PHYSICS COLLEGE

Nichrome wire, often used for heating elements, has resistivity of 1.0 × 10-6 Ω ∙ m at room temperature. What length of No. 30 wire (of diameter 0.250 mm) is needed to wind a resistor that has 50 ohms at room temperature?

Answers

Answer 1

Answer:

Answer:

Length = 2.453 m

Explanation:

Given:

Resistivity of the wire (ρ) = 1 × 10⁻⁶ Ω-m

Diameter of the wire (d) = 0.250 mm = 0.250 × 10⁻³ m

Resistance of the wire (R) = 50 Ω

Length of the wire (L) = ?

The area of cross section is given as:

$A=(1)/(4)\pi d^2\n\nA=(1)/(4)*\ 3.14* (0.250* 10^(-3))^2\n\nA=0.785* 6.25* 10^(-8)\n\nA=4.906* 10^(-8)\ m^2$

We know that, for a constant temperature, the resistance of a wire is directly proportional to its length and inversely proportional to its area of cross section. The constant of proportionality is called the resistivity of the wire. Therefore,

$R=\rho (L)/(A)$

Expressing the above in terms of length 'L', we get:

$L=(RA)/(\rho)$

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

$L=(50* 4.906* 10^(-8))/(1* 10^(-6))\ m\n\nL=(2.453)/(1)=2.453\ m$

Therefore, length of No. 30 wire (of diameter 0.250 mm) is 2.453 m.

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A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net work done on the object if its velocity changes to 1 8.00 i ^ 1 4.00 j ^

Answers

(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
$v= √((6.00m/s)^2+(2.00 m/s)^2)=6.32 m/s$
And so, the kinetic energy of the object is
$K= (1)/(2)mv^2= (1)/(2)(3.00 kg)(6.32 m/s)^2=60 J$

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
$v= √((8.00 m/s)^2+(4.00 m/s)^2)=8.94 m/s$
And so the new kinetic energy is
$K= (1)/(2)mv^2= (1)/(2)(3.00 kg)(8.94 m/s)^2=120 J$

So, the work done on the object is the variation of kinetic energy of the object:
$W=\Delta K=120 J-60 J=60 J$

Final answer:

The initial kinetic energy of the 3.00-kg object traveling at a velocity of 2.00 m/s is 6.00 Joules. When the object's velocity changed to 4.47 m/s, its kinetic energy became 30.02 Joules. Hence, the net work done on the object is 24.02 Joules.

Explanation:

The kinetic energy of any object can be calculated using the formula KE = 0.5 * m * v^2, where m is the object's mass and v is its velocity. For the 3.00-kg object with a velocity of 6.00 i ^ 2 and 2.00 j ^2 m/s, its velocity magnitude would be the square root of (6.00^2 + 2.00^2), which is 2.00 m/s. Plugging the values into the formula, the kinetic energy (a) would be 0.5 * 3.00 * 2.00^2 = 6.00 Joules.

The net work done on an object (b) can be obtained by finding the change in kinetic energy when the object’s velocity changes to 8.00 I and 4.00 j. The final velocity's magnitude would be the square root of (8.00^2 + 4.00^2), which is 4.47 m/s. Hence, the final kinetic energy is 0.5 * 3.00 * 4.47^2 = 30.02 Joules. Therefore, the net work done equals the change in kinetic energy, which is 30.02 - 6.00 = 24.02 Joules.

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OFFERING 60 POINTS IF YOU CAN SHOW THE WORK!!!!A 1000 kg roller coaster begins on a 10 m tall hill with an initial velocity of 6m/s and travels down before traveling up a second hill. As the coaster moves from its initial height to its lowest position, 1700J of energy is transformed to thermal energy by friction.

Answers

Answer; 10.6 i think

Explanation:

(a) At the top of the hill, the coaster has total energy (potential and kinetic)

E = (1000 kg) g (10 m) + 1/2 (1000 kg) (6 m/s)² = 116,000 J

As it reaches its lowest position, its potential energy is converted to kinetic energy, and some is lost to friction, making its speed v such that

1/2 (1000 kg) v ² = 116,000 J - 1700 J = 114,300 J

===> v ≈ 15.2 m/s

If no energy is lost to friction as the coaster makes its way up the second hill, all of its kinetic energy would be converted to potential energy at the maximum possible height H.

1/2 (1000 kg) (15.2 m/s)² = (1000 kg) gH

===> H ≈ 11.7 m

(b) At the top of the second hill with minimum height h, and with maximum speed 4.6 m/s, the coaster has energy

E = P + K = (1000 kg) gh + 1/2 (1000 kg) (4.6 m/s)²

Assuming friction isn't a factor again, the energy here should match the energy at the lowest point in part (a), 114,300 J.

