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Pani-rosa [81]
3 years ago
11

Trina makes a diagram to summarize how to use the right-hand rule to determine the direction of the magnetic force on each type

of charge.
Which labels belong in the regions marked X, Y, and Z?

A) X: Thumb, for velocity
Y: Fingers, for magnetic field
Z: Palm of hand, for force

B) X: Back of hand, for force
Y: Thumb, for velocity
Z: Palm of hand, for force

C)X: Palm of hand, for force
Y: Thumb, for velocity
Z: Fingers, for magnetic field

D) X: Palm of hand, for force
Y: Fingers, for magnetic field
Z: Back of hand, for force

Physics
2 answers:
Tems11 [23]3 years ago
7 0

Labels that belong in regions X, Y, Z;

B) X: Back of hand, for force

Y: Thumb, for velocity

Z: Palm of hand, for force

mel-nik [20]3 years ago
6 0

Answer: A)

B) X: Back of hand, for force

Y: Thumb, for velocity

Z: Palm of hand, for force

Explanation:

Right-hand rule is used to find direction of magnetic force. When a charge enters magnetic field, magnetic force acts perpendicular to both magnetic field and velocity of charged particle and the charge moves in a curved path.

F = q v × B

The direction of the magnetic force is given by the right hand:

The fingers should be point in the direction of the magnetic field. Then the direction of palm gives the direction of magnetic force acting on positive charge and back of hand gives direction for magnetic force for negative charge.

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tatyana61 [14]

Ill provide the answer choices here, assuming its from edge.

A) Sasha’s monthly expenses would be less for buying than for renting.


B) The extra expenses in the mortgage payment cover all maintenance and repairs.


C) Sasha’s down payment will likely be less if she decided to buy.


D) Sasha will own the house and earn equity as its value increases.

the correct answer is D) Sasha will own the house and earn equity as its value increases.

6 0
3 years ago
Read 2 more answers
Stress distributed over an area is best described as: a) External force b) Axial force c) Radial force d) Internal resistive for
Anit [1.1K]

Answer:

Option D is the correct answer.

Explanation:

Stress is the force per unit area that tend to change the shape of body.

Stress is defined as internal resistive force per unit area.

         \texttt{Stress}=\frac{\texttt{Internal resistive force}}{\texttt{Area}}

         \sigma =\frac{F}{A}

So, so stress distributed over an area is best described as internal resistive force.

Option D is the correct answer.

8 0
3 years ago
Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

5 0
3 years ago
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.1 rad/s in 2.99 s.(a) fi
elena55 [62]
The angular acceleration of a rotating object is given by
\alpha =  \frac{\omega_f - \omega_i}{\Delta t}
where
\omega_f is the final angular speed of the object
\omega_i is its initial angular speed
\Delta t is the time taken to accelerate

For the wheel in our problem, \omega_f=11.1 rad/s, \omega_i = 0 and \Delta t=2.99 s, so its angular acceleration is
\alpha= \frac{11.1 rad/s-0}{2.99 s}=3.71 rad/s^2
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2 years ago
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icang [17]

Answer:

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4 0
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