A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 V. If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answers

Answer 1

Answer:

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

The additional resistance is $R_z = 4.4 \Omega$

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

The rate at which internal energy increase in the battery is $Z_1 = 32 W$

The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

So this implies that

The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = (E)/(R_T)$

Substituting values

$4 = (28)/(R_z + 2.6)$

Making $R_z$ the subject of the formula

So $R_z = (28 - 10.4)/(4)$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

$Z_1 = 4^2 * 2$

$Z_1 = 32 W$

The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

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A 0.060 ???????? tennis ball, moving with a speed of 5.28 m/???? , has a head-on collision with a 0.080 ???????? ball initially moving in the same direction at a speed of 3.00 m/ ???? . Assume that the collision is perfectly elastic. Determine the velocity (speed and direction) of both the balls after the collision.

Answers

Explanation:

It is given that,

Mass of the tennis ball, $m_1=0.06\ kg$

Initial speed of tennis ball, $u_1=5.28\ m/s$

Mass of ball, $m_2=0.08\ kg$

Initial speed of ball, $u_2=3\ m/s$

In case of elastic collision, the momentum remains conserved. The momentum equation is given by :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$v_1\ and\ v_2$ are final speed of tennis ball and the ball respectively.

$0.06* 5.28+0.08* 3=0.06v_1+0.08v_2$

$0.06v_1+0.08v_2=0.5568$ ..............(1)

We know that the coefficient of restitution is equal to 1. It is given by :

$(v_2-v_1)/(u_1-u_2)=1$

$(v_2-v_1)/(5.28-3)=1$

${v_2-v_1}=2.28$ .................(2)

On solving equation (1) and (2) to find the values of velocities after collision.

$v_1=5.28\ m/s$

$v_2=3\ m/s$

So, the speed of both balls are 5.28 m/s and 3 m/s respectively. Hence, this is the required solution.

When a tennis ball is spun around in a circle on a string and the string breaks the tennis ballwill be pulled in a curved path away from the center because of Centrifugal force
True or false

Answers

Answer:

False

whenever the string breaks, the ball will follow the straight line tangential path

Explanation:

No, the ball will not follow a curved path after the string breaks. Since, the the direction of velocity is tangential to each point of the circular motion. Therefore, it changes at every point. This produces an acceleration in the circle called centripetal acceleration. There is also a tangential component of acceleration acting on the ball during this motion.

So, whenever the string breaks, the ball will follow the straight line tangential path. Hence, the given statement is false.

An electric current in a conductor varies with time according to the expression i(t) = 110 sin (120πt), where i is in amperes and t is in seconds. what is the total charge passing a given point in the conductor from t = 0 to t = 1/180 s?

Answers

As we know that current is defined as rate of flow of charge

$i = (dq)/(dt)$

so by rearranging the equation we can say

$q = \int i dt$

here we know that

$i(t) = 110 sin(120\pi t)$

here we will substitute it in the above equation

$q = \int 110 sin(120\pi t) dt$

$q = 110 [- (cos(120\pi t))/(120\pi)]$

now here limits of time is from t = 0 to t = 1/180s

so here it will be given as

$q = (110)/(120\pi)( -cos0 + cos((2\pi)/(3)))$

$q = 0.44 C$

so total charge flow will be 0.44 C

Answer:

The total charge passing a given point in the conductor is 0.438 C.

Explanation:

Given that,

The expression of current is

$i(t)=110\sin(120\pi t)$

$(dq(t))/(t)=110\sin(120\pi t)$

$dq(t)=110\sin(120\pi t)dt$ ....(I)

We need to calculate the total charge

On integrating both side of equation (I)

$\int_(0)^(q)dq(t)=\int_(0)^{(1)/(180)}110\sin(120\pi t)dt$

$q=110((-\cos(120\pi t))/(120\pi))_(0)^{(1)/(180)}$

$q=-(110)/(120\pi)(cos(120\pi((1)/(180)))-\cos120\pi(0))$

$q=-0.2918(-(1)/(2)-1)$

$q=0.438\ C$

Hence, The total charge passing a given point in the conductor is 0.438 C.

The sun does not normally affect the tides to a noticeable degree. however, under certain circumstances the gravitational pull of the sun does affect the tides. under what two (2) conditions.the sun does not normally affect the tides to a noticeable degree. however, under certain circumstances the gravitational pull of the sun does affect the tides. under what two (2) conditions.

Answers

The answers are :
1) when the sun, moon, and earth are in a line only
2) when the gravitational forces of the Moon and the Sun are
perpendicular to one another with respect to the Earth.

Answer:

High and low tides are result of combined effect of gravitational pull of the sun and the moon. When the two align in a straight line, the range of tides is maximum. This happens on new moon and full moon day.

On the other hand, when the sun and the moon align at right angles, the effect of gravity is minimum and the range of the tides is minimum.

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Answers

All of that is fascinating information. Thank you for sharing.

A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratio of their speeds, vproton/valpha is:

Answers

Explanation:

Charge on proton, q₁ = e

Charge on alpha particles, q₂ = 2e

The magnetic force is given by :

$F=qvB\ sin\theta$

Here, $\theta=90=sin(90) = 1$

For proton, $F_p=ev_pB$ ..........(1)

For alpha particle, $F_a=2ev_aB$ ..........(2)

Since, a proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle. So,

$ev_pB=2ev_aB$

$(v_p)/(v_a)=(2)/(1)$

So, the ratio of the speed of proton to the alpha particle is 2 : 1 .Hence, this is the required solution.

Final answer:

If a proton and an alpha particle experience the same force in a magnetic field, the proton must be traveling at twice the speed of the alpha particle. This is because the force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field.

Explanation:

The force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field. Given that a proton (charge e) and alpha particle (charge 2e) experience the same force in the same magnetic field, we can create an equation to solve for their speed ratio.

The force on a particle due to a magnetic field is given by F = qvB where q is the charge, v is the speed, and B is the magnetic field. Since the force on the proton and alpha particle are the same, we can set their force equations equal to each other.

This means that e * v_proton * B = 2e * v_alpha * B. Simplifying, the ratio v_proton/v_alpha = 2.

Therefore, the proton is moving twice as fast as the alpha particle.

Learn more about Forces in Magnetic Fields here:

brainly.com/question/3160109

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