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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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Answer:

Let d be the density of fluid.

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After the level reaches 50cm^3

Final reading of balance , F2 =50dg N

Given that difference between final and initial reading is 30g

i.e, F2 −F1

=30 g

⟹50dg−30dg=30g

⟹20dg=30g

⟹d=30g/20g

⟹d=1.5g/cm^3

So, density of fluid is 1.5g/cm^3

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3 years ago
a man is walking inside a bus which is travelling at 56.2 km/h.if the speed of the man relative to the ground is 55.8 km/h,is he
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6 0
2 years ago
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3 years ago
How Does Earth's gravitational force field act on objects that aren't touching Earth's surface?
natali 33 [55]

Answer:

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8 0
2 years ago
An electric furnace runs 13 hours a day to heat a house during January (31 days). The heating element has a resistance of 7.2 an
liq [111]

Answer:

cost of running the furnace during January is $5619.62

Explanation:

given data

runs a day = 13 hours

January days = 31 days

resistance = 7.2 ohm

current = 16.7 A

cost of electricity = $0.10/kWh

to find out

cost of running the furnace during January

solution

first we get her power consumed by furnace that is

Power consumed = \frac{I^2}{R}  ........1

put here value we get

Power consumed = \frac{16.7^2}{7.2}

Power consumed = 38.7347 W

and

Power consumed by furnace in one hour is

Power consumed by furnace in one hour is = Power consumed × 3600

Power consumed by furnace in one hour is = 38.7347 × 3600  

Power consumed by furnace in one hour is 139.445kWh

and

Power consumed by furnace in the month of January is

Power consumed by furnace in the month of January = 139.445kWh × 13 hours × 31 days

Power consumed by furnace in the month of January = 56196.335 kWh

so

cost of running the furnace during January is = $0.10/kWh × 56196.335 kWh

cost of running the furnace during January is $5619.62

4 0
3 years ago
Read 2 more answers
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