(1000 kg) g h + 1/2 (1000 kg) (4.6 m/s)² = 114,300 J

===> h ≈ 10.6 m

A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -7000, 0> kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is <-4000, -780, 0> N.Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.6 seconds after the probe is launched to 22.9 seconds after the launch?

Answers

Answer:

The change in momentum is $\Delta p = <-1200 , -234 , 0> kg \cdot m/s$

Explanation:

From the question we are told that

The mass of the probe is $m = 170 kg$

The location of the prob at time t = 22.9 s is $A = <5600, 7200,0>$

The momentum at time t = 22.9 s is $p = < 51000, -7000, 0> kg m/s$

The net force on the probe is $F = <-4000 , -780 , 0> N$

Generally the change in momentum is mathematically represented as

$\Delta p = F * \Delta t$

The initial time is 22.6 s

The final time is 22.9 s

Substituting values

$\Delta p = <-4000 , -780 , 0> * (22.9 - 22.6)$

$\Delta p = <-4000 , -780 , 0> * (0.3)$

$\Delta p = <-1200 , -234 , 0> kg \cdot m/s$

An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The valve connecting this container to the air-conditioning system is now opened until the mass in the container is 0.25 kg, at which time the valve is closed. During this time, only liquid R-134a flows from the container. Presuming that the process is isothermal while the valve is open, determine the final quality of the R-134a in the container and the total heat transfer. Use data from the tables.

Answers

Answer:

Answer

The Final Quality of teh R-134a in the container is 0.5056

The Total Heat transfer is $Q_(in) = 22.62 KJ$

Explanation:

Explanation is in the following attachments

Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate from 2.00 to 6.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.

Answers

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

W = $F_(total) .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) .d =(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) = ((1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i))/(d)$

$F_(total=) ((1)/(2) X 62 X6^(2) -(1)/(2) X 62 X2^(2) )/(25)$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_(sprinter) = F_(total) + F_(wind) = 39.7 + 30 = 69.68 N$

Answer:

Force exerted by sprinter = 69.68 N

Explanation:

From work energy theorem, we know that, work done is equal to change in kinetic energy.

Thus,

W = ΔK = Kf - Ki = (1/2)m•(v_f)² - (1/2)m•(v_i)² - - - - eq(1)

Now,

Work done is also;

W = Force x Distance = F•d - - - (2)

From the question, we are given ;

v_f = 6 m/s

v_i = 2 m/s

d = 25m

m = 62 kg

Equating equation 1 and 2,we get;

(1/2)m•(v_f)² - (1/2)m•(v_i)² = F•d

Plugging in the relevant values to obtain ;

(1/2)(62)[(6)² - (2)²] = F x 25

31(36 - 4) = 25F

992 = 25F

F = 39.68 N

The force the sprinter exerts backward on the track will be the sum of this force and the headwind force.

Thus,

Force of sprinter = 39.68 + 30 = 69.68N

A 8.00-μF capacitor that is initially uncharged is connected in series with a 3.00-Ω resistor and an emf source with E = 70.0 V and negligible internal resistance. At the instant when the resistor is dissipating electrical energy at a rate of 300 W, how much energy has been stored in the capacitor?

Answers

Answer:

The energy stored is $E_s = 0.0064 \ J$

Explanation:

From the question we are told that

The capacitance is $C = 8 \ \mu F = 8*10^(-6) \ F$

The resistance is R = 3.00-Ω

The emf is $E_t = 70.0 V$

The power is P = 300 W

Generally the total emf is mathematically represented as

$E_t = E_c + E_r$

Here $E_c$ is the emf across that capacitor which is mathematically represented as

$E_c = (q)/(C)$

and $E_r$ is the emf across the resistor which is mathematically represented as

$E_r = √(P R)$

$E_t = √(PR) + (q)/(C)$

=> $q = C[E_t - √(PR) ]$

Generally the energy stored in a capacitor is mathematically represented as

$E_s = (q^2)/(2C)$

=> $E_s = ([C [ E_t - √(PR) ]]^2)/(2C)$

=> $E_s = ([8.0*10^(-6) [ 70 - √(300 * 3))/(2 *(8.0*10^(-6)))$

=> $E_s = 0.0064 \ J$

Final answer:

The energy stored in the capacitor is 0 J.

Explanation:

When a 8.00-μF capacitor that is initially uncharged is connected in series with a 3.00-Ω resistor and an emf source with E = 70.0 V

At the instant when the resistor is dissipating electrical energy at a rate of 300 W, we can calculate the current flowing through the circuit using Ohm's law: I = V/R = 70.0 V / 3.00 Ω = 23.33 A.

The energy stored in a capacitor can be calculated using the formula: E = 1/2 * C * V^2, where C is the capacitance and V is the voltage across the capacitor.

Since the capacitor is initially uncharged, the voltage across it is also zero. So the energy stored in the capacitor is 0.5 * 8.00 x 10^-6 F * (0 V)^2 = 0 J.

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