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4 years ago

Explain how a transverse wave transmits energy.

1 answer:

6 0

''Surface waves are neither longitudinal nor transverse. In longitudinal and transverse waves, all the particles in the entire bulk of the medium move in a parallel and a perpendicular direction (respectively) relative to the direction of energy transport. ... Any wave moving through a medium has a source.''

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A 25-kg child runs at 4.0 m/s and jumps onto a shopping cart and holds on for dear life. The cart has mass 15 kg. Assuming the c

VLD [36.1K]

Linear momentum (mass x speed) has to be conserved.

-- Momentum before the jump:

(boy's mass) x (boy's speed) = (25 kg) x (4.0 m/s) = 100 kg-m/s

(cart's mass) x (cart's speed) = (15 kg) x (zero) = zero

Total momentum before the jump: (100 kg-m/s) + (zero) = (100 kg-m/s)

-- Momentum after the jump:

(mass of boy+cart) x (speed of boy+cart) = (40 kg) x (speed)

-- Momentum after the jump = momentum before the jump

(40 kg) x (speed) = 100 kg-m/s

Divide each side by 40 kg:

Speed = (100 kg-m/s) / (40 kg)

Speed = 2.5 m/s (d)

5 0

4 years ago

An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the accelerati

rosijanka [135]

Answer:

A) $\frac{g}{b}(1-e^{-bt})$

Explanation:

Since a = g - bv,

We can substitute a = dv/dt into the equation.

Then, the equation will be like dv/dt = g - bv.

So we got first order differential equation.

As known, v = 0 at t = 0, and v = g/b at t = ∞.

Since $\frac{dv}{dt}= g - bv = b( \frac{g}{b} - v)$ ⇒ $\frac{dv}{ \frac{g}{b} - v}= bdt$

So take the integral of both side.

$- ln (\frac{g}{b} - v) = bt + C$

Since for t=0, v = 0 ⇒ $C =- ln (\frac{g}{b})$

$v = \frac{g}{b} + e^{-bt-ln(\frac{g}{b})} = \frac{g}{b}- \frac{g}{b}e^{-bt} = \frac{g}{b}(1-e^{-bt})$

5 0

3 years ago

The light intensity incident on a metallic surface with a work function of 1.88 eV produces photoelectrons with a maximum kineti

ivanzaharov [21]

Answer:

KE = KE (incidental) - KE of emitted photons

or KE = h * f - Wf

So h * f = KE + Wf = 1.2 + 1.88 = 3.08 incident energy

If you double the frequency then h * f = 6.16

KE = 6.16 - 1.2 = 4.96 eV

7 0

3 years ago

1. The volume of a given mass of gas is 20cm when its

Pavel [41]

Answer:

Explanation:

The way to show a cubed substance is either like this³ or like this x^3. The small three is found at the bottom toolbar at the bottom of the question space marked by the Ω symbol.

100 mmHg

Givens

V1 = 20 cm^3

V2 = 80 cm^3

P1 = 400 mmHg

P2 = ?

Formula

V1 * P1 = V2 * P2

Solution

20 * 400 = 80 * P2 Divide by 80

20 * 400/80 = P2

P2 = 8000 / 80

P2 = 100 mmHg

5 0

3 years ago

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

tia_tia [17]

Answer:

a) $a=33.73mm/s^{2}$

b) mg>N

c) $\%_{change}=0.343\%$

d) $a=24.07mm/s^{2}$

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

$a_{c}=\omega ^{2}r$

where:

$\omega=\frac{2\pi}{T}$

we know the period of rotation of the earth is about 24 hours, so:

$T=24hr*\frac{3600s}{1hr}=86400s$

so we can now find the angular speed:

$\omega=\frac{2\pi}{86400s}$

$\omega=72.72x10^{-6} rad/s^{2}$

So the centripetal acceleration will be:

$a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)$

which yields:

$a_{c}=33.73mm/s^{2}$

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

$\sum F=0$

so we get that:

$N-mg+ma_{c} = 0$

and solve for the normal force:

$N=mg-ma_{c}$

In this case, we can clearly see that:

$mg>mg-ma_{c}$

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

$mg=(60kg)(9.81m/s^{2})=588.6N$