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3 years ago

Solids in which the atoms have no particular order or pattern are called

Physics

1 answer:

posledela3 years ago

8 0

Answer:

Amorphous solids are composed of atoms or molecules that are in no particular order. Each particle is in a particular spot, but the particles are in no organized pattern. Examples include rubber and wax. Crystalline solids have a very orderly, three-dimensional arrangement of atoms or molecules

Explanation:

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An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.80 kW. A receiving antenna 7

Alecsey [184]

Answer:0.1759 v

Explanation:

Intensity of wave at receiver end is

I= $\frac{P_{avg}}{A}$

I= $\frac{3.80\times 10^3}{4\times \pi \times \left ( 4\times 1609.34\right )^2}$

I= $7.296\times 10^{-6} W/m^2$

Amplitude of electric field at receiver end

$E_{max}=\sqrt{2I\mu _0c}$

Amplitude of induced emf

= $E_{max}d$

= $\sqrt{2\times 7.29\times 10-6\times 4\pi \times 3\times 10^8}\times 0.75$

= $17.591\times 10^{-2}=0.1759 v$

7 0

3 years ago

Two forces that are not equal in size are

Bumek [7]

They are unbalanced forces ..... Hope this helps :3

4 0

3 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

6 0

3 years ago

30. A box with mass 20kg is on a cement floor. The coefficient of static friction between the box and floor is 0.25. A man is pu

Mnenie [13.5K]

Answer:

84.05

Explanation:

F=mg×0.25
F=20x9.81×0.25
f=49.05N
F=35N

F=f+F

F=49.05+35

=84.05

7 0

3 years ago

A boy kicks a football from ground level. The ball takes 3 seconds to reach its maximum height. What is the angle of the initial

ruslelena [56]

<h2>The angle of the initial velocity with respect to the horizontal is 85.14°</h2>

Explanation:

Given that the ball takes 3 seconds to reach its maximum height.

Consider the vertical motion of ball till maximum height.

We have equation of motion v = u + at

Acceleration, a = -9.81 m/s²

Final velocity, v = 0 m/s

Time, t = 3 s

Substituting

v = u + at

0 = u + -9.81 x 3

u = 29.43 m/s

Initial vertical velocity is 29.43 m/s.

Now consider horizontal motion of ball.

Time of flight of ball = 2 x Time to reach maximum height = 2 x 3 = 6 s

Displacement = 15 m

We have equation of motion s = ut + 0.5 at²

Displacement, s = 15 m

Acceleration, a = 0 m/s²

Time, t = 6 s

Substituting

s = ut + 0.5 at²

15 = u x 6 + 0.5 x 0 x 6²

u = 2.5 m/s

Initial horizontal velocity is 2.5 m/s

Let r be the initial velocity and θ be the angle with horizontal

Initial vertical velocity = rsinθ = 29.43 m/s

Initial horizontal velocity = rcosθ = 2.5 m/s

Dividing

$\frac{rsin\theta }{rcos\theta }=\frac{29.43}{2.5}\\\\tan\theta=11.77\\\\\theta=85.14^0$

The angle of the initial velocity with respect to the horizontal is 85.14°

4 0

3 years ago