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nadya68 [22]
3 years ago
13

A machine part is undergoing SHM with a frequency of 5.00Hz and amplitude 1.80cm . How long does it take the part to gofrom x=0

to x= -1.80 cm?
Physics
1 answer:
garik1379 [7]3 years ago
6 0

To solve this exercise we will use the expression of the displacement in the aromatic movement in the two given points. Thus we will find the value of time as a function of frequency. In the end we will replace that value and find the time taken,

The expression for the displacement of the object in simple harmonic motion is

x = Acos(2\pi ft)

At the initial position the value of x is zero,

0 = Acost (2\pi ft_1)

cost (2\pi ft_1) = 0

2\pi ft_1 = \frac{\pi}{2}

t _1 = \frac{1}{4f}

Now the expression for the displacement of the object in simple harmonic motion is

x = Acos (2\pi ft)

At the final position the value of x is -1.8cm

Replacing we have that

-1.8=1.8cos(2\pi ft_2)

-1 = cos(2\pi ft_2)

t_2 = \frac{1}{2f}

The total change in time will be

t = t_2-t_1

t = \frac{1}{2f}-\frac{1}{4f}

t = \frac{1}{4f}

Replacing the value of the frequency we have that

t = \frac{1}{4*5}

t = 0.05s

Therefore the time taken for the displacement in the interval of x=0 to x=-1.8 is 0.05s

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Answer:

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Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

            Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

            λ = q_ {int} /l

            q_ {int} = l λ

we substitute

            E A = l λ /ε₀

is area of ​​cylinder is

           A = 2π r l

we substitute

            E = \frac{ l \ \lambda}{ \epsilon_o \ 2\pi  \ r \ l }

             E = \frac{\lambda}{ 2\pi  \epsilon_o \ r}

the amount

            k = 1 / 4πε₀

            E = 2k  \frac{\lambda}{ r}

5 0
3 years ago
Light does not pass through some materials. What do you think happens
Hatshy [7]

Answer:

When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. One way of thinking about this higher energy state is to imagine that the electron is now moving faster, (it has just been "hit" by a rapidly moving photon)

A photon is a quantum of EM radiation. Its energy is given by E = hf and is related to the frequency f and wavelength λ of the radiation by. E=hf=hcλ(energy of a photon) E = h f = h c λ (energy of a photon) , where E is the energy of a single photon and c is the speed of light.

7 0
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Suppose a 2.0kg bird is flying at a speed of of 1m/s. It’s kinetic energy would be what?
Ganezh [65]

As we know that the formula of kinetic energy will be

KE = \frac{1}{2} mv^2

now here we know that

m = 2 kg

v = 1 m/s

so from the above equation we have

KE = \frac{1}{2}(2)(1^2)

KE = 1 J

7 0
3 years ago
If the bonds in the reactants of Figure 7-3 contained 432 kJ of chemical energy and the bonds in the
Yuri [45]

Answer:

  • <u>The energy change would be 46kJ</u>
  • <u>The energy would be absorbed</u>

Explanation:

The <em>energy change </em>during a chemical reation, i.e. the reaction energy, is equal to the chemical energy stored in the<em> bonds of the products </em>less the chemical energy stored in the <em>bonds of the reactants</em>.

Hence:

  • <em>Energy change</em> = 478 kJ - 432kJ = 46kJ

The change is positive, this is, the chemical energy of the products is greater than the chemical energy of the reactants.

That corresponds to the second graph, where the level of the energy of the products in the graph is higher than the level of the energy of the reactants. Therefore, the conclusion is that the reaction <em>absorbed energy</em> and it is endothermic.

4 0
3 years ago
A 70 ft rope hangs from a helicopter above this room. The rope has a mass per unit length of 2 lb/ft. In order to be rescued fro
Mrac [35]

Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope  = 2 lb/ft x 70 ft

                                         = 140 lbs.

Total mass to be pulled to the helicopter, M = 120 lbs  + 140 lbs  

                                                                       = 260 lbs

The work done is calculated from work-energy theorem as follows;

W = Mgh

where;

g is acceleration due gravity = 32.17 ft/s²

h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s²  x 70 ft

W = 585494 lb.ft²/s²

1 lb.ft²/s² = 0.0421 J

W = 585494 lb.ft²/s²  = 2.47 X 10⁴ J.

Therefore, the work done to get you safely away from the test is  2.47 X 10⁴ J.

4 0
2 years ago
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