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3 years ago

Parts of a transverse wave include:

Physics

2 answers:

iris [78.8K]3 years ago

8 0

Answer:

Parts of a transverse wave include:

Crest - the highest point on the wave

Trough - the lowest point on the wave

Wavelength - the distance from one crest to the next crest, or one trough to the next trough

Amplitude - the displacement of the wave from the midpoint to the highest point (crest) or the lowest point (trough)

Explanation:

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abruzzese [7]3 years ago

4 0

Answer:

Parts of a transverse wave include:

Crest - the highest point on the wave

Trough - the lowest point on the wave

the answer is: WAVELENGTH - the distance from one crest to the next crest, or one trough to the next trough

Amplitude - the displacement of the wave from the midpoint to the highest point (crest) or the lowest point (trough)

Explanation:

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An what would be needed to change an objects state of motion

VladimirAG [237]

Inertia: tendency of an object to resist changes in its velocity. An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force.

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4 years ago

what is the shortest possible time in which a bacterium could travel a distence of 8.4 cm across a petri dish at a constant spee

Inessa [10]

D = RT

D is 8.4 cm or 84mm, R is 3.5mm (units of time, assuming seconds for now)

84 = 3.5T

T = 24 seconds

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3 years ago

What is the slit spacing of a diffraction necessary for a 600 nm light to have a first order principal maximum at 25.0°?

Sauron [17]

Explanation:

Given that,

Wavelength of light, $\lambda=600\ nm=6\times 10^{-7}\ m$

Angle, $\theta=25^{\circ}$

We need to find the slit spacing for diffraction. For a diffraction, the first order principal maximum is given by :

$d\sin\theta=n\lambda$

n is 1 here

d is slit spacing

$d=\dfrac{\lambda}{\sin\theta}\\\\d=\dfrac{6\times 10^{-7}}{\sin(25)}\\\\d=1.41\times 10^{-6}\ m\\\\d=1.41\ \mu m$

So, the slit spacing is $1.41\ \mu m$ .

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4 years ago

Assume that a scale is in an elevator on Earth.

Ne4ueva [31]

★ Remember :

The reading of scale of a man in an elevator :

When the elevator moves upward with acceleration a, R = m(g + a)
When the elevator moves downwards with acceleration a , R = m(g - a)
When the elevator falls freely, we take a = g so, R = m(g - g) = 0
When the lift is at rest or moves with uniform velocity, a = 0, so R = m(g - 0) = mg

A). The elevator moves up at a constant speed $.$

∴ Acceleration of elevator = 0

➠ R = mg

➠ 53 × 10

➠ R = 530N

B). It slows at 2m/s² while moving upward.

∴ Acceleration of elevator = -2m/s²

[Negative sign shows that speed decreases with time]

➠ R = m(g + a)

➠ R = 53(10 + (-2))

➠ R = 53 × 8

➠ R = 424N

C). It speeds up at 2m/s² while moving downward.

∴ Acceleration of elevator = 2m/s²

➠ R = m(g - a)

➠ R = 53(10 - 2)

➠ R = 53 × 8

➠ R = 424N

D). It moves downward at constant speed.

∴ Acceleration of elevator = 0

➠ R = mg

➠ R = 53 × 10

➠ R = 530N

<h3>Hope It Helps!</h3>

4 0

3 years ago

Assuming an engine operates as a reversible heat engine calculate the outlet temperature (temperature of rejected heat) that ach

Serjik [45]

Answer:

$64.8^{\circ}C$

Explanation:

The efficiency of a reversible heat engine is given by:

$\eta = 1 -\frac{T_C}{T_H}$

where

Tc is the cold temperature (outlet temperature)

Th is the hot temperature (temperature of the source)

In this problem, we know:

$\eta = 63\% = 0.63$

$T_H = 640^{\circ}C=913 K$

So, we can calculate the outlet temperature by re-arranging the formula:

$T_C = T_H ( 1 -\eta)=(913 K)(1- 0.63)=337.8 K=64.8^{\circ}C$

8 0

3 years ago