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elena55 [62]
3 years ago
13

In a waterfall, much of the energy of the falling water is converted into heat. If all the mechanical energy is converted into h

eat that stays in the water, how much of a rise in temperature occurs in a 100 m waterfall? For water, c = 4.186 × 103 J/(kg • C°).
Physics
1 answer:
tiny-mole [99]3 years ago
6 0
It’s A... hope these helped
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A charge Q is distributed uniformly on a non-conducting ring of radius R and mass M. The ring is dropped from rest from a height
mihalych1998 [28]

Answer:

Check below for the explanation

Explanation:

Since it is stated that the ring is dropped from a height, h, through a non uniform magnetic field, two kinds of force will act on the ring, namely:

  • A magnetic force (that is non uniform since the field is  non uniform)
  • Gravitational force

A certain amount of torque is provided by the non uniform magnetic force on the ring while the force gravity pulls it down. Due to the downward pull by the force of gravity on the ring and the torque acting on it as a result of the non uniform magnetic force, the ring begins to rotate.

5 0
3 years ago
Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed
likoan [24]

Answer:

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let t represents the time in seconds and v the speed in meters-per-second.

For 0 < x \le 1:

  • Initial value of v: \rm 2\;m\cdot s^{-1} at t = 0; Hence the point on the segment: (0, 2).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 2\; m\cdot s^{-2}.
  • Find the equation of this segment in slope-point form: v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1.

Similarly, for 1 < x \le 2:

  • Initial value of v is the same as the final value of v in the previous equation at t = 1: t = 2t + 2 = 4; Hence the point on the segment: (1, 4).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 1\; m\cdot s^{-2}.
  • Find the equation of this segment in slope-point form: v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2.

For 2 < x \le 3:

  • Initial value of v is the same as the final value of v in the previous equation at t = 2: t = t + 3 = 5; Hence the point on the segment: (2, 5).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 0\; m\cdot s^{-2}. There's no acceleration. In other words, the velocity is constant.
  • Find the equation of this segment in slope-point form: v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3.

For 3 < x \le 4:

  • Initial value of v is the same as the final value of v in the previous equation at t = 3: t = 5; Hence the point on the segment: (3, 5).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm -3\; m\cdot s^{-2}. In other words, the velocity is decreasing.
  • Find the equation of this segment in slope-point form: v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4.

For 4 < x \le 5:

  • Initial value of v is the same as the final value of v in the previous equation at t = 4: t = -3t + 14; Hence the point on the segment: (4, 2).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 0\; m\cdot s^{-2}. In other words, the velocity is once again constant.
  • Find the equation of this segment in slope-point form: v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5.

t = \rm 3.49\;s is in the interval 3 < x \le 4. Apply the equation for that interval: v = -3t +14 = \rm 3.53\; m \cdot s^{-1}.

4 0
3 years ago
Can you make sunglasses from new 3D-glasses???
katen-ka-za [31]
Yes you can make sunglasses from 3d glasses

5 0
3 years ago
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Which type of radiation on the electromagnetic spectrum has the greatest energy and why?
nika2105 [10]

Answer:

Gamma rays

Explanation:

Have the highest energies. The shortest wavelengths, and the highest frequencies.

4 0
3 years ago
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A cello string vibrates in its fundamental mode with a frequency of 335 1/s. The vibrating segment is 28.5 cm long and has a mas
Inga [223]

Answer:

The tension in string is found to be 188.06 N

Explanation:

For the vibrating string the fundamental frequency is given as:

f1 = v/2L

where,

f1 = fundamental frequency = 335 Hz

v = speed of wave

L = length of string = 28.5 cm = 0.285 m

Therefore,

v = f1 2L

v = (335 Hz)(2)(0.285)

v = 190.95 m/s

Now, for the tension:

v = √T/μ

v² = T/μ

T = v² μ

where,

T = Tension

v = speed = 190.95 m/s

μ = linear mass density of string = mass/L = 0.00147 kg/0.285 m = 5.15 x 10^-3 kg/m

Therefore,

T = (190.95 m/s)²(5.15 x 10^-3 kg/m)

<u>T = 188.06 N</u>

4 0
3 years ago
Read 2 more answers
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