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kaheart [24]
3 years ago
5

Which of the following best describes transverse waves?

Physics
2 answers:
gladu [14]3 years ago
7 0

Answer:

Displacement of particles is back and forth, perpendicular to the direction of wave motion.

den301095 [7]3 years ago
5 0

Best Answer: perpendicular to the direction of wave motion

think of you and a friend holding different ends of a rope. you shake your end side to side and the wave travels down to your friend's end. if you pay attention to a given segment of the rope, it moves, but it moves to the right and left while the wave travels forward. in a transverse wave, the constituents of the wave move perpendicular to the direction of motion of the wave itself.

a longitudinal wave, on the other hand, is the opposite. the constituents of the wave alternate moving in the same or opposite direction as the wave moves. this one is more like if you and your friend were holding onto a slinky. you give your end a push towards your friend, and you can see the wave move towards the other end.

hope that helps :D
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The answer is B.

The planet cannot be too hot or too cold it has to be the right distance from its sun to maintain life. 
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3 years ago
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The current theory of the structure of the
Mariana [72]

Answers:

a) 2.82(10)^{21} kg

b) 1410 J

c) 36.62 m/s

Explanation:

<h3>a) Mass of the continent</h3>

Density \rho  is defined as a relation between mass m and volume V:

\rho=\frac{m}{V} (1)

Where:

\rho=2720 kg/m^{3} is the average density of the continent

m is the mass of the continent

V is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}

Finding the mass:

m=\rho V (2)

m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3}) (3)

m=2.82(10)^{21} kg (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy K is given by the following equation:

K=\frac{1}{2}mv^{2} (5)

Where:

m=2.82(10)^{21} kg is the mass of the continent

v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s is the velocity of the continent

K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2} (6)

K=1410 J (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass m=77 kg and the same kinetic energy as that of the continent 1413 J, we can find its velocity by isolating v from (5):

v=\sqrt{\frac{2 K}{m}} (6)

v=\sqrt{\frac{2 (1413 J)}{77 kg}}

Finally:

v=36.62 m/s This is the speed of the jogger

5 0
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What is the purpose of each of the components (each shown with leader and line) of the circuit shown is the diagram?
NemiM [27]

The wires is what is needed to put together the whole thing, kinda like glue when you're gluing a piece of paper on it.

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That energy that comes from the battery, thanks to the wires, it can transfer that said energy to both the switch and light bulb.

And as you flick the switch, it depends of how you put it together, there's two options, turning the light bulb on or turning it off.

Though it doesn't mean that since the light bulb is connected to the battery makes the bulb turn on no matter what since the switch can cancel the main source's energy.

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5 0
3 years ago
what is the approximate weight of a 20-kg cannonball on the moon if the acceleration due to gravity is 1.6m/s^2
monitta
On Earth, a cannonball with a mass of 20 kg would weigh 196 Newtons.
With the formula F=mg, where F is the weight in Newtons, m is the mass, and g is the acceleration due to gravity on the Earth which is 9.8m/s^2.
F=20kg x 9.8m/s^2= 196 Newtons

BUT on the moon, acceleration due to gravity is 1.6 m/s^2,
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5 0
3 years ago
When the mass of the bottle is 0.125 kg, the KE is______ kg m2/s2.
DiKsa [7]
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2


Explanation:

(1) Given mass = 0.125 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2

(2) Given mass = 0.250 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2

(3) Given mass = 0.375 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2

(4) Given mass = 0.500 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
3 0
3 years ago
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