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IceJOKER [234]
3 years ago
12

A toy of mass 0.170-kg is undergoing SHM on the end of a horizontal spring with force constant k = 250 N/m . When the toy is a d

istance 0.0110 m from its equilibrium position, it is observed to have a speed of 0.400 m/s .
A)What is the toy's total energy at any point of its motion? Express your answer with the appropriate units.
B)What is the toy's amplitude of the motion? Express your answer with the appropriate units.
C)What is the toy's maximum speed during its motion? Express your answer with the appropriate units.
Physics
1 answer:
MariettaO [177]3 years ago
7 0

Answer:

a) Em = 2.87 10⁻² J , b) A = 0.0152 m , c) v = 0.581 m / s

Explanation:

a) For this problem we will use the energy ratio. Mechanical energy is conserved or we can calculate it in one place. We fear energy for all places. Caicule the mechanical energy in the place where they give data

    Em = K + K_{e}

    Em = ½ m v² + ½ k x²

    Em = ½ 0.170 0.400² + ½ 250 0.0110²

    Em = 0.0136 + 0.0151

    Em = 0.0287 J

    Em = 2.87 10⁻² J

b) to calculate the amplitude we use the energy

    Em = ½ k A²

    A = √ 2 Em / k

    A = √(2 0.0287 / 250)

    A = 0.01515 m

    A = 0.0152 m

c) the maximum speed

At the point where the elongation is zero, midpoint of the movement, all energy is kinetic and this is the maximum value of the velocity

    Em = ½ m v²

    v = √2 Em / m

    v = √(2 0.0287 / 0.170)

    v = 0.581 m / s

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5213 mg is the answer
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Assuming the average rate of heat energy flowing outwards from earth to be 0.063 W/m2 calculate the energy available at a hot sp
MaRussiya [10]

Answer:

"13.48 Kwhr" is the right solution.

Explanation:

The given values are:

Average rate of heat energy,

= 0.063 W/m²

Diameter,

= 8m

Efficiency of conversion,

= 50%

Now,

The area of hotspot will be:

⇒  A=\frac{\pi}{4} d^2

On substituting the values, we get

⇒      =\frac{3.14}{4} (8)^2

⇒      =0.785\times 64

⇒      =50.24 \ m^2

Total heat generation rate will be:

⇒  Q=q\times A

        =0.063\times 50.24

        =3.16 \ W

hence,

The electricity generation capacity will be:

⇒  P=\eta Q t

On substituting the values, we get

⇒      =0.5\times 3.16\times 8760

⇒      =13840.8 \ Whr

On converting into Kwhr, we get

⇒      =13.84 \ Kwhr

6 0
3 years ago
What is the magnitude of the magnetic dipole moment of the bar magnet
Annette [7]

The magnitude of the magnetic dipole moment of the bar magnet is 1.2 Am²

<h3> Magnetic dipole moment of the bar magnet</h3>

The magnitude of the magnetic dipole moment of the bar magnet at distance from its axis is calculated as follows;

B = \frac{2\mu_0m}{4\pi r^3} \\\\m = \frac{4\pi r^3 B}{2\mu_0}

where;

  • B is magnetic field
  • m is dipole moment
  • μ is permeability of free space

m = (4π x 0.1³ x 2.4 x 10⁻⁴)/(2 x 4π x 10⁻⁷)

m = 1.2 Am²

The complete question is below:

What is the magnitude of the magnetic dipole moment of the bar magnet from 0.1 m of its axis and magnetic field strength of 2.4 x 10⁻⁴ T.

Learn more about dipole moment here: brainly.com/question/27590192

#SPJ11

6 0
2 years ago
A soccer ball with a mass of 0.45 kg is kicked and is moving at 8.9 m/s. Find the kinetic
strojnjashka [21]

Answer:

17.82J

Explanation:

Kinetic energy = 1/2 mv^2

Given

Mass M = 0.45kg

Velocity v = 8.9m/s

Therefore,

K.E. = 1/2 x 0.45 x (8.9)^2

= 1/2 x 0.45 x (8.9 x 8.9)

= 1/2 x 0.45 x 79.21

Multiply through

= 35.6445/2

= 17.82J

The kinetic energy of the ball is 17.82J

3 0
3 years ago
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
Lemur [1.5K]

Answer:

a)   v = 0.9167 m / s,  b)  A = 0.350 m,  c)  v = 0.9167 m / s, d)  A = 0.250 m

Explanation:

a) to find the velocity of the wave let us use the relation

          v = λ f

the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength

           λ = x

           λ = x

the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period

          T / 2 = t

           T = 2t

period and frequency are related

           f = 1 / T

           f = 1 / 2t

we substitute

           v = x / 2t

           v = 5.50 / 2 3

           v = 0.9167 m / s

b) the amplitude is the distance from a maximum to zero

          2A = y

           A = y / 2

           A = 0.700 / 2

           A = 0.350 m

c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same

      v = 0.9167 m / s

d) the amplitude is

           A = 0.500 / 2

           A = 0.250 m

4 0
3 years ago
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