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ziro4ka
9 days ago
13

The operator of a space station observes a space vehicle approaching at a constant speed v. The operator sends a light signal at

speed c toward the space vehicle. What is the speed of the light signal as viewed from the space vehicle
Physics
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A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at
Ostrovityanka [3204]

Answer

Given:

Wavelength = λ = 18.7 cm

                  = 0.187 m

Amplitude, A = 2.34 cm

Velocity, v = 0.38 m/s

A)  Calculate the angular frequency.

     f = \dfrac{v}{\lambda}

     f = \dfrac{0.38}{0.187}

     f =2.03\ Hz

Angular frequency,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) Calculate the wave number:

      K = \dfrac{2\pi}{\lambda}

     K= \dfrac{2\pi}{0.187}

    K =33.59\ m^{-1}

C)

Since the wave is traveling in the -x direction, the sign is positive between x and t

y (x, t) = A sin(k x - ω t)

y (x, t) = 2.34 sin(33.59 x - 12.75 t)

4 0
3 months ago
A charge of 4.9 x 10-11 C is to be stored on each plate of a parallel-plate capacitor having an area of 150 mm2 and a plate sepa
Keith_Richards [3271]

Answer:2.53*10^-10F

Explanation:

C=£o£r*A/d

Where £ represents the permittivity constant

£o= 8.85*10^-12f/m

£r=6.3

A=150mm^2=0.015m^2

d=3.3mm= 0.0033m

C=8.85*10^-12*6.3*0.015/0.0033

C=8.85*6.3*10^-12*0.015/0.0033

C=55.755*0.015^-12/0.003

C=8.36/3.3*10^-13+3

C=2.53*10^-10F

7 0
3 months ago
Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by th
Yuliya22 [3333]
U = 1794.005 × 10⁶ J. Explanation: Information provided indicates that the capacitance of the original capacitor is C = 1.27 F, and the potential difference applied to it is V = 59.9 kV, or 59.9 × 10³ V. The potential energy (U) for the capacitor is determined by the formula: U = (1/2) × C × V². Substituting the respective values, we find U = (1/2) × 1.27 × (59.9 × 10³)², resulting in U = 1794.005 × 10⁶ J.
7 0
2 months ago
A student wants to determine the coefficient of static friction μ between a block of wood and an adjustable inclined plane. Of t
serg [3582]

Response:

A protractor to gauge the angle between the inclined plane and the horizontal

Explanation:

The student must elevate the free end of the adjustable inclined plane until the object just begins to slide and record the angle at that precise moment. At this juncture, the frictional force is balanced by the weight component aligned with the incline. That is:

f=\mu\,* N = \mu * m g\, cos(\theta)

and  w_{//}= m\,g\,sin(\theta)

Consequently, the coefficient of static friction can be entirely established by calculating the tangent of the angle formed by the incline with the horizontal.

f = w_{//}\\\mu *\,m \,g\,cos(\theta) = m\,g\,sin(\theta)\\\mu = tan(\theta)

For this, the sole additional tool needed is a protractor for angle measurement.

7 0
2 months ago
A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn
Ostrovityanka [3204]
The alteration in potential energy is  \Delta PE = - 3.8*10^{-16} \ J

In the query, it is stated that

  The intensity of the uniform electric field equals E = 950 \ N/C

     The distance the electron covers is  x = 2.50 \ m

Typically, the force exerted on this electron is expressed mathematically as

     F = qE

Where F signifies the force and  q represents the charge of the electron, which is a fixed value of q = 1.60*10^{-19} \ C

    Thus  

      F = 950 * 1.60 **10^{-19}

      F = 1.52 *10^{-16} \ N

Generally, the work-energy theorem is mathematically framed as

          W = \Delta KE

Where W denotes the work done on the electron by the electric field and  \Delta KE  is the change in kinetic energy

Additionally, work done on the electron can also be described as

        W = F* x *cos( \theta )

Where  \theta = 0 ^o assuming that the electron's movement aligns with the x-axis  

        So

             \Delta KE = F * x cos (0)

Inserting values

         \Delta KE = 1.52 *10^{-16} * 2.50 cos (0)

          \Delta KE = 3.8*10^{-16} J

According to the conservation of energy

       \Delta PE = - \Delta KE

Where \Delta PE signifies the change  in  potential energy  

Thus  

        \Delta PE = - 3.8*10^{-16} \ J

               

7 0
2 months ago
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