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Nikitich
15 days ago
11

Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. H

e tells you that the force is always pointed away a definite point in space, which we can call the force center. The magnitude of the force turns out to be proportional to B/r3, where r is the distance from the force center to any other point. Your friend says that it has been determined that the constant of proportionality has been determined to be B= 2 (in units to be determined later), so that the magnitude of the force on a particle (in newtons) can be written as 2r^3, when the particle is at a distance r from the force center.
Required:
Write an expression of potential energy.
Physics
1 answer:
ValentinkaMS [2.4K]15 days ago
5 0

Answer:

             U = 1 / r²

Explanation:

In this problem, the task does not require calculating potential energy via the force equation since these two variables are interconnected.

             

         F = - dU / dr

This derivative represents a gradient, meaning it indicates direction, leading us to write

          dU = - F. dr

The formula for force becomes

         F = B / r³

Now, let’s apply this in the integral:

          ∫ dU = - ∫ B / r³  dr

Here, the force aligns with the displacement, simplifying the scalar product to the product of magnitudes.

Now, we can solve the integrals:

            U - Uo = -B (- / 2r² + 1 / 2r₀²)

To finalize the calculations, a reference point for energy must be designated; commonly, potential energy is set to zero (Uo = 0) at infinity (r = ∞).

             U = B / 2r²

Substituting B = 2, we arrive at:

             U = 1 / r²

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Resistance of rod is 1 ohm. It is bent in the form of square. The resistance across adjoint corners is.​
Keith_Richards [2263]

Answer: The result to the query is 0.25 ohms

Explanation:

R = u x/A.......1

where u represents the resistivity of the

rod, A is the cross-sectional area, and x denotes

the length of the rod.

Let R* represent the resistance across the adjacent sections of the rod.

Then, R* = u1/4.......2

By comparing equation 1 with equation 2, we find that

R* = 1/4

which equals 0.25 ohms.

3 0
19 days ago
If you accidentally touch the "hot" wire connected to the 120 V line, how much current will pass through your body?
serg [2593]

Complete Question:

Picture yourself on an aluminum ladder on the ground, attempting to fix an electrical connection with a metal screwdriver featuring a metallic handle. Since you are sweating profusely, your body has a resistance of 1.60 kΩ.

(a) If you accidentally contact the "hot" wire from the 120 V power line, what current will flow through your body?

(b) What is the amount of electrical power transferred to your body?

Answer:

(a) 0.075A

(b) 9W

Explanation:

According to Ohm's law, the voltage (V) applied to or passing through a body corresponds to the current (I) via the relationship:

V ∝ I

=> V = I x R                 ----------------------(i)

Where;

R denotes the resistance of the body

(a) As mentioned;

Due to wet conditions, the body will conduct electricity, and possesses the following values;

V = supplied voltage = 120V

R = resistance of the wet body = 1.60kΩ = 1.6 x 1000Ω = 1600Ω

Substituting these values into equation(i):

120 = I x 1600

To find I;

I = \frac{120}{1600}

I = 0.075A

Thus the current passing through your body is 0.075A

(b) Electrical power (P), which is expressed in Watts (W), delivered to the body is the product of current (I) and voltage (V) received. Thus:

P = I x V           ---------------------(ii)

Where;

I equates to 0.075A   [as derived above]

V is 120V     [as outlined in the question]

Plugging these values into equation (ii):

P = 0.075 x 120

P = 9W

Hence, the electrical power received by your body is 9W

7 0
14 days ago
When jumping, a flea reaches a takeoff speed of 1.0 m/s over a distance of 0.47 mm .What is the flea's acceleration during the j
Ostrovityanka [2204]
Assuming constant acceleration for simplicity, even if not entirely realistic, we have the final speed, vf, at 1.0 m/s, and we cover a distance, d, of 0.47 mm, which converts to 0.00047 m (as 1 m equals 1000 mm). At the start of its jump, we can set the initial velocity, vi, to be 0. Using the formula vf^2 = vi^2 + 2ad, we solve for acceleration, a, as follows: a = (vf^2 - vi^2)/(2d) = (1.0^2 - 0^2)/(2*0.00047) = 1,064 m/s^2, which is quite impressive for a flea!
8 0
27 days ago
Read 2 more answers
While spinning down from 500.0 rpm to rest, a solid uniform flywheel does 5.1 kJ of work. If the radius of the disk is 1.2 m, wh
Keith_Richards [2263]

