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jek_recluse [69]
3 years ago
15

In a simple electric circuit, ohm's law states that v=irv=ir, where vv is the voltage in volts, ii is the current in amperes, an

d rr is the resistance in ohms. assume that, as the battery wears out, the voltage decreases at 0.030.03 volts per second and, as the resistor heats up, the resistance is increasing at 0.040.04 ohms per second. when the resistance is 200200 ohms and the current is 0.040.04 amperes, at what rate is the current changing?
Physics
1 answer:
Tju [1.3M]3 years ago
8 0
We take the derivative of Ohm's law with respect to time: V = IR
Using the product rule:
dV/dt = I(dR/dt) + R(dI/dt)
We are given that voltage is decreasing at 0.03 V/s, resistance is increasing at 0.04 ohm/s, resistance itself is 200 ohms, and current is 0.04 A. Substituting:
-0.03 V/s = (0.04 A)(0.04 ohm/s) + (200 ohms)(dI/dt)
dI/dt = -0.000158 = -1.58 x 10^-4 A/s
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You glide 28 units to the +y-direction with respect to the +x-axis, then move 32 units with respect to the +x-direction. What is
NNADVOKAT [17]

This question can be solved by using Pythagora's Theorem.

The resultant magnitude of the movement is "42.5 units".

The x and y components of the movement are given. We can use Pythagora's Theorem to find the resultant of these movements. Hence, applying the Pythagora's Theorem<em>:</em>

d = \sqrt{d_x^2+d_y^2}

where,

d = resultant movement = ?

d_x = movement in x direction = 32 units

d_y = movement in y direction = 28 units

Therefore,

d = \sqrt{(32\ units)^2+(28\ units)^2}

<u>d = 42.5 units</u>

Learn more about Pythagora's Theorem here:

brainly.com/question/343682?referrer=searchResults

The attached picture shows Pythagora's Theorem<em>.</em>

4 0
2 years ago
Children are sled riding on a hill One little girl pulls her sled back up the hill and does 379.5 joules of work while pulling i
Serggg [28]

Answer:

2.2N

Explanation:

Given parameters:

Work done  = 379.5J

Height  = 173m

Unknown:

Amount of force exerted on the sled  = ?

Solution:

The amount of force she exerted on the sled is the same as her weight.

Work done is the force applied to move a body through a distance.

      Work done  = mgh

m is the mass

g is the acceleration due to gravity

h is the height

     mg  = weight;

        Work done  = weight x h

           379.5 = weight  x  173

           weight  = \frac{379.5}{173}   = 2.2N

4 0
2 years ago
Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.3
MAXImum [283]

Answer:

Δθ = 15747.37 rad.

Explanation:

  • The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.
  • Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:

       \omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)

  • Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:

       \omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)

  • Solving for Δθ in (2):

       \theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)

  • The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:

       \theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)

  • Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:

       \omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)

  • Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:

      \theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)

  • The total angular displacement is just the sum of (3), (4) and (6):
  • Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad
  • ⇒ Δθ = 15747.37 rad.
4 0
2 years ago
A football is thrown due north across a 40 meter river and it takes 2 seconds to cover that distance.
Alexxx [7]

Answer:

I. Speed = 20m/s

II. Velocity = 20m/s due North.

Explanation:

<u>Given the following data;</u>

Distance = 40m

Time = 2secs

To find the speed;

Mathematically, speed is given by the formula;

Speed = \frac{distance}{time}

Substituting into the equation, we have;

Speed = \frac{40}{2}

<em>Speed = 20m/s.</em>

In physics, we use the same formula for calculating speed and velocity. The only difference is that speed is a scalar quantity and as such has magnitude but no direction while velocity is a vector quantity and as such it has both magnitude and direction.

Velocity = \frac{distance}{time}

<em>Therefore, the velocity is 20m/s due North</em>.

6 0
3 years ago
Energy conservation
balandron [24]

Answer:

7.328m/s

Explanation:

Given parameters:

height of table = 0.68m

final velocity of the ball = 6m/s

Unknown:

Initial velocity of ball = ?

Solution:

To solve this problem, we are going to employ the appropriate motion equation.

We must understand that this fall occurs in the presence of gravity;

            V = U + 2gH

Where;

 V is the final velocity

 U is the initial velocity

 g is the acceleration due to gravity

 H is the height of the pool table

    Since U is the unknown, let us make it the subject of the expression;

           U = V - 2gH

        U = 6 - (2 x 9.8 x 0.68)  = 7.328m/s(deceleration)

7 0
3 years ago
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