All categories

3 years ago

Which of the following results when a crest and trough meet?

1 answer:

6 0

D. Destructive interference. An easy way to think about it is the waves are opposite each other, so they essentially cancel each other out, or make an effort to.

You might be interested in

Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

Andrej [43]

(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

where

$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

Here we have

$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

$K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J$

3 0

2 years ago

Challenge: Some but not all of the gasoline used by car's enegine is transformed into kinetic energy. Where might some of the en

Levart [38]

Most of the excess energy is released as waste heat into the air surrounding the engine. Small amounts of excess energy are also released as sound energy, and as electrical energy generated by the alternator in a car's engine.

8 0

3 years ago

a mass stands on a platform which vibrates simple harmonically in a vertical Direction and with a frequency of 5 Hertz. show tha

Nastasia [14]

shm

displacement is A*sin 2pi f t

vel =( A*2*pi*f*)*cos 2 pi f t

accn =-( A*2*pi*f*)^2*sin 2 pi f t

accn = (0.01*2*3*5)

accn = (0.01x30)^2=9

4 0

3 years ago

find the gravitational force this shell exerts on a 1.60 kg point mass placed at the distance 5.01 m from the center of the shel

RideAnS [48]

The gravitational force of the shell exerts is 4.25m x 10¯¹² N.

We need to know about gravitational force to solve this problem. The gravitational force is the force caused by two masses of objects. The magnitude of gravitational force can be determined as

F = G.m1.m2 / R²

where F is the gravitational force, G is the gravitational constant (6.674 × 10¯¹¹ Nm²/kg²), m1 and m2 are the mass of the object and R is the radius.

From the question above, we know that

m1 = 1.6 kg

m2 = m

R = 5.01 m

By substituting the following parameters, we get

F = G.m1.m2 / R²

F = 6.674 × 10¯¹¹ . 1.6 . m / 5.01²

F = 4.25m x 10¯¹² N

where m is the mass of the shell

For more on gravitational force at: brainly.com/question/19050897

#SPJ4

7 0

1 year ago

Tell me the hearing range of all animals please. 55 POINTS AVAILABLE!!!!! This means a lot thanks.

WINSTONCH [101]

Answer:

Hearing range describes the range of frequencies that can be heard by humans or other animals, though it can also refer to the range of levels. The human range is commonly given as 20 to 20,000 Hz, although there is considerable variation between individuals, especially at high frequencies, and a gradual loss of sensitivity to higher frequencies with age is considered normal. Sensitivity also varies with frequency, as shown by equal-loudness contours. Routine investigation for hearing loss usually involves an audiogram which shows threshold levels relative to a normal.

Several animal species are able to hear frequencies well beyond the human hearing range. Some dolphins and bats, for example, can hear frequencies up to 100,000 Hz. Elephants can hear sounds at 14–16 Hz, while some whales can hear infrasonic sounds as low as 7 Hz (in water).

Explanation:

HOPE THIS HELPS!!!

6 0

2 years ago

Read 2 more answers