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What is the difference between a compressional wave with a large amplitude and one with a small amplitude?

rosijanka [135]

A larger amplitude means a smaller sound and a smaller amplitude means a larger sound.

6 0

3 years ago

How many units measure one wavelength? a 16 b 8 c 2 d 4

pochemuha

B it makes more sense

8 0

4 years ago

Cause and Effect :

bazaltina [42]

Answer:

The thyroid gland produces thyroxine (referred to as T4), which is a relatively inactive prohormone. The highly active hormone is triiodothyronine (referred to as T3).

Key actions of adrenaline include increasing the heart rate, increasing blood pressure, expanding the air passages of the lungs, enlarging the pupil in the eye (see photo), redistributing blood to the muscles and altering the body's metabolism, so as to maximise blood glucose levels (primarily for the brain).

7 0

3 years ago

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In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let's find out by doing a

MaRussiya [10]

Answer:

$v= \frac{10\times10^{-3} V}{70}$ m/s

Explanation:

Assumption: bullet leaves the muzzle at a speed of V m/s

and velocity of push received by the man be v m/s

According to newton's third law to every action there is always an equal and opposite reaction.

therefore,

mass of man× velocity = mass of bullet×its velocity

⇒70×v= 10×10^-3 ×V

solving the above eqaution we get

therefore $v= \frac{10\times10^{-3} V}{70}$ m/s

7 0

4 years ago

A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to

aksik [14]

Answer:

The answer is below

Explanation:

a) The vertical displacement = Δy = 21.5 m - 1.5 m = 20 m

The horizontal displacement = Δx = 69 m wide

Using the formula:

$\Delta y = u_yt+ \frac{1}{2}a_yt^2\\ \\u_y=initial\ velocity\ of \ car\ in\ y\ direction = 0,a_y=g=acceleration\ due\ to\ gravity\\=10m/s^2\\\\\Delta y = \frac{1}{2}a_yt^2\\\\\Delta y=\frac{1}{2}a_yt^2\\\\t=\sqrt{\frac{2\Delta y}{a_y} }=\sqrt{\frac{2*20}{10} } =2\ m/s$

Also:

$\Delta x = u_xt+ \frac{1}{2}a_xt^2\\ \\u_x=initial\ velocity\ of \ car\ in\ x\ direction = 0,a_x=acceleration=0\\\\\Delta x = u_xt\\\\u_x=\frac{\Delta x}{t}=\frac{69}{2} =34.5\ m/s$

b)The car is moving at a constant speed in the horizontal direction, hence the initial velocity = final velocity

$v_x=u_x=34.5\ m/s\\\\v_y=u_y+a_yt\\\\v_y=0+gt\\\\v_y=10(2)=20\ m/s\\\\v=\sqrt{v_x^2+v_y^2}=\sqrt{34.5^2+20^2}=39.9\ m/s\\ v=39.9\ m/s$

4 0

4 years ago