PHYSICS COLLEGE

A long wire carries a current density proportional to the distance from its center, J=(Jo/ro)•r, where Jo and ro are constants appropriate units. Determine the magnetic field vector inside this wire.

Answers

Answer 1

Answer:

Answer:

$B = \mu_0((1)/(3) (J_0)/(r_0) r^2)$

Explanation:

As the current density is given as

$J = (J_0)/(r_0)r$

now we have current inside wire given as

$i = \int J(2\pi r)dr$

$i = \int (J_0)/(r_0) r(2\pi r)dr$

$i = 2\pi (J_0)/(r_0) \int r^2 dr$

$i = (2)/(3) \pi (J_0)/(r_0) r^3$

Now by Ampere's law we will have

$\int B. dl = \mu_0 i$

$B. (2\pi r) = \mu_0((2)/(3) \pi (J_0)/(r_0) r^3)$

$B = \mu_0((1)/(3) (J_0)/(r_0) r^2)$

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Help me with my physics, please

Answers

The right answer would be

-20t+ 80

A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°C. What will be the volume of the air in the bag if you take it with you, still sealed, to the mountains where the temperature is 7.00°C and atmospheric pressure is 70.0 kPa

Answers

The volume of the air in the bag of potato chips to the mountains which is still sealed, 2.766 liters.

What is the gas law?

The gas law is used to show the relationship between the pressure and the temperature of the gases. It can be given as,

$PV=nrT$

Here, (n) and (r) are the constant. Therefore,

$(PV)/(T)=\rm Constant$

For the initial and final values, the gas law can be given as,

$(P_1V_1)/(T_1)=(P_2V_2)/(T_2)$

Here, (subscript 1,and 2) is used for the initial and final amount of pressure and temperature.

The initial values of the bag of potato chips as volume of 2.00 L, pressure of 1.00 ATM and a temperature of 20.0°C. It is known that the value of 1 ATM is equal to the 101.325 kPa.

The final temperature of the pack is 7.00°C and atmospheric pressure is 70.0 kPa

Put the values in the above formula as,

$(101.325*2)/(293)=(70* V_2)/(280)\nV_2=2.766\rm \; liters$

Hence, the volume of the air in the bag of potato chips to the mountains which is still sealed, 2.766 liters.

Learn more about the gas law here;

brainly.com/question/25290815

Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume $V_(1) = 2.00\ L$

Pressure $P_(1)= 1.00\ atm$

Pressure $P_(2)= 70.0\ kPa$

Temperature $T_(1)= 20.0°C = 293\ K$

Temperature $T_(2)= 7.00°C = 280\ K$

We need to calculate the volume at mountains

Using gas law

$(PV)/(T)=\ Constant$

For both temperature,

$(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))$

Put the value into the formula

$(101.325*2)/(293)=(70* V_(2))/(280)$

$V_(2)=(101.325*2*280)/(293*70)$

$V_(2)=2.766\ litre$

Hence, The volume at mountains is 2.766 L.

A 1000-kg car rolling on a smooth horizontal surface ( no friction) has speed of 20 m/s when it strikes a horizontal spring and is brought to rest in a distance of 2 m What is the spring’s stiffness constant?

Answers

Explanation:

kinetic energy was converted to potential energy in the spring.

the answer is in the above image

A pilot performs an evasive maneuver by diving vertically at 270. If he can withstand an acceleration of 9.0 's without blacking out, at what altitude must he begin to pull out of the dive to avoid crashing into the sea? y= ?m

Answers

Answer:

The pilot must be began at an altitude of 826.53 m to avoid crash into the sea.

Explanation:

Given that,

Velocity = 270 m/s

Acceleration = 9.0g s

We need to calculate the altitude

Using formula of centripetal acceleration

$a_(c)=(v^2)/(r)$

$r=(v^2)/(a_(c))$

Where, v = velocity

r = altitude

a = acceleration

Put the value into the formula

$r=(270^2)/(9.0*9.8)$

$r=826.53\ m$

Hence, The pilot must be began at an altitude of 826.53 m to avoid crash into the sea.

