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3 years ago

11

A wave is a disturbance that transfers energy through medium from one place to another. Do the particles in the medium travel w

ith the wave?

1 answer:

amm18123 years ago

6 0

The internet said no it does not

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Suppose the half-life of an element is 10 years. How many half-lives will it take before only about 6% of the original sample re

sergiy2304 [10]

You don't need to worry about the 10 year bit with this question. Just grab a calculator and divide 100/2, then the answer to that (50) by 2 etc and keep dividing by 2 until you get down to 6.25.

The answer ends up being 4 half lives :)

If you don't understand what a half life is please let me know :)

4 0

4 years ago

Read 2 more answers

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:

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

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

3 years ago

If Earth were completely blanketed with clouds and we couldn’t see the sky, could we learn about the realm beyond the clouds? Wh

Fudgin [204]

The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.

Through these waves, you can know that there is beyond the clouds.

Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.

Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.

4 0

3 years ago

If you saw a waxing gibbous moon what phase would you expect one week later?

Leona [35]

Seven days after a waxing gibbous moon, the moon will be waning gibbous, and at some point during that seven days, it will have been Full.

6 0

3 years ago

Can concave lens be used to make hand lens .why?

mario62 [17]

Answer:

No concave lens cannot be used to make hand lens because in hand Lens we use convex lens so as to converge the rays of the light After refraction and helps to produce a magnified image of an object.

Hope it will help you :)❤

8 0

3 years ago