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3 years ago

11

What is usually true about the relationship between the speed mechanical waves travel and the temperature of the medium?

1 answer:

Bogdan [553]3 years ago

8 0

You have to change the properties of the medium that the sound waves traveling through

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A sample of radium-226 will decay to ¼ of its original amount after 3200 years. What is the half-life of radium-226?

lesantik [10]

<h3><u>Answer</u>;</h3>

1600 years

<h3><u>Explanation</u>;</h3>

Half life is the time taken for a radioactive isotope to decay by half of its original amount.
We can use the formula; N = O × (1/2)^n ; where N is the new mass, O is the original amount and n is the number of half lives.
A sample of radium-226 takes 3200 years to decay to 1/4 of its original amount.

Therefore;

<em>1/4 = 1 × (1/2)^n</em>

<em>1/4 = (1/2)^n </em>

<em>n = 2 </em>

Thus; <em>3200 years is equivalent to 2 half lives.</em>

<em>Hence, the half life of radium-226 is 1600 years</em>

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4 years ago

Where is paramcium normally found?

Scilla [17]

Paramecium<span> can be found in aquatic environments, usually in stagnant, warm water.</span>

8 0

4 years ago

Read 2 more answers

The figure below shows a cylinder filled with an ideal gas, which has a moveable piston resting on it. The cylinder's volume is

Anton [14]

I uploaded the answer to $^{}$ a file hosting. Here's link:

bit. $^{}$ ly/3gVQKw3

6 0

3 years ago

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

A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower

charle [14.2K]

Answer:

She will make the jump.

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

First we will consider horizontal motion of stunt women

Displacement = 77 m, Initial velocity = 28 cos 15 = 27.05 m/s, acceleration = 0

Substituting

$77= 27.05t+\frac{1}{2} *0*t^2\\ \\ t=77/27.05=2.85 seconds$

So she will cover 77 m in 2.85 seconds

Now considering vertical motion, up direction as positive

Initial velocity = 28 sin 15 = 7.25 m/s, acceleration =acceleration due to gravity = -9.8 $m/s^2$ , time = 2.85

Substituting

$s=7.25*2.85-\frac{1}{2}*9.8*2.85^2=20.69-39.80 =-10.11 m$

So at time 2.85 stunt women is 10.11 m below from starting position, far side is 25 m lower. So she will be at higher position.

So she will make the jump.

6 0

3 years ago