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alisha [4.7K]
3 years ago
13

If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w

ith the appropriate units. ΔEΔ E = nothing nothing Request Answer Provide Feedback
Physics
1 answer:
marshall27 [118]3 years ago
5 0

Answer:

P(3600)=593.247W

Explanation:

First, let's find the voltage through the resistor using ohm's law:

V=IR=20*8=160V

AC power as function of time can be calculated as:

P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)  (1)

Where:

\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency

Because of the problem doesn't give us additional information, let's assume:

\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W

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How many significant figures does the number 91010.0 have?
kirza4 [7]

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Can someone help me with this question
Mademuasel [1]

Answer:

Net force: 20 N to the right

mass of the bag: 20.489 kg

acceleration:  0.976  m/s^2

Explanation:

Since the normal force and the weight are equal in magnitude but opposite in direction, they add up to zero in the vertical direction. In the horizontal direction, the 195 N tension to the right minus the 175 force of friction to the left render a net force towards the right of magnitude:

195 N - 175 N = 20 N

So net force on the bag is 20 N to the right.

The mass of the bag can be found using the value of the weight force: 201 N:

mass = Weight/g = 201 / 9.81 = 20.489 kg

and the acceleration of the bag can be found as the net force divided by the mass we just found:

acceleration = 20 N / 20.489 kg = 0.976  m/s^2

8 0
3 years ago
The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: =E−Ryn2 In this equation Ry stands for
skelet666 [1.2K]

Answer:

  λ = 162 10⁻⁷ m

Explanation:

Bohr's model for the hydrogen atom gives energy by the equation

         E_{n} = - k²e² / 2m (1 / n²)

Where k is the Coulomb constant, e and m the charge and mass of the electron respectively and n is an integer

The Planck equation

           E = h f

The speed of light is

          c = λ f

          E = h c /λ

For a transition between two states we have

          E_{n} - E_{m} = - k²e² / 2m (1 / n_{f}² -1 / n_{i}²)

           h c / λ = -k² e² / 2m (1 / n_{f}² - 1/ n_{i}²)

           1 / λ = (- k² e² / 2m h c) (1 / n_{f}² - 1/n_{i}²)

The Rydberg constant with a value of 1,097 107 m-1 is the result of the constant in parentheses

Let's calculate the emission of the transition

            1 /λ = 1.097 10⁷ (1/10² - 1/8²)

            1 / λ = 1.097 10⁷ (0.01 - 0.015625)

            1 /λ = 0.006170625 10⁷

            λ = 162 10⁻⁷ m

3 0
4 years ago
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x= 5.0 cos
lakkis [162]

Answer:

a)   x = 4.33 m ,   b)  w = 2 rad / s ,  f = 0.318 Hz ,  c) a = - 17.31 cm / s²,  

d) T =  3.15 s,  e)  A = 5.0 cm

Explanation:

In this exercise on simple harmonic motion we are given the expression for motion

          x = 5 cos (2t + π / 6)

they ask us for t = 0

a) the position of the particle

      x = 5 cos (π / 6)

      x = 4.33 m

remember angles are in radians

 

b) The general form of the equation is

          x = A cos (w t + Ф)

when comparing the two equations

         w = 2 rad / s

angular velocity and frequency are related

          w = 2π f

           f = w / 2π

           f = 2 / 2pi

           f = 0.318 Hz

c) the acceleration is defined by

      a == d²x / dt²

      a = - A w² cos (wt + Ф)

for t = 0 ,  we substitute

      a = - 5,0 2² cos (π / 6)

      a = - 17.31 cm / s²

d) El period is

          T = 1/f

         T= 1/0.318

         T =  3.15 s

e) the amplitude

        A = 5.0 cm

3 0
3 years ago
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