Answer:

The mass will be calculated as 5.173 kg

Explanation:

The energy from the rotation of the uniform flywheel is given by E = 5.1 KJ = 5100 J

Angular speed \omega =500rpm=\frac{2\times 3.14\times 500}{60}=52.33rad/sec

Radius r = 1.2 m

The rotational kinetic energy of the flywheel is E=\frac{1}{2}I\omega ^2

Therefore, 5100=\frac{1}{2}\times I\times 52.33 ^2

I=3.724kgm^2

The moment of inertia for a solid flywheel is I=\frac{1}{2}mr^2

Thus, the mass will equal 5.173 kg 3.724=\frac{1}{2}\times m\times 1.2^2

m=5.173kg

7 0
4 days ago
1. Suppose a tank filled with water has a liquid column with a height of 10 meters. If the area is 2 square meters (m²), what's
ValentinkaMS [2425]

Answers:


1. To begin with, we need to clarify that Pressure P is defined as the Force applied F divided by the area A. This can be mathematically formulated as follows:


P=\frac{F}{A}   (1)


The unit for Pressure is Pascal (Pa), which corresponds to \frac{kg}{ms^{2}} and is also equivalent to \frac{N}{m^{2} }


Another formulation for Pressure considers the density d  of the liquid, its height h within the container, and the force due to gravity g:


P=d*h*g     (2)


In this scenario, the liquid in question is water, with a known density of roughly:


d=1000kg/m^{3}


Thus, we need to input the values into equation (2) to find the pressure (make sure to check the units):


P=1000\frac{kg}{m^{3}}*10m*9.8\frac{m}{s^{2}}


P=98000Pa


Next, substitute this value into equation (1) to isolate F:


F=P*A


In conclusion:

F=196000N



2. For this problem, we will apply equation (1) to determine the Pressure. We already have the area A and the force applied by the water in the container F:


P=\frac{F}{A}=\frac{900N}{3m^{2}}


P=300Pa


3. Here, equation (2) is ideal for calculating the hydrostatic pressure at any point located at the bottom of the tank (be cautious with your units):


P=d*h*g      

P=1000\frac{kg}{m^{3}}*7.5m*9.8\frac{m}{s^{2}}      


P=73500Pa


4. In this situation, it's critical to note that in fluids (specifically water here), higher fluid levels result in reduced pressure. Therefore, if P_{1} and P_{2} denote the pressures at heights h_{1} and h_{2}, respectively, and recognizing that the density of water and gravitational force remain constant, we can employ the following equation for our solution:


P_{2}- P_{1} =d*g(h_{2}- h_{1})   (3)


Where:


P_{1}=1.5 kPa at h_{1}=2m


It's noteworthy that 1kPa=1*1000 Pa


And P_{2}=? is the variable at a specific height h_{2}=6m


Now we can substitute the known values into equation (3) to compute P_{2}:


P_{2}-1500Pa=1000\frac{kg}{m^{3}}*9.8\frac{m}{s^{2}} (6m-2m)    


In conclusion: P_{2} =40700Pa    


5. In this scenario, we have the area A=0.75m^{2} and the mass of the piston m=200kg, and our goal is to find the pressure P.


We'll utilize equation (1):  

P=\frac{F}{A}


However, do you recall that we mentioned earlier that pressure is defined as the force exerted over an area?

In this instance, we will apply the following equation that incorporates both the gravitational force and the mass of an object to determine F:


F=m*g=200kg*9.8\frac{m}{s^{2}}


Subsequently:


F=1960N


Now we are able to perform the calculation for P:


P=\frac{1960N}{0.75m^{2}}


P=2613.33Pa



5 0
4 days ago
Read 2 more answers
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