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Bond [772]
2 years ago
6

How does changing the amount of copper wire affect an electromagnet’s strength?

Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
6 0

Answer:copper is the #1 electricity conducter in the world say you have 20,000 volts you would have to have a decent size piece of copper which means the higher the voltage the thicker the copper has to be

Explanation:

You might be interested in
Can someone please help a struggling physics student?
zhenek [66]
<h3><u>Part A:</u></h3>

<u><em>What is the maximum height the ball will reach in the air?</em></u>

Kinematics equation used:

  • v_f^2=v_i^2+2ad, where v_f is final velocity, v_i is initial velocity, a is acceleration, and d is distance travelled. From SI units, velocity should be in m/s, acceleration should be in m/s^2, and distance should be in m

We're given that the initial velocity is 12.0 m/s in the y-direction. At the maximum height, the vertical velocity of the ball will be 0 m/s, otherwise it would not be at maximum height. This is our final velocity.

The only acceleration in the system is acceleration due to gravity, which is approximately 9.8\:\mathrm{ m/s^2}. However, the acceleration is acting down, whereas the ball is moving up. To express its direction, acceleration should be plugged in as -9.8\:\mathrm{m/s^2}. We have three variables, and we are solving for the fourth, which is distance travelled. This will be the maximum height of the ball.

Substitute v_i=12, v_f=0, a=-9.8 to solve for d:

0^2=12^2+2(-9.8)(d),\\0=144-19.6d,\\-19.6d=-144,\\d=\frac{-144}{-19.6}=7.34693877551\approx \boxed{7.35\text{ m}}

<u><em>What is the velocity of the ball when it hits the ground?</em></u>

This question tests a physics concept rather than a physics formula. The vertical velocity of the ball when it hits the ground is equal in magnitude but opposite in direction to the ball's initial vertical velocity. This is because the ball spends equal time travelling to its max height as it does travelling from max height to the ground (ball is accelerating from initial velocity to 0 and then from 0 to some velocity over the same distance and time). Since the ball has an initial vertical velocity of +12.0 m/s, its velocity when it hits the ground will be \boxed{-12.0\text{ m/s}}. (The negative sign represents the direction. Because velocity is a vector, it is required.)

<h3><u>Part B:</u></h3>

<u>**Since my initial answer exceeds the character limit, I've attached the first question to Part B as an image. Please refer to the attached image for the answer and explanation to the first question of Part B. Apologies for the inconvenience.**</u>

<u><em>What is the direction of the velocity of the ball when it hits the ground? Express your answer in terms of the angle (in degrees ) of the ball's velocity with respect to the horizontal direction (see figure).</em></u>

This question uses a similar concept as the second question of Part A. The vertical velocity of the ball at launch is equal in magnitude but opposite in direction to the ball's final velocity. The horizontal component is equal in both magnitude and direction throughout the entire launch, since there are no horizontal forces acting on the system. Therefore, the angle below the horizontal of the ball's velocity when it hits the ground is equal to the angle of the ball to the horizontal at launch.

To find this, we need to use basic trigonometry for a right triangle. In any right triangle, the tangent/tan of an angle is equal to its opposite side divided by its adjacent side.

Let the angle to the horizontal at launch be \theta. The angle's opposite side is represented by the vertical velocity at launch (12.0 m/s) and the angle's adjacent side is represented by the horizontal velocity at launch (2.3 m/s). Therefore, we have the following equation:

\tan \theta=\frac{12.0\text{m/s}}{2.3\text{ m/s}}

Take the inverse tangent of both sides:

\arctan (\tan \theta)=\arctan (\frac{12.0}{2.3})

Simplify using \arctan(\tan \theta)=\theta \text{ for }\theta \in (-90^{\circ}, 90^{\circ}):

\theta=\arctan(\frac{12.3}{2.3}),\\\theta =79.14989537\approx \boxed{79.15^{\circ}}

We can express our answer by saying that the direction of the velocity of the ball when it hits the ground is \boxed{\text{approximately }79.15^{\circ} \text{ below the horizontal}} or \boxed{\text{approximately }-79.15^{\circ} \text{ to the horizontal}}.

5 0
3 years ago
Q: Why are distances in space often measured in light years?
marin [14]

Answer:

the answer is b

Explanation:

Hope it helps :)

6 0
3 years ago
Read 2 more answers
The position vector of a particle of mass 1.65 kg as a function of time is given by = (6.00 î + 4.15 t ĵ), where is in meters an
SashulF [63]

Answer:

 L = 41.09 Kg m2 / s      The angular momentum does not depend on the time

Explanation:

The definition of angular momentum is

        L = r x p

Where blacks indicate vectors

Let's apply this definition our case. Linear momentum

      p = m v

Let's replace

      L = m r x v

The given function is

      x = 6.00 i ^ + 4.15 t j ^

We look for speed

     v = dx / dt

     v = 0 + 4.15 j ^

To evaluate the angular momentum one of the best ways is to use determinants

     L = m \left[\begin{array}{ccc}i&j&k\\6&4.15t&0\\0&4.15&0\end{array}\right]

      L = m 6 4.15 k ^

The other products give zero

Let's calculate

      L = 1.65 6 4.15 k ^

      L = 41.09 Kg m2 / s

The angular momentum does not depend on the time

7 0
3 years ago
_____ is the frictional force needed to slow an object in motion
s2008m [1.1K]

Answer:

<u>Drag force</u> is the frictional force needed to slow an object in motion

Explanation:

8 0
3 years ago
Read 2 more answers
1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp
Tanzania [10]

Answer with explanation:

We are given that  

Mass of ball,m_1=75 g=\frac{75}{1000}=0075kg

1 kg=1000 g

Height,h_1=1.6 m

h_2=0.6 m

Horizontal velocity,v_x=2 m/s

Mass of platem_2=400 g=\frac{400}{1000}=0.4 kg

a.Initial velocity of plate,u_2=0

Velocity before impact=u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s

Where g=9.8 m/s^2

Velocity after impact,v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s

According to law of conservation of momentum  

m_1u_1+m_2u_1=-m_1v_1+m_2v_2

Substitute the values  

0.075\times 5.6+0=-0.075\times 3.4+0.4v_2

0.4v_2=0.075\times 5.6+0.075\times 3.4

v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s

Velocity of plate=1.69 m/s

b.Initial energy=\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J

Final energy=\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2

Final energy=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0
2 years ago
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