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ehidna [41]
3 years ago
5

Why does the rider continue to go the same height on the ramp every time and would continue to do that forever?

Physics
1 answer:
Aleks [24]3 years ago
3 0
It’s called conservation of energy. The system (the rider) can’t randomly gain energy without something else giving it energy because energy can’t be created or destroyed, only transferred. The pendulum example is beyond popular and you see physics professors doing it all the time. I think your teacher wants you to say it’ll continue forever, but I don’t think it will. The system will most likely lose energy to its surroundings. And there’s things like air resistance which will, overtime, slow it down. So no, it won’t go on forever
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A piece of amber is charged by rubbing with a piece of fur. If the net excess charge on the fur is +8.0 nC (+8.0 10-9 C), how ma
Ann [662]

Answer:

Number of electron on the fur will be 5\times 10^{10}electron

Explanation:

We have given net charge on the fur q=8nC=8\times 10^{-9}C

Charge on one electron e=1.6\times 10^{-19}C

Let there are n number of electrons

So ne=8\times 10^{-9}

n\times 1.6\times 10^{-19}=8\times 10^{-9}

n=5\times 10^{10}electron

So number of electron on the fur will be 5\times 10^{10}electron

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(URGENT PLS HELP)The diagram shows a wire between the poles of a magnet.
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You're driving down the highway late one night at 20 m/s when a deer steps onto the road 39 m in front of you. Your reaction tim
miss Akunina [59]

Answer:

a) 10.8 m

b) 24.3 m/s

Explanation:

a)

  • In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.
  • The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.
  • We can find this distance simply applying the definition of average velocity, as follows:

       \Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)

  • The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.
  • We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):

       v_{1f} ^{2}  - v_{1o} ^{2} = 2* a* \Delta x  (2)

  • where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².
  • Solving for Δx, we get:

       \Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)

  • So, the total distance traveled was the sum of (1) and (3):
  • Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)
  • Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.

b)

  • We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.
  • The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.
  • ⇒Δx = Δx₁ + Δx₂ = 39 m. (5)
  • Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:

       \Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)

  • Δx₂, is the distance traveled while decelerating, and can be obtained  using (2):

        v_{omax} ^{2} = 2* a* \Delta x_{2} (7)

  • Solving for Δx₂, we get:

       \Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a}  = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)

  • Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.
  • Taking the positive root in the quadratic formula, we get the following value for vomax:
  • v₀max = 24.3 m/s.
6 0
3 years ago
The fourth harmonic on a string fixed at both ends shows
Charra [1.4K]

Answer:

I believe the answer is B) Two wavelengths

8 0
3 years ago
A spherical balloon is being inflated. Find the rate of increase of the surface area (S = 4Ï€r2) with respect to the radius r wh
erastovalidia [21]

Answer:

A) 8π ft²/ft

B) 24π ft²/ft

C) 48π ft²/ft

Explanation:

Surface area of the spherical balloon is not clear here but it is supposed to be;

S = 4πr²

where:

r is the radius of the spherical balloon

So thus, the rate of change of the surface area of the spherical balloon by its radius will be:

dS/dr = 8πr

A) at r = 1ft;

dS/dr = 8 × π × 1

dS/dr = 8π ft²/ft

B) at r = 3 ft;

dS/dr = 8 × π × 3

dS/dr = 24π ft²/ft

C) at r = 6ft;

dS/dr = 8 × π × 6

dS/dr = 48π ft²/ft

8 0
3 years ago
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