Can someone help me with this question

A real power supply can be modeled as an ideal EMF of 60 Volts in series with an internal resistance. The voltage across the ter

lara31 [8.8K]

Answer:

5 ohms

Explanation:

Given:

EMF of the ideal battery (E) = 60 V

Voltage across the terminals of the battery (V) = 40 V

Current across the terminals (I) = 4 A

Let the internal resistance be 'r'.

Now, we know that, the voltage drop in the battery is given as:

$V_d=Ir$

Therefore, the voltage across the terminals of the battery is given as:

$V= E-V_d\\\\V=E-Ir$

Now, rewriting in terms of 'r', we get:

$Ir=E-V\\\\r=\frac{E-V}{I}$

Plug in the given values and solve for 'r'. This gives,

$r=\frac{60-40}{4}\\\\r=\frac{20}{4}\\\\r=5\ ohms$

Therefore, the internal resistance of the battery is 5 ohms.

5 0

3 years ago

Using a horizontal force of 60 N, a wagon is pushed horizontally across the floor a distance of 12 meters at a constant speed of

RideAnS [48]

Answer:

a) The work done by the force is 720 joules, b) The power supplied by the force is 138 watts.

Explanation:

a) Since force is uniform and parallel to the direction of motion, the work ( $W$ ), in joules, done by the force ( $F$ ), in newtons, is defined by this formula:

$W = F\cdot s$ (1)

Where $s$ is the travelled distance, in meters.

If we know that $F = 60\,N$ and $s = 12\,m$ , then the work done by the force is:

$W = F\cdot s$

$W = (60\,N)\cdot (12\,m)$

$W = 720\,J$

The work done by the force is 720 joules.

b) And an expression for the power supplied by the force ( $\dot W$ ), in watts, is concieved by differentiating (1) in time:

$\dot W = F\cdot \dot s$

Where $\dot s$ is the speed of the wagon, in meters per second.

If we know that $F = 60\,N$ and $\dot s = 2.3\,\frac{m}{s}$ , then the power supplied by the force is:

$\dot W = F\cdot \dot s$

$\dot W = (60\,N)\cdot \left(2.3\,\frac{m}{s} \right)$

$\dot W = 138\,W$

The power supplied by the force is 138 watts.

3 0

2 years ago

What is 2 plus 2 and 1 plus 1

Stels [109]

Answer: 6

Explanation:

Have an awesome amazing day

8 0

2 years ago

Read 2 more answers

A stone is thrown vertically upward with a speed of 18.0 m/s. How fast is it moving when it reaches a height of 11.0m? How long

steposvetlana [31]

Given: v0= 18.0 m/s, y0=0m, yf=11m, g=-9.81 m/s^2

v0= initial velocity, vf= final velocity, y0= initial height, yf= final height, g= gravity, sqrt()= square root, ^2=squared

vf^2=v0^2 + (2)(g)(yf-y0)
vf^2=(18.0 m/s)^2+(2)(-9.81 m/s^2)(11 m-0m)
vf^2=18.0 m/s)^2 + (-19.62 m/s^2)(11 m)
vf^2=(324 m^2/s^2) - (215.82 m^2/s^2)
vf^2=108.18 m^2/s^2
vf=sqrt(108.18 m^2/s^2)
vf=10.4 m/s

8 0

3 years ago

At t= 0, a particle moving in the xy plane with constant acceleration has a velocity of Vi= (3.00i -2.00j) m/s and is at the ori

ivanzaharov [21]

Answer:

(a) a = (2i + 4.5j) m/s^2

(b) r = ro + vot + (1/2)at^2

Explanation:

(a) The acceleration of the particle is given by:

$\vec{a}=\frac{\vec{v}-\vec{v_o}}{t}\\\\$

vo: initial velocity = (3.00i -2.00j) m/s

v: final velocity = (9.00i + 7.00j) m/s

t = 3s

by replacing the values of the vectors and time you obtain:

$\vec{a}=\frac{1}{3s}[(9.00-3.00)\hat{i}+(7.00-(-2.00))\hat{j}]\\\\\vec{a}=(2\hat{i}+4.5\hat{j})m/s^2$

(b) The position vector is given by:

$\vec{r}=\vec{r_o}+\vec{v_o}t+\frac{1}{2}\vec{a}t^2$

where vo = (3.00i -2.00j) m/s and a = (2.00i + 4.50j)m/s^2

4 0

2 years ago