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Maslowich
3 years ago
13

As waves pass by a duck floating on a lake, the duck bobs up and down but remains in essentially one place. explain why the duck

is not carried along by the wave motion.
Physics
1 answer:
Rudiy273 years ago
3 0
The duck is not carried by the wave motion because <span>Waves move energy through a medium but don't transfer particles.  So when the waves pass by a duck floating on a lake, the duck just bobs up and down, but remains in its place.</span>
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an object of mass 8 kg is whriled round in a vertical circle of radius 2m with a constant speed of 6m/s .Then the maximum and mi
algol13

Answer:

Maximum Tension=224N

Minimum tension= 64N

Explanation:

Given

mass =8 kg

constant speed = 6m/s .

g=10m/s^2

Maximum Tension= [(mv^2/ r) + (mg)]

Minimum tension= [(mv^2/ r) - (mg)]

Then substitute the values,

Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N

Minimum tension= [8 × 6^2)/2 -(8×9.8)]

=64N

Hence, Minimum tension and maximum Tension are =64N and 2224N respectively

5 0
2 years ago
You add 50 mL of water at 20C to 200 mL of water at 70C. What is the most likely final temperature of the mixture?
Veseljchak [2.6K]
100*20 + 200*80 = 300* T
T = 60
5 0
3 years ago
Read 2 more answers
Two books with different masses fall off the same bookshelf. As they fall,which has more kinetic energy and why
Alex
Kinetic energy E = m * v^2 

<span>Since the acceleration of both books will be -neglecting air resistance - the same, the kinetic energy will be directly proportional to the mass of the book.</span>
5 0
3 years ago
A golf ball is dropped from rest from a height of 9.5m.  It hits the pavement then bounces back up rising  just 5.7 m before fal
Oksanka [162]

Answer: 3.4s

Explanation:

There are three stages in the motion of the ball, so you have to calculate the times for every stage.

1) Ball dropping from 9.5m: free fall

d = Vo + gt² / 2

Vo = 0 ⇒ d = gt² / 2 ⇒ t² = 2d / g = 2 × 9.5 m / 9.81 m/s² = 1.94 s²

⇒ t = √ (1.94 s²) = 1.39s

2) Ball rising 5.7m (vertical rise)

i) Determine the initial speed:

Vf² = Vo² - 2gd

Vf² = 0 ⇒ Vo² = 2gd = 2 × 9.81 m/s² × 5.7m = 111.8 m²/s²

⇒ Vo = 10.6 m/s

ii) time rising

Vf = Vo - gt

Vf = 0 ⇒ Vo = gt ⇒

t = Vo / g = 10.6 m/s / 9.81 m/s² = 1.08 s

3) Ball dropping from 5.7 m to 1.20m above the pavement (free fall)

i) d = 5.7m - 1.20m = 4.5m

ii) d = gt² / 2 ⇒ t² = 2d / g = 2 × 4.5 m / 9.81 m/s² = 0.92 s²

⇒ t = √ (0.92 s²) = 0.96s

4) Total time

t = 1.39s + 1.08s + 0.96s = 3.43s ≈ 3.4s

4 0
3 years ago
Read 2 more answers
How much heat is required to convert 2.55g of water at 28 degrees c to steam?
pantera1 [17]
There are two different processes here:
1) we must add heat in order to bring the temperature of the water from 28^{\circ}C to 100^{\circ}C (the temperature at which the water evaporates)
2) other heat must be added to make the water evaporates

1) The heat needed for process 1) is
Q_1=m C_s \Delta T
where 
m=2.55 g is the water mass
C_s = 4.18 g/J^{\circ}C is the water specific heat
\Delta T=100^{\circ}C-28^{\circ}C=72^{\circ}C is the variation of temperature of the water
If we plug the numbers into the equation, we find
Q_1 = (2.55 g)(4.18 J/g^{\circ}C)(72^{\circ}C)=767.4 J

2) The heat needed for process 2) is
Q=m L_e
where 
m=2.55 g is the water mass
L_e = 2264.7 J/g is the latent heat of evaporation of water
If we plug the numbers into the equation, we find
Q_2=(2.55 g)(2264.7 J/g)=5775.0 J

So, the total heat needed for the whole process is
Q=Q_1+Q_2=767.4 J+5775.0 J=6542.4 J
4 0
3 years ago
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