An object is being acted upon by three forces and moves with a constant velocity. One force is 60.0 N along the x-axis, the second in 75.0 N along the y-axis. What is the magnitude of the third force?

Answers

Answer:

96.05 N

Explanation:

From Vector,

The two forces acting along the x and y axis are perpendicular,

Fr = √(60²+75²) .............. Equation 1

Where Fr is the result of the two forces

Fr = √(3600+5625)

Fr = √(9225)

Fr = 96.05 N.

Note: Since the object moves with a constant velocity when it is acted upon by the three forces, The acceleration is zero and as such the resultant of the forces is equal to zero.

Therefore,

Ft = Fr+F3................... Equation 2

Where Ft = Total resistance of the three forces, F3 = magnitude of the third force.

make F3 the subject of the equation,

F3 = Ft-Fr

Given: Ft = 0 N, Fr = 96.05 N.

Substitute into equation 2

F3 = 0-96.05

F3 = -96.05 N.

Sometimes, in an intense battle, gunfire is so intense that bullets from opposite sides collide in midair. Suppose that one (with mass M = 5.12 g moving to the right at a speed V = [08]____________________ m/s directed 21.3° above the horizontal) collides and fuses with another with mass m = 3.05 g moving to the left at a speed v = 282 m/s directed 15.4° above the horizontal. a. What is the magnitude of their common velocity (m/s) immediately after the collision? b. What is the direction of their common velocity immediately after the collision? (Measure this angle in degrees from the horizontal.) c. What fraction of the original kinetic energy was lost in the collision?

Answers

The magnitude of the speed is 83.0325 m\s, the direction is 62.7 degrees, and the fraction of kinetic energy lost is 0.895.

What is collision?

The collision is the phenomenon when two objects come in direct contact with each other. Then both the bodies exert forces on each other.

The mass, angle, and velocity of the first object are 5.12 g, 21.3°, and 239 m/s.

And the mass, angle, and velocity of the second object be 3.05 g, 15.4°, and 282 m/s.

The momentum (P₁) before a collision will be

$\rm P_1 = (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) \hat{x} + (m_1 u_1 sin \theta _1+ m_2 u_2 sin \theta _2) \hat{y}$

The momentum (P₂) after a collision will be

$\rm P_2 = (m_1 + m_2) u \ cos\ \theta \ \hat{x} \ + (m_1 + m_2) u \ sin \ \theta \ \hat{y}$

Applying momentum conservation, we have

$\rm (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) = (m_1 + m_2) u \ cos\ \theta \ \n\n$ ...1

$\rm (m_1 u_1 sin \theta _1+m_2 u_2 sin \theta _2) \ =(m_1 + m_2) u \ sin \ \theta$ ...2

From equations 1 and 2, we have

$\rm \theta = tan \ ^(-1) ( (m_1 u_1 cos \theta _1 +m_2 u_2cos \theta _2))/( (m_1 u_1 sin \theta _1 - m_2 u_2 sin \theta _2))\n\n\n\theta = tan \ ^(-1) (5.12*239*cos21.3+3.05*282*cos15.4)/(5.12*239*sin21.3-3.05*282*sin15.4)\n\n\n\theta = 62.7^o$

From equation 1, we have

$\rm u = ((m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) )/( (m_1 + m_2) \ cos\ \theta ) \n\n\nu = (5.12*239*cos21.3 - 3.05*282*cos15.4)/((5.12+3.05)cos62.2)\n\n\nu = 83.0325 m/s$

Then the change in kinetic energy, we have

$\rm \Delta KE = (1)/(2)m_1u_1^2+(1)/(2)m_2u_2^2-(1)/(2)(m_1+m_2)u^2\n\n\n\Delta KE = (1)/(2) * 5.12 * 239^2 + (1)/(2)*3.05*282^2 - (1)/(2)(5.12+3.05)*83.032^2\n\n\n\Delta KE = 239.34 \ J$

The fraction of kinetic energy lost will be

$\rm Energy \ lost = (239.34)/(267.5) = 0.895$

More about the collision link is given below.

brainly.com/question/13876829

Answer:

Detailed solution is given below

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