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Marianna [84]
2 years ago
11

In a wave, what do you call the difference from crest to crest or trough to trough

Physics
1 answer:
UNO [17]2 years ago
7 0
The correct answer is "wavelength".

In fact, this is exactly the definition of wavelength: in a wave, the wavelength is defined as the distance between two succesive crests or two successive troughs, therefore the correct answer is 
D) wavelength.
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Murrr4er [49]
Yeah lowkey i think it’s B
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3 years ago
PLEASE HELP ASAP!!!<br> GIVING BRAINLIEST!!!<br> 40 points!!
Nana76 [90]
W = Fd = 4(2100) = 8400 J

So the answer is A) 8400 J


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4 0
3 years ago
Read 2 more answers
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

         \frac{y_{sphere}} {y_bolck} = 5/7

3 0
3 years ago
Calculate the accleration of a car if its velocity increases from 15m/s to 75m/s in 5 second​
andrezito [222]

Answer:

I think the acceleration is 12m/s

3 0
3 years ago
Need help on this please
marin [14]
The answer is D, the amount of energy stays the same.
3 0
3 years ago
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