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Charra [1.4K]
3 years ago
8

The frequency of a sound wave in water is 15,000 Hz, and the sound wave travels through water at a speed of 1,500 m/s. What is t

he wavelength?
Physics
1 answer:
motikmotik3 years ago
8 0
I think the answer would be 150 m/s
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A student touches sphere x and moves it close to, but not touching sphere y. What are the natures of the charges left on the two
e-lub [12.9K]
No charge I know this because
7 0
3 years ago
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon
aleksklad [387]

Answer:

230 N

Explanation:

At the lowest position , the velocity is maximum hence at this point, maximum support force  T  is given by the branch.

The swinging motion of the ape on a vertical circular path , will require

a centripetal force  in upward direction . This is related to weight as follows

T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T =  mg -  mv² / R

T = 8.5 X 9.8 + 8.5 X 3.2² / .60  { R is length of hand of ape. }

T = 83.3 + 145.06

= 228.36

= 230 N ( approximately )

5 0
2 years ago
What is the angular displacement of a minute hand of a clock after 3 minutes?​
Inessa [10]

Answer:

π/10 rads

Explanation:

It takes an hour (60 minutes) for the minute's hand to turn a full circle or achieve an angular rotation of

2πl rad.

Now, number of periods of 3 minutes in an hour is;

Number of periods = 60/3 = 20 periods

Thus, 3 minutes rotation accounts for 1/20 of 2π the rotation of the minute's hand in an hour.

Thus;

Angular displacement = (1/20) * 2π = π/10 rads

6 0
1 year ago
Magnet A and B are of equal magnetic strength and which position will magnets A&B have the greatest attraction for toward ea
77julia77 [94]
When they are facing each other, almost touching
3 0
3 years ago
Read 2 more answers
Lloyd is standing on a scaffolding 12 meters above the ground to clean the windows of a tall building. His bucket, which has a m
Sever21 [200]

Answer:

U₂ = 20 J

KE₂ = 40 J

v= 12.64 m/s

Explanation:  

Given that

H= 12 m

m = 0.5 kg

h= 4 m

The potential energy at position 1

U₁ = m g H

U₁ = 0.5 x 10 x 12        ( take g= 10 m/s²)

U₁ = 60 J

The potential energy at position 2

U₂ = m g h

U ₂= 0.5 x 10 x 4        ( take g= 10 m/s²)

U₂ = 20 J

The kinetic energy at position 1

KE= 0

The kinetic energy at position 2

KE= 1/2 m V²

From energy conservation

U₁+KE₁=U₂+KE₂

By putting the values

60 - 20 = KE₂

KE₂ = 40 J

lets take final velocity is v m/s

KE₂= 1/2 m v²

By putting the values

40 = 1/2 x 0.5 x v²

160 = v²

v= 12.64 m/s

3 0
2 years ago
Read 2 more answers
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