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natulia [17]
2 years ago
13

1) My 14V car battery could be used to charge my laptop, but I need to use an inverter to first convert it to a standard 120V. T

hen my laptop charge cord has a resistor inside which ensures that 1.6A of current flow through to the laptop. What is the resistance in the charge cord?
2) My laptop charge cord actually has another inverter inside tat converts the electricity to 19.5V and 3.33A. What is the resistance in the cord that accomplishes this?
Physics
1 answer:
LenaWriter [7]2 years ago
8 0

Answer:

1) Charge chord resistance is 75 Ω

2) Charge chord resistance is 6.33 Ω

Explanation:

1) To answer the question, we note that the the formula voltage is found as follows;

V = IR

Therefore,

R = \frac{V}{I} =  \frac{120}{1.6} = 75  \, \Omega

2) Where the voltage, V = 19.5 V and the current, I = 3.33 A, we have;

Initial resistance R₁ = 19.5 V/(3.33 A) = 5.86 Ω

However, to reduce the current to 1.6 A, we have;

R_T = \frac{19.5}{1.6} = 12.1875 \ \Omega

Therefore, where the resistance is found by the sum of the total resistance we have;

R_T = R₁ + Charge chord resistance

∴ 12.1875 = 5.86 + Charge chord resistance

Hence, charge chord resistance = 6.33 Ω

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A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-N pull
Elden [556K]

Answer:

(A) 374.4 J

(B) -332.8 J

(C) 0 J

(D) 41.6 J

(E)  351.8 J

Explanation:

weight of carton (w) = 128 N

angle of inclination (θ) = 30 degrees

force (f) = 72 N

distance (s) = 5.2 m

(A) calculate the work done by the rope

  • work done = force x distance x cos θ
  • since the rope is parallel to the ramp the angle between the rope and

        the ramp θ will be 0

       work done = 72 x 5.2 x cos 0

       work done by the rope = 374.4 J

(B) calculate the work done by gravity

  • the work done by gravity = weight of carton x distance x cos θ
  • The weight of the carton = force exerted by the mass of the carton = m x g
  • the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.

      work done by gravity = 128 x 5.2  x cos 120

      work done by gravity = -332.8 J

(C) find the work done by the normal force acting on the ramp

  • work done by the normal force = force x distance x cos θ
  • the angle between the normal force and the ramp is 90 degrees

       

         work done by the normal force = Fn x distance x cos θ

         work done by the normal force = Fn x 5.2 x cos 90

         work done by the normal force = Fn x 5.2 x 0

         work done by the normal force = 0 J

(D)  what is the net work done ?

  • The net work done is the addition of the work done by the rope,       gravitational force and the normal force

     net work done = 374.4 - 332.8 + 0 =  41.6 J  

(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal

  • work done by the rope= force x distance x cos θ
  • the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp

work done = 72 x 5.2 x cos 20

work done = 351.8 J

5 0
2 years ago
A block is projected with speed v across a horizontal surface and slides to a stop due to friction. The same block is then proje
natka813 [3]

Answer: C. The case on the inclined surface had the least decrease intotal mechanical energy.

Explanation:

First and foremost, it should be noted that the mechanical energy is the addition of the potential and the kinetic energy.

From the information given, it should be known that when the block is projected with the same speed v up an incline where is slides to a stop due to friction, the box will lose its kinetic energy but there'll be na increase in the potential energy as a result of the veritcal height. This then brings about an increase in the mechanical energy.

Therefore, the total mechanical energy of the block will decrease the least when the case on the inclined surface had the least decrease intotal mechanical energy.

8 0
1 year ago
3) A defense football player on one team tackles the other team’s quarterback, who is running down the field. The quarterback is
alisha [4.7K]
Newton's first law says that an object at rest tends to stay at rest while an object in motion stays in motion at a constant velocity unless acted upon by an outside force so the amount of force behind the defensive football player (N) was greater than the quarterback's so he was able to over power him which is also called unbalanced forces
5 0
2 years ago
Two sources emit beams of light of wavelength 550 nm. The light from source A has an intensity of 10 μW/m2, and the light from s
alexira [117]

Answer:

<em>A) Beam B carries twice as many photons per second as beam A.</em>

Explanation:

If we have two waves with the same wavelength, then their intensity is proportional to their power, or the energy per unit time.

We also know that the amount of photon present in an electromagnetic beam is proportional to the energy of the beam, hence the amount of beam per second is proportional to the power.

With these two facts, we can say that the intensity is a measure of the amount of photon per second in an electromagnetic beam. So we can say that <em>beam B carries twice as more power than beam A, or Beam B carries twice as many photons per second as beam A.</em>

3 0
2 years ago
How, if at all, would the equations written in Parts C and E change if the projectile was thrown from the cliff at an angle abov
sveta [45]

Answer:

x = v₀ cos θ   t ,   y = y₀ + v₀ sin θ t - ½ g t2

Explanation:

This is a projectile launch exercise, in this case we will write the equations for the x and y axes

Let's use trigonometry to find the components of the initial velocity

              sin θ = v_{oy} / v₀

              cos θ = v₀ₓ / v₀

              v_{y} = v_{oy} sin θ

              v₀ₓ = vo cos θ

now let's write the equations of motion

X axis

         x = v₀ₓ t

         x = v₀ cos θ   t

        vₓ = v₀ cos θ

Y axis

        y = y₀ + v_{oy} t - ½ g t2

        y = y₀ + v₀ sin θ t - ½ g t2

        v_{y} = v₀ - g t

       v_{y} = v₀  sin θ - gt

        v_{y}^{2} = v_{oy}^2 sin² θ - 2 g y

As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components

7 0
2 years ago
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