Diffraction supports the:<br><br> A. wave theory of light.<br><br> B. particle theory of light.

A doctor counts 68 heartbeats in 1.0 minute. What are the corresponding period and frequency of the heart rhythm

sergejj [24]

Answer:

$f=1.13s^{-1}=1.13Hz$

Explanation:

Hello,

In this case, a frequency stands for a rate in which some action is done per unit of time. In this case, for the heartbeat, since 68 actions (heartbeats) occur in 1.0, the frequency turns out:

$f=\frac{68}{1.0min}=68min^{-1}$

Or as most commonly used in Hz ( $s^{-1}$ ):

$f=68\frac{1}{min} *\frac{1min}{60s}=1.13s^{-1}=1.13Hz$

Best regards.

8 0

3 years ago

In 8,450 seconds, the number of radioactive nuclei decreases to 1/16 of the number present initially. What is the half-life (in

almond37 [142]

Answer:

<h2>2113 seconds</h2>

Explanation:

The general decay equation is given as $N = N_0e^{-\lambda t} \\\\$ , then;

$\dfrac{N}{N_0} = e^{-\lambda t} \\$ where;

$N/N_0$ is the fraction of the radioactive substance present = 1/16

$\lambda$ is the decay constant

t is the time taken for decay to occur = 8,450s

Before we can find the half life of the material, we need to get the decay constant first.

Substituting the given values into the formula above, we will have;

$\frac{1}{16} = e^{-\lambda(8450)} \\\\Taking\ ln\ of \both \ sides\\\\ln(\frac{1}{16} ) = ln(e^{-\lambda(8450)}) \\\\\\ln (\frac{1}{16} ) = -8450 \lambda\\\\\lambda = \frac{-2.7726}{-8450}\\ \\\lambda = 0.000328$

Half life f the material is expressed as $t_{1/2} = \frac{0.693}{\lambda}$

$t_{1/2} = \frac{0.693}{0.000328}$

$t_{1/2} = 2,112.8 secs$

Hence, the half life of the material is approximately 2113 seconds

7 0

3 years ago

What is the relationship between the frequency and the pitch of a sound?

andrezito [222]

Choice-C is the correct one.

3 0

3 years ago

Read 2 more answers

A car is moving with speed 60 m/s and acceleration 4 m/s2 at a given instant. (a) using a second-degree taylor polynomial, estim

pantera1 [17]

Answer:

Explanation:

Using second degree taylor polynomials

let $S(t)$ be position function and set $S(0)=0$

where S(0) is the initial position

Then $v(t) = S^i(t)$ and $a(t)=S^{ii}(t)$

we have $v(0) = 60m/s$ , $a(0) = 4m/s^2$

so $T_{2}(t) =S(0)+v(0)t+\frac{a(0)}{2} t^2\\\\\\=0+60t+\frac{4}{2} t^2\\\\T_{2}(t)=60t+2t^2\\\\S(1)=T_{2}(1)=60+2=62m$

b.) yes

6 0

3 years ago

A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,

n200080 [17]

Answer:

a. $t = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$ b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

$t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0} +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁ = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0

3 years ago

Diffraction supports the: A. wave theory of light. B. particle theory of light.