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lukranit [14]
2 years ago
6

Given a circuit consisting of a DC battery of voltage 200 volts connected to a single resistor. If the electric current through

the resistor is 50000 mA then which of these is the resistance of the resistor in SI units?A)16B)24C)20D)4E)12
Physics
1 answer:
sertanlavr [38]2 years ago
7 0

Given:

The battery voltage, V=200 V

The current through the resistor, I=50000 mA=50000×10⁻³ A

To find:

The resistance of the resistor.

Explanation:

From Ohm's law, the voltage across a circuit is directly proportional to the current through the circuit.

Thus the voltage across the resistor is given by,

V=IR

On substituting the known values in the above equation,

\begin{gathered} 200=50000\times10^{-3}\times R \\ \Rightarrow R=\frac{200}{50000\times10^{-3}} \\ =4\text{ }\Omega \end{gathered}

Final answer:

The resistance of the resistor is 4 Ω

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The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p
beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

v(t)=ate^{-6t}   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a

hence, the maximum speed is v_max = ((1/6)e^-1)a

5 0
3 years ago
A potential energy function is given by U(x)=(3.00J)x+(1.00J/m^2)x^3. What is the force function F(x) (in newtons) that is assoc
snow_lady [41]

Answer:

F(x)=-3 N -(3N/m^2) x^2

Explanation:

The force is equal to the negative of the derivative of the potential energy function:

F=-\frac{dU}{dx}

In this problem, the potential energy function is

U(x)=3 x + 1 x^3

Therefore, the force function is:

F(x)=-\frac{dU}{dx}=-\frac{d}{dx}(3x+1x^3)=-3-3x^2

So, adding the units of measurement,

F(x)=-3 N -(3N/m^2) x^2

4 0
4 years ago
Please answerASAP
Illusion [34]

Answer:

8.9 m/s^2

Explanation:

The period of a simple pendulum is given by the equation

T=2\pi \sqrt{\frac{L}{g}}

where

L is the lenght of the pendulum

g is the acceleration due to gravity at the location of the pendulum

We notice from the formula that the period of a pendulum does not depend on the mass of the system

In this problem:

-The pendulum comes back to the point of release exactly 2.4 seconds after the release. --> this means that the period of the pendulum is

T = 2.4 s

- The length of the pendulum is

L = 1.3 m

Re-arranging the equation for g, we can find the acceleration due to gravity on the planet:

g=(\frac{2\pi}{T})^2 L=(\frac{2\pi}{2.4})^2(1.3)=8.9 m/s^2

7 0
4 years ago
A person standing on a scale feels a normal force of 655 N pushing up on him. What is his mass? (Unit=kg)
dimulka [17.4K]

Answer:

Approximately 66,8kg

Hope this help :D

Explanation:

F=mg. So 655N= 9.8 m/s² × mass.

Mass = 655 / 9.8

3 0
3 years ago
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Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support
alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

r_{M\perp}(F_M)-r_{W\perp}(W)=0

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

r_{M\perp}(F_M)=r_{W\perp}(W)

Now, we divide by the distance of the muscle:

(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}

Now, we substitute the values:

F_M=\frac{(2.4cm)(50N)}{5.1cm}

Now, we solve the operations:

F_M=23.53N

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

\Sigma F_v=F_j-F_M-W

Since there is no vertical movement the sum of vertical forces is zero:

F_j-F_M-W=0

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

F_j=F_M+W

Now, we plug in the values:

F_j=23.53N+50N

Solving the operations:

F_j=73.53N

Therefore, the force is 73.53 Newtons.

8 0
1 year ago
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