An open beaker of pure water has a water potential of.

Why do planets speed up as they get closer to the sun?:

Serggg [28]

Answer:

C

Explanation:

Gravity is the main reason that make our planets to pull each other

5 0

2 years ago

A circular wire loop of radius 15.0 cm carries a current of 2.60

Korolek [52]

Part (a): Magnetic dipole moment

Magnetic dipole moment = IA, I = Current, A = Area of the loop
Then,
Magnetic dipole moment = 2.6*π*0.15^2 = 0.184 Am^2

Part (b): Torque acting on the loop
T = IAB SinФ, where B = Magnetic field, Ф = Angle
Then,
T = Magnetic dipole moment*B*SinФ = 0.184*12*Sin 41 = 1.447 Nm

5 0

3 years ago

A straight segment of a current-carrying wire has a current element IL where I = 2.70 A and L = 3.20 cm i + 4.30 cm j. The segme

myrzilka [38]

The component of the force in negative z-direction is -0.144 N.

The given parameters;

current in the wire, I = 2.7 A
length of the wire, L = (3.2 i + 4.3j) cm
magnetic filed, B = 1.24 i

The force on the segment of the wire is calculated as follows;

$F = ILBsin(\theta)$

where;

θ is the angle wire and magnetic field

The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.

The angle between the wire and the magnetic field is calculated as follows;

$\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{4.3}{3.2} )\\\\\theta = 53.3 \ ^0$

The magnitude of the wire length is calculated as follows;

$|l | = \sqrt{3.2^2 + 4.3^2} = 5.36 \ cm = 0.0536 \ m$

The component of the force in negative z-direction is calculated as;

$F_z = -ILB sin(\theta)\\\\F_z = -2.7 \times 0.0536 \times 1.24 \times sin(53.3)\\\\F_z = -0.144 \ N$

Thus, the component of the force in negative z-direction is -0.144 N.

Learn more here:brainly.com/question/22719779

6 0

2 years ago

A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio

natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

$\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}$

For the maximum velocity of the ball at the top of the vertical circular motion,

$v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}$

where g is the acceleration due to gravity,

Substituting the known values,

$\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}$

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0

1 year ago

A copper sphere was moving at 40 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f

USPshnik [31]

Answer:

Temperature increase = 2.1 [C]

Explanation:

We need to identify the initial data of the problem.

v = velocity of the copper sphere = 40 [m/s]

Cp = heat capacity = 387 [J/kg*C]

The most important data given is the fact that when the shock occurs kinetic energy is transformed into thermal energy, therefore it will have to be:

$E_{k}=Q\\ E_{k}= kinetic energy [J]\\Q=thermal energy [J]\\Re-employment values and equalizing equations\\\\\frac{1}{2} *m*v^{2}=m*C_{p}*dT \\The masses are canceled \\\\dT=\frac{v^{2}}{C_{p} *2} \\dT=2.1 [C]$

8 0

3 years ago