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You have a battery marked " 6.00 V ." When you draw a current of 0.361 A from it, the potential difference between its terminals is 5.07 V . What is the potential difference when you draw 0.591 A

Answers

Answer 1

Answer:

Explanation:

Battery voltage is 6V

A current of 0.361A is draw the voltage reduces to, 5.07V

This shows that the appliances resistance that draws the currents is

Using KVL

The battery has an internal resistance r

V=Vr+Va

Vr is internal resistance voltage

Va is appliance voltage

6=5.07+Va

Va=6-5.07

Va=0.93

Using ohms law to the resistance of the appliance

Va=iR

R=Va/i

R=0.93/0.361

R=2.58ohms

Then if the circuit draws a current of 0.591A

Then the voltage across the load is

V=iR

Va=0.591×2.58

Va=1.52V

Then the voltage drop at the internal resistance is

V=Vr+Va

Vr=V-Va

Vr=6-1.52

Vr=4.48V

Answer 2

Answer:

Answer:

V = 4.48 V

Explanation:

• As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.

• We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_(rint) = I* r_(int)$

• The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_(b) - V_(rint) = 5.07 V = 6.00 V - 0.361 A * r_(int)$

• We can solve for rint, as follows:

$r_(int) =(V_(b)- V_(rint) )/(I) = (6.00 V - 5.07 V)/(0.361A) = 2.58 \Omega$

• When the circuit draws from battery a current I of 0.591A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_(b) - V_(rint) = 6.00 V - 0.591 A * 2.58 \Omega = 4.48 V$

• As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

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Since the density of air decreases with an increase in temperature, but the bulk modulus B is nearly independent of temperature, how would you expect the speed of sound waves in air to vary with temperature?

Answers

To develop this problem it is necessary to apply the concept related to the speed of sound waves in fluids.

By definition we know that the speed would be given by

$v=\sqrt{(\beta)/(\rho)}$

$\beta =$ Bulk modulus

$\rho =$ Density of air

From the expression shown above we can realize that the speed of sound is inversely proportional to the fluid in which it is found, in this case the air. When the density increases, the speed of sound decreases and vice versa.

According to the statement then, if the density of the air decreases due to an increase in temperature, we can conclude that the speed of sound increases when the temperature increases. They are directly proportional.

A Lincoln Continental is twice as long as a VW Beetle, when they are at rest. On a 2-lane road, the Continental driver passes the VW Beetle. Unfortunately for the Continental driver, a stationary policeman has set up a speed trap, and the policeman observes that the Continental and the Beetle have the same length. The VW is going at half the speed of light. How fast is the Lincoln going ? (Express your answer as a multiple of c).

Answers

Answer:

V(t) = √13/4c

Explanation:

See attachment

A neutron star has a mass of 2.0 × 1030 kg (about the mass of our sun) and a radius of 5.0 × 103 m (about the height of a good-sized mountain). Suppose an object falls from rest near the surface of such a star. How fast would this object be moving after it had fallen a distance of 0.025 m? (Assume that the gravitational force is constant over the distance of the fall and that the star is not rotating.)

Answers

Answer:

$v=516526.9m/s$

Explanation:

The force to which the object of mass m is attracted to a star of mass M while being at a distance r is:

$F=(GMm)/(r^2)$

Where $G=6.67*10^(-11)Nm^2/Kg^2$ is the gravitational constant.

Also, Newton's 2nd Law tells us that this object subject by that force will experiment an acceleration given by F=ma.

We have then:

$ma=(GMm)/(r^2)$

Which means:

$a=(GM)/(r^2)$

The object departs from rest ( $v_0=0m/s$ ) and travels a distance d, under an acceleration a, we can calculate its final velocity with the formula $v^2=v_0^2+2ad$ , which for our case will be:

$v^2=2ad=(2GMd)/(r^2)$

$v=\sqrt{(2GMd)/(r^2)}$

We assume a constant on the vecinity of the surface because d=0.025m is nothing compared with $r=5*10^3m$ . With our values then we have:

$v=\sqrt{(2GMd)/(r^2)}=\sqrt{(2(6.67*10^(-11)Nm^2/Kg^2)(2*10^(30)Kg)(0.025m))/((5*10^3m)^2)}=516526.9m/s$

When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of a star's mass outward, in the form of a rapidly expanding spherical shell. As a simple model of the supernova process, assume that the star is a solid sphere of radius R that is initially rotating at 2.4 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is 4.3R. Assume that all of the star's original mass is contained in the shell.

