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vodomira [7]
3 years ago
6

A student swings a container of water in a vertical circle of radius 1.0 m. Calculate the minimum speed of the container so that

the water remains in the container at the top of the circle. ANS. 3.1 m/s E-4, P-12?​
Physics
1 answer:
12345 [234]3 years ago
6 0

Answer:

Explanation:

The centripetal acceleration requirement must equal gravity at the top of the circle

mg = mv²/R

  v = √Rg

  v = √(1.0(9.8))

  v = 3.1304951...

  v = 3.1 m/s

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Describe two sources of earth's energy that are not produced
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6 0
2 years ago
-A 180 kg hippo is riding a bicycle at a speed of 6.0
Vlad1618 [11]

Answer:

0.0675 seconds

Explanation:

From the question,

We apply newton's second law of motion

F = m(v-u)/t.................... Equation 1

Where F = force exert by the brake, v = final speed, u = initial speed m = mass of the bicycle, t = time.

make t the subject of the equation

t = m(v-u)/F................... Equation 2

Given: m = 180 kg, u = 6.0 m/s, v = 0 m/s (comes to stop), F = -1600 N ( agianst the dirction of motion)

Substitute these value into equation 2

t = 180(0-6.0)/-1600

t = -1080/-1600

t = 0.0675 seconds.

8 0
2 years ago
A light source emits light with dominant wavelengths in the range of 650 to 690 nm. what is the principle color of light emitted
dsp73

Answer:

Please see below as the answer is self-explanatory.

Explanation:

  • The visible range extends roughly from 400 nm (violet) to 700 nm (red).
  • Below the violet is the ultra-violet spectrum (with higher energy) and above red, we have the infra-red spectrum.
  • The wavelengths in the range of 650 to 690 nm have red as the dominant color.
6 0
2 years ago
A metal cylinder with a mass of 4.20 kg is attached to a spring and is able to oscillate horizontally with negligible friction.
kherson [118]

Answer:

a) k = 120 N / m

, b)    f = 0.851 Hz

, c)  v = 1,069 m / s

, d)  x = 0

, e)  a = 5.71 m / s²

, f)   x = 0.200 m

, g)  Em = 2.4 J

, h) v = -1.01 m / s

Explanation:

a) Hooke's law is

         F = k x

         k = F / x

          k = 24.0 / 0.200

          k = 120 N / m

b) the angular velocity of the simple harmonic movement is

        w = √ k / m

        w = √ (120 / 4.2)

        w = 5,345 rad / s

Angular velocity and frequency are related.

       w = 2π f

        f = w / 2π

        f = 5.345 / 2π

        f = 0.851 Hz

c) the equation that describes the movement is

        x = A cos (wt + Ф)

As the body is released without initial velocity, Ф = 0

        x = 0.2 cos wt

Speed ​​is

       v = dx / dt

       v = -A w sin wt

The speed is maximum for sin wt = ±1

       v = A w

       v = 0.200 5.345

       v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

       x = A cos wt = 0

       x = 0

e) the acceleration is

       a = d²x / dt² = dv / dt

       a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

       a = A w²

        a = 0.2   5.345

        a = 5.71 m / s²

f) the position for this acceleration is

       x = A cos wt

       x = A

       x = 0.200 m

g) Mechanical energy is

        Em = ½ k A²

        Em = ½ 120 0.2²

       Em = 2.4 J

h) the position is

         x = 1/3 A

Let's calculate the time to reach this point

         x = A cos wt

        1/3 A = A cos 5.345t

         t = 1 / w cos⁻¹(1/3)

The angles are in radians

t = 1.23 / 5,345

t = 0.2301 s

Speed ​​is

v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

v = -1.01 m / s

i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

      a = -1.91 m / s²

5 0
3 years ago
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