All categories

2 years ago

Which feature of a longitudinal wave corresponds to a trough in a transverse wave?

Physics

2 answers:

Tanya [424]2 years ago

5 0

Answer:

Crest

Explanation:

The crest is the peak at the top of the wave which peaks at the top of the wave on the longitudinal wave. Compared to a trough in a transverse wave, where it is the lowest point in the wave where the medium is at it lowest.

Therefore, they correspond

Hope this helps :)

Have a nice day!

Rama09 [41]2 years ago

3 0

It’s the crest, the crest is the top part of the wave and the trough is the bottom so they correspond

You might be interested in

What type of circuit is illustrated?

Reil [10]

Answer:

I beleive its B

Explanation:

If not then A but I'm positive its B

6 0

2 years ago

A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic fie

Nutka1998 [239]

Answer:

i = 0.5 A

Explanation:

As we know that magnetic flux is given as

$\phi = NBA$

here we know that

N = number of turns

B = magnetic field

A = area of the loop

now we know that rate of change in magnetic flux will induce EMF in the coil

so we have

$EMF = NA\frac{dB}{dt}$

now plug in all values to find induced EMF

$EMF = (20)(50 \times 10^{-4})(\frac{6 - 2}{2})$

$EMF = 0.2 volts$

now by ohm's law we have

$current = \frac{EMF}{Resistance}$

$i = \frac{0.2}{0.40} = 0.5 A$

5 0

2 years ago

) Water falls from a height of 60m at the rate of 15kg/s to operate a turbine. The losses due to frictional force are 10% of ene

Angelina_Jolie [31]

Answer:

8100W

Explanation:

Let g = 10m/s2

As water is falling from 60m high, its potential energy from 60m high would convert to power. So the rate of change in potential energy is

$P = \dot{E} = \dot{m}gh = 15*10*60 = 9000 J/s$ or 9000W

Since 10% of this is lost to friction, we take the remaining 90 %

P = 9000*90% = 8100 W

3 0

2 years ago

Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th

aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0

2 years ago

a cyclist coasting down a 5.0 ◦ incline at a constant speed of 6.0 km/h because of air resistance. If the total mass of the bicy

Dvinal [7]

Answer:

$F_{net}= 85.41\ N$

Explanation:

mass of the bicycle + cyclist = 50 kg

constant speed = 6 km/h

a cyclist coasting down a 5.0° incline

the downward velocity is constant, so net acceleration must be zero

the air drag must be equal to gravitational force downward along the ramp

$F_a = mg sin \theta$

now for upward motion

$F_{net} = mg sin \theta + air\ drag$

$F_{net} = mg sin \theta + mg sin \theta$

$F_{net} = 2 mg sin \theta$

$F_{net} = 2\times 50 \times 9.8 sin 5^0$

$F_{net}= 85.41\ N$

3 0

3 years ago