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3 years ago

Which describes a high frequency wave?

Physics

1 answer:

pav-90 [236]3 years ago

5 0

<span>B.
many wavelengths pass a given point in one second
</span>

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How has psychology developed from its prescientific roots in early understandings of mind and body to the beginnings of modern

bonufazy [111]

Answer:

By the experiences, ideas of origin, power and modern empiricism.

Explanation:

The route of the psychology has been recorded by the history of India, Middle East, Europe and China. The Confucius and Buddha focused on the ideas origin and the power. The ancient Hebrews, Aristotle and Pluto tells the idea whether the body with the mind is connected or it is distinct and this connection or distinct idea is the result of experiences.

Locke offers his theories about the mind that mind as the blank state, on which the experience writes. The modern empiricism idea has been developed by Bacon and Locke

3 0

3 years ago

The coil of a generator has a radius of 0.14 m. When this coil is unwound, the wire from which it is made has a length of 10.0 m

adell [148]

Answer:

5.565 V

Explanation:

Radius of coil of generator=r=0.14 m

Length of wire=l=10 m

Magnetic field,B=0.24 T

Angular speed, $\omega=34rad/s$

We have to find the peak emf of the generator.

$N=\frac{l}{2\pi r}=\frac{10}{2\pi\times 0.14}=11$

$A=\pi r^2=\pi (0.14)^2=0.062m^2$

Peak(maximum) induced emf of generator= $E_{max}=NBA\omega$

Using the formula

$E_{max}=11\times 0.24\times 0.062\times 34$

$E_{max}=5.565 V$

3 0

3 years ago

A 1.5 kg orange falls from a tree and hits the ground in 0.75s. What is the speed of the orange just before it hits the ground?

Olenka [21]

The final speed of the orange is 7.35 m/s

Explanation:

The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration $g=9.8 m/s^2$ towards the ground. So we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time elapsed

For the orange in this problem, we have

u = 0 (it is dropped from rest)

$a=g=9.8 m/s^2$ is the acceleration

Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:

$v=0+(9.8)(0.75)=7.35 m/s$

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

8 0

3 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$