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1 day ago

A 35 g steel ball is held by a ceiling-mounted electromagnet 3.8 m above the floor. A compressed-air cannon sits on the floor, 4

.8 m to one side of the point directly under the ball. When a button is pressed, the ball drops and, simultaneously, the cannon fires a 25 g plastic ball. The two balls collide 1.4 m above the floor. What was the launch speed of the plastic ball?

Physics

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An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft

kicyunya [3294]

Response:

a) 80 V

b) The electric field has a strength of 100 N/C, directed from point B toward point A, where the charge is negative.

Clarification:

Given:

An object with a charge of q = -6.00 x 10^-9 C starts from rest at point A, making its kinetic energy zero ( $K_{A}$ = 0) and moving to point B at a distance l = 0.500m where its kinetic energy is ( $K_{B}$ = 5.00 x 10^-7J). Additionally, the electric potential of q at point A is VA = +30.0 v.

Required:

(a) We seek to find the electric potential VB

(b) We need to compute the magnitude and orientation of the electric field E.

Solution

(a) Utilizing the given values for VA, $K_{B}$ and q, we derive a relationship among the three parameters and VB to compute VB.

At points A and B, the charge moves from A to B due to the electric field. The mechanical energy of the object remains conserved throughout this journey, allowing us to apply eq(1) in this context:

$K_{A} +U_{A} =K_{B} +U_{B}$ .........................................(1)

Where $K_{A}$ = 0, and the potential energy U of the charge is defined as U = q V

In this equation, V represents the electric potential. Thus, equation (1) can be expressed as:

0+qVA= $K_{B}$ +qVB (Dividing by q)

VA= $K_{B}$ /q + VB (Restructuring for VB)

VB=VA- $K_{B}$ /q.......................................(2)

We now have the relation between VB, VA, and $K_{B}$ , allowing us to substitute our values for VA, $K_{B}$ , and q into equation (2) to obtain VB

VB=VA- $K_{B}$ /q

=30V-(5.00 x 10^-7J)/(-6.00 x 10^-9)

=80 V

(b) After calculating VB, we may use equation a to derive the electric field E affecting the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between these points

VA-VB = $\int\limits^1_0 {E} \, dl$ ...................................(a)

$=E\int\limits^1_0 {} \, dl$

VA-VB=E $l$ (Restructuring for E)

E= VA-VB/ $l$ ..................................(3)

Now, substituting our values for VA, Vs, and l into equation (3) allows us to compute the electric field E

E= VA-VB/ $l$

=-100 N/C

The electric field's magnitude equals 100 N/C and it directs from point B to point A towards the negative charge.

5 0

2 months ago

A small house was built on an island off a perfectly straight shoreline. The point B on the shoreline that is closest to the isl

Sav [3153]

Q should be positioned 4.8 miles east of point B. Explanation: From the diagram, we can define the construction cost as a function of angle θ (as illustrated). The underwater pipe length (marked in blue) equals 6/cos θ, while the land pipe length (marked in brown) is (8 - 6*tan θ). The total construction cost formula is: Construction Cost = (6/cos θ)(6000) + (8 - 6*tan θ)(3750). This formula is represented in terms of θ, which can vary from 0 degrees to 53.13 degrees according to the diagram. To find the angle θ that minimizes the construction cost, we differentiate the Construction Cost function with respect to θ and set it to zero. The derivative yields: -4500(5*sec θ – 8*tan θ)(sec θ) = 0, leading to θ = 38.68 degrees. By substituting θ, we can determine the distance of Q from B, which equals 6*tan θ. This calculates to a distance of 4.8 miles.

7 0

2 months ago

A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

serg [3582]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

The car's position over time t can be described by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

To find the average velocity, we divide the displacement by the elapsed time:

$v=\frac{\Delta x}{\Delta t}$

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 2.00 s, the position is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

This leads us to the displacement of

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The duration for this interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

Therefore, the average velocity during this period is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 4.00 s, the position is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

Thus, the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

This yields an average velocity of

$v=\frac{20.8}{4.0}=5.2 m/s$

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

And at t = 4 s it is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

This gives us a displacement of

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the resulting average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

Find out more about average velocity:

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3 months ago

Describe several uses of plastic, and explain why plastic is a good choice for these products

Maru [3345]

Numerous items, including bags, toys, and various goods, incorporate plastic. Its remarkable durability and flexibility are key characteristics.:) I hope this information is beneficial to you! c;

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1 month ago