Answers

Answer:

0.0768 revolutions per day

Explanation:

R = Radius

$\omega$ = Angular velocity

As the mass is conserved the angular momentum is conserved

$I_1\omega_1=I_2\omega_2\n\Rightarrow (I_1)/(I_2)=(\omega_2)/(\omega_1)$

Moment of intertia for solid sphere

$I_1=(2)/(5)MR^2\n\Rightarrow I_1=0.4MR^2$

Moment of intertia for hollow sphere

$I_2=(2)/(3)M(4.3R)^2\n\Rightarrow I_2=12.327MR^2$

Dividing the moment of inertia

$(I_1)/(I_1)=(0.4MR^2)/(12.327MR^2)\n\Rightarrow (I_1)/(I_2)=0.032$

From the first equation

$\omega_2=\omega_1(I_1)/(I_2)\n\Rightarrow \omega_2=2.4* 0.032\n\Rightarrow \omega_2=0.0768\ rev\day$

The angular velocity, in revolutions per day, of the expanding supernova shell is 0.0768 revolutions per day

Final answer:

To find the angular velocity of the expanding supernova shell, we can use the principle of conservation of angular momentum. The initial angular momentum of the star can be equated to the final angular momentum of the shell. By substituting the given information and solving the equation, we can find the angular velocity of the shell.

Explanation:

When a star undergoes a supernova explosion, a large amount of its mass is blown outward in the form of a rapidly expanding shell. To find the angular velocity of the expanding shell, we can use the principle of conservation of angular momentum. Assuming that all of the star's original mass is contained in the shell, we can equate the initial angular momentum of the star to the final angular momentum of the shell.

The angular velocity of the star before the explosion can be calculated using the equation:

angular velocity before = 2 * pi * initial frequency

where the initial frequency is given as 2.4 revolutions per day.

After the explosion, the radius of the expanding shell is given as 4.3 times the radius of the star. Using the principle of conservation of angular momentum, we can set the initial angular momentum of the star equal to the final angular momentum of the shell:

initial angular momentum of the star = final angular momentum of the shell

Since the final angular momentum of the shell is given by:

final angular momentum of the shell = moment of inertia of the shell * angular velocity of the shell

where the moment of inertia of the shell is given by:

moment of inertia of the shell = 2/5 * mass of the shell * (radius of the shell)^2

and the angular velocity of the shell is what we are trying to find, we can rewrite the equation as:

initial angular momentum of the star = 2/5 * mass of the shell * (radius of the shell)^2 * angular velocity of the shell

By substituting the expression for the initial angular momentum of the star and solving for the angular velocity of the shell, we can find the answer.

Learn more about Angular velocity of a supernova shell here:

brainly.com/question/13896592

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A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used to launch a mass m along the frictionless surface. What compression of the spring would result in the mass attaining double the kinetic energy received in the above situation?

Answers

Answer:

The compression is $√(2) \ d$ .

Explanation:

A Hooke's law spring compressed has a potential energy

$E_(potential) = (1)/(2) k (\Delta x)^2$

where k is the spring constant and $\Delta x$ the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

$E_(kinetic) = (1)/(2) m v^2$ .

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity $v_1$ . Knowing that the energy is constant.

$(1)/(2) m v_1^2 = (1)/(2) k d^2$

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

$2 * ((1)/(2) m v_1^2) = (1)/(2) k D^2$

But, in the left side we can use the previous equation to obtain:

$2 * ((1)/(2) k d^2) = (1)/(2) k D^2$

$D^2 = (2 \ ((1)/(2) k d^2))/((1)/(2) k)$

$D^2 = 2 \ d^2$

$D = √(2 \ d^2)$

$D = √(2) \ d$

And this is the compression we are looking for

Answer:

$d'=√(2) d$

Explanation:

By hooke's law we have that the potential energy can be defined as:

$U=(kd^(2) )/(2)$

Where k is the spring constant and d is the compression distance, the kinetic energy can be written as

$K=(mv^(2) )/(2)$

By conservation of energy we have:

$(mv^(2) )/(2)=(kd^(2) )/(2)$ (1)

If we double the kinetic energy

$2((mv^(2) )/(2))=(kd'^(2) )/(2)$ (2)

where d' is the new compression, now if we input (1) in (2) we have

$2((kd^(2) )/(2))=(kd'^(2) )/(2)$

$2((d^(2) )/(2))=(d'^(2) )/(2)$

$d'=√(2) d$

Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has drafted an up-and-coming, left-handed pitcher who can consistently pitch at 42.91 m/s (96.00 mph) . Assuming a pitched ball has a mass of 0.1434 kg and has this speed just before a batter makes contact with it, how much kinetic energy does the ball have?

Answers

Answer: 132.02 J

Explanation:

By definition, the kinetic energy is written as follows:

KE = 1/2 m v²

In our question, we know from the question, the following information:

m = 0.1434 Kg

v= 42.91 m/s

Replacing in the equation for KE, we have:

KE = 1/2 . 0.1434 Kg. (42.91)² m²/s² ⇒ KE = 132.02 N. m = 132.02 J

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