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AnnZ [28]
3 years ago
6

Your pulse is caused by pressure of the blood on the artery wall, and it corresponds to your heart beat.

Physics
2 answers:
solong [7]3 years ago
8 0

weeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee true eeeeeeeeeeeeeeeeeeeee

r-ruslan [8.4K]3 years ago
6 0

Answer:

true

Explanation:

sorry if im wrong

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For a given amount of gas at a constant temperature, the volume of a gas varies inversely with its pressure is a statement of __
Ksivusya [100]

Answer:

d. Boyle's

Explanation:

Boyle's Law: States that the volume of a fixed mass of gas is inversely proportional proportional to its pressure, provided temperature remains constant.

Stating this mathematically. this implies that:

V∝1/P

V = k/P, Where k is the constant of proportionality

PV = k

P₁V₁ = P₂V₂

Where P₁ and P₂ are the initial and final pressure respectively, V₁ and V₂ are the the initial and final volume respectively.

Hence the right option is d. Boyle's

8 0
3 years ago
Cooling causes a material to
e-lub [12.9K]

Answer:

whats the question?

Explanation:

4 0
3 years ago
Read 2 more answers
(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit
Ainat [17]

Answer:

a

The number of fringe is  z  = 3 fringes

b

The  ratio is I = 0.2545I_o

Explanation:

a

 From the question we are told that

        The wavelength is  \lambda = 600 nm

        The distance between the slit is  d = 0.117mm = 0.117 *10^{-3} m

        The width of the slit is  a = 35.7 \mu m = 35.7 *10^{-6}m

let  z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is  and this mathematically represented as

             z = \frac{d}{a}

Substituting values

             z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}  

             z  = 3 fringes

b

   From the question  we are told that the order  of the bright fringe is  n = 3

   Generally the intensity of  a pattern  is mathematically represented as

                 I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]

Where I_o is the intensity  of the  central fringe

 And  Generally  sin \theta = \frac{n \lambda }{d}

               I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]

               I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]

               I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]

                I = I_o (1)(0.2545)

                  I = 0.2545I_o

6 0
3 years ago
If the box is a distance 1.81 m from the rear of the truck when the truck starts, how much time elapses before the box falls off
alexira [117]
If the box is a distance 1.81 m from the rear of the truck when the truck starts,<span> ... Force of Friction = mu_s * Normal Force( </span>M<span> * G) ... The </span>box starts<span> moving! ... Now that the </span>box<span> is moving, the bed of the </span>truck<span> pulls at it with 17.4 ... out how </span>long<span> it will take the </span>box<span> to reach the back of the </span>truck<span>. ... T^2 = 2 * </span>1.81<span> / .64</span>
4 0
3 years ago
A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se
zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

\vec{E'} = 1800\ N/C     (towards)

Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

7 0
3 years ago
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