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blondinia [14]
2 years ago
6

When a particle is a distance r from the origin, its potential energy function is given by the equation U(r)=kr, where k is a co

nstant and r=x2+y2+z2−−−−−−−−−−√
(a) What are the SI units of k?

Part B (b) Find a mathematical expression in terms of x, y, and z for the y component of the force on the particle.

Part C (c) If U=3.00 J when the particle is 2.00 m from the origin, find the numerical value of the y component of the force on this particle when it is at the point (-1.00 m, 2.00 m, 3.00 m).

Physics
1 answer:
Reika [66]2 years ago
5 0

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

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A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen
Svetllana [295]

Answer:

h_p = 30.46\ m

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

v_f=g\cdot t

Where g = 9.8 m/s^2

Similarly, the distance y the object has traveled is calculated as follows:

\displaystyle y=\frac{g\cdot t^2}{2}

If we know the height h from which the object was dropped, we can solve the above equation for t:

\displaystyle t=\sqrt{\frac{2\cdot y}{g}}

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

t_2=2.56 - 2 = 0.56\ sec

Therefore, it has traveled down a distance:

\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m

Thus, the height of the pen is:

h_p = 32 - 1.54\Rightarrow h_p=30.46\ m

8 0
2 years ago
A wooden piece is made in different shapes take length (l) = radius (r) = 2m Calculate its volume as a:
In-s [12.5K]

This question deals with the volume of different shapes.

a) volume of the sphere is "33.51 m³".

b) volume of the cylinder is "25.13 m³".

a)

The volume of a sphere is given by the following formula:

Volume = \frac{4}{3}\pi r^3\\\\Volume = \frac{4}{3}\pi (2\ m)^3

<u>Volume = 33.51 m³</u>

<u />

b)

The volume of a cylinder is given by the following formula:

Volume = \pi r^2l\\\\Volume =\pi (2\ m)^2(2\ m)

<u>Volume = 25.13 m³</u>

<u />

Learn more about <em>volume </em>here:

brainly.com/question/16686115?referrer=searchResults

The attached picture shows the formulae of the <em>volume</em> of different shapes.

7 0
2 years ago
A police officer draws a sketch of the scene of an accident, as shown.
iren2701 [21]
I would have to say that it is Y
5 0
2 years ago
Read 2 more answers
A toroidal coil has a mean radius of 16 cm and a cross-sectional area of 0.25 cm2; it is wound uniformly with 1000 turns. A seco
Roman55 [17]

Answer:

Explanation:

Mutual inductance is equal to magnetic flux induced in the secondary coli due to unit current in the primary coil .

magnetic field in a torroid  B = μ₀ n I , n is number of turns per unit length and I is current .

B = 4π x 10⁻⁷ x (1000 / 2π x .16  )x 1 ( current = 1 A)

flux in the secondary coil

= B x area of face of coil x no of turns of secondary

= 4π x 10⁻⁷ x (1000 /2π x .16  ) .25 x 10⁻⁴ x 750

= 2 x 1000 x .25 x( 750 /.16) x 10⁻¹¹

2343.75 x 10⁻⁸

= 23.43 x 0⁻⁶ H.

.

6 0
2 years ago
Read 2 more answers
Fields of Point Charges Two point charges are fixed in the x-y plane. At the origin is q1 = -6.00 nC . and at a point on the x-a
My name is Ann [436]

Answer:

Part A) Electric fields at the point due to q₁ and q₂:

E₁ = 33.75*10³ N/C (-j) , E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C

Part B) Net electric field at P (Ep)

Ep=   (6.48*10³ (-i)+25.11 10³ (-j) )N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence

1nC= 10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

q₁ = -6.00 nC = -6 *10⁻⁹C

q₂ = +3.00 nC = +3*10⁻⁹C

d₁ = 4cm = 4 *10⁻²m

d_{2} =\sqrt{(4*10^{-2})^{2}+((3*10^{-2})^{2} }

d₂ = 5 *10⁻²m

Part A) Calculation of the electric fields at the point due to q₁ and q₂

Look at the attached graphic:

E₁: Electric Field at point  P(0,4) cm due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge

E₂: Electric Field at point  P(0,4) cm  due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge

E₁ = k*q₁/d₁² = 9*10⁹ *6 *10⁻⁹/ (4 *10⁻²)² = 33.75*10³ N/C

E₂ = k*q₂/d₂²= 9*10⁹ *3*10⁻⁹/(5 *10⁻²)² =  10.8*10³ N/C

E₁ = 33.75*10³ N/C (-j)

E₂x=E₂cosβ = 10.8*(3/5) = 6.48*10³ N/C

E₂y=E₂sinβ = 10.8*(4/5) =  8.64*10³ N/C

E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C

Part B) Calculation of the net electric field at P (Ep)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Ep=Epx (i) + Epy (j)

Epx= E₂x= 6.48*10³ N/C (-i)

Epy= E₁y+E₂y= (33.75*10³ (-j) + 8.64*10³ (+j) ) N/C=25.11 10³ (-j) N/C

Ep=   (6.48*10³ (-i)+25.11 10³ (-j) )N/C

Ep=   (6.48*10³ (-i)+25.11 10³ (-j) )N/C

3 0
2 years ago
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