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3 years ago

In the waves lab, you changed ____ to see how it would affect wave speed.

Physics

1 answer:

ikadub [295]3 years ago

6 0

Hi there!

I believe the answer is transversal or transverse.

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A boulder on the mythical planet mongo drops off a cliff and falls from rest 1000 m in 10.0 s. (A) what's the initial speed of t

Amiraneli [1.4K]

At rest, initial speed zero

x=v(initial) t+ 1/2 at^2
-1000m=0(10) + 1/2 a 10^2
-1000m=50a
a = -20 m/s^2

6 0

2 years ago

Can someone help me with science:

Oliga [24]

Answer:

trur

Explanation:

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2 years ago

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A roller coaster car is elevated to a height of 30 m and released from rest to roll along a track. At a certain time T it is at

Naddika [18.5K]

Explanation:

Initial energy = final energy + work done by friction

PE = PE + KE + W

mgH = mgh + 1/2 mv² + W

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000

v = 22.1 m/s

Without friction:

PE = PE + KE

mgH = mgh + 1/2 mv²

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²

v = 23.4 m/s

4 0

2 years ago

You drop your frozen rock from a green bridge. The frozen rock starts from rest (initial velocity = 0ms). The rock takes 4.3s to

valentinak56 [21]

Answer:

The velocity of the frozen rock at $t = 1.5\,s$ is -14.711 meters per second.

Explanation:

The frozen rock experiments a free fall, which is a type of uniform accelerated motion due to gravity and air viscosity and earth's rotation effect are neglected. In this case, we need to find the final velocity ( $v$ ), measured in meters per second, of the frozen rock at given instant and whose kinematic formula is:

$v = v_{o} + g\cdot t$ (Eq. 1)

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$g$ - Gravity acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we get that $v_{o} = 0\,\frac{m}{s}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $1.5\,s$ , then final velocity is:

$v = 0\,\frac{m}{s}+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (1.5\,s)$

$v = -14.711\,\frac{m}{s}$

The velocity of the frozen rock at $t = 1.5\,s$ is -14.711 meters per second.

5 0

2 years ago

Two parallel conducting plates are separated by 9.2 cm, and one of them is taken to be at a potential of zero volts.What is the

True [87]

Answer:

$E=54V/cm$

$\Delta V=496.8V$ between the plates.

Explanation:

The equation for change of voltage between two points separated a distance d inside parallel conducting plates (<em>which have between them constant electric field</em>) is:

$\Delta V=Ed$

So to calculate our electric field strength we use the fact that the potential 8.8 cm from the zero volt plate is 475 V:

$E=\frac{\Delta V}{d}=\frac{475 V}{8.8cm}=54V/cm$

And we use the fact that the plates are 9.2cm apart to calculate the voltage between them:

$\Delta V=Ed=(54V/cm)(9.2cm)=496.8V$

8 0

3 years ago