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alexandr1967 [171]
2 years ago
10

Which example best illustrates the transfer of energy between two waves?

Physics
1 answer:
FromTheMoon [43]2 years ago
7 0

Answer:

A) A buoy rises in the water as a boat speeds past.

Explanation:

The passing boat transfers energy in the form of a wave. Other options illustrate other physics concepts like gravity (falling egg) or Newton's law (for every action, there is an equal and opposite reaction).

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The element in an incandescent light bulb that releases light energy is
Leviafan [203]
A thin tungsten filament
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2 years ago
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Elza [17]
1.cool down
2.activity log
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6 0
3 years ago
The engineer of a passenger train traveling at 25.0m/s sights a freight train whose caboose is 200m ahead on the same track. The
Vika [28.1K]

Answer:

No, there won't be a collision.

Explanation:

We will use the constant acceleration formulas to calculate,

v = u + a*t

0 = 25 + (-0.1)*t

t = 250 seconds (the time taken for the passenger train to stop)

v^2 = u^2 + 2*a*s

0 = (25)^2  + 2*(-0.1)*s

s = 3125 m (distance traveled by passenger train to stop)

If the distance traveled by freight train in 250 seconds is less than (3125-200=2925 m) than the collision will occur

Speed*time = distance

Distance = (15)*(250)

Distance = 3750 m

As the distance is way more, there won’t be a collision

5 0
3 years ago
A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The
nordsb [41]

Answer:

t=6.96s

Explanation:

From this exercise, our knowable variables are <u>hight and initial velocity </u>

v_{oy}=96ft/s

y_{o}=112ft

To find how much time does the <u>ball strike the ground</u>, we need to know that the final position of the ball is y=0ft

y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}

0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}

Solving for t using quadratic formula

t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}

a=-\frac{1}{2} (32.2)\\b=96\\c=112

t=-0.999s or t=6.96s

<u><em>Since time can't be negative the answer is t=6.96s</em></u>

7 0
3 years ago
A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.
matrenka [14]

Answer:

Part a)

T = 42 N

Part b)

v_f = 11.8 m/s

Part c)

t = 1.7 s

Part d)

F = 159.7 N

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

mg - T = ma

for uniform cylinder we will have

TR = I\alpha

so we have

T = \frac{1}{2}MR^2(\frac{a}{R^2})

T = \frac{1}{2}Ma

now we have

mg = (\frac{M}{2} + m)a

a = \frac{mg}{(\frac{M}{2} + m)}

a = \frac{15 \times 9.81}{(6 + 15)}

a = 7 m/s^2

now we have

T = \frac{12 \times 7}{2}

T = 42 N

Part b)

speed of the bucket can be found using kinematics

so we have

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(7)(10)

v_f = 11.8 m/s

Part c)

now in order to find the time of fall we can use another equation

v_f - v_i = at

11.8 - 0 = 7 t

t = 1.7 s

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

F = T + Mg

F = 42 + (12 \times 9.81)

F = 159.7 N

5 0
2 years ago
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