/1000 = 0.02kg

speed v = 50m / s

P.E = K.E = ½mv²

P.E = ½ × 0.02 × 50²

P.E = 25 J

work done = P.E = 25J

Qestion 2

A 80 kg athlete stood on a trampoline with a coefficient of elasticity of k = 2 kN / m. As far as the edge of the trampoline lowers.

force of elasticity

F = -kx

x = F / k

in our case F will be the force of pressure or gravity

F = mg

g is gravitational acceleration, and according to Newton's second law, acceleration is force through mass - unit of force N, unit of mass kg. Acceleration either in m / s ^ 2 or N / kg

F = 80kg * 10N / kg = 800 N

x = 800N / -2000N = -0.4

The trampoline will lower, so from the level by 0.4 meters and hence this minus

7 0

3 years ago

In a certain process, the energy of the system decreases by 250 kJ. The process involves 480 kJ of work done on the system. Find

gregori [183]

Answer:

Q = - 730 KJ

730 KJ is transferred out of the system

Explanation:

According to the first law of thermodynamics, energy can neither be created nor destroyed, but can be transformed from one form to another.

For a particular process/system, the first law is interpreted as

ΔU = Q + W (depending on convention, some textbooks give this relation as ΔU = Q - W)

ΔU is the change in internal energy of the system, in my convention, it is positive if the internal energy increases and negative otherwise.

Q = Heat transferred into or out of the system. Q is positive for heat transferred into the system and negative for heat transferred out of the system.

W = workdone by the system or work done on the system. W is positive for workdone on the system and W is negative for when work is done by the system.

ΔU decreases by 250 KJ, that is, ΔU = - 250 KJ

Q = ?

W = 480 KJ (Work is done on the system)

- 250 = Q + 480

Q = - 250 - 480 = - 730 KJ

The heat is transferred out of the system.

6 0

3 years ago

Read 2 more answers