Based on the measured force between objects that are 10 meters apart, how can you find the force between objects that are any distance apart ?

Answers

Answer 1

Answer:

The force between objects that are any distance apart is expressed as $P'=(100P)/(r^2)$

According to the gravitational law, the force acting on an object is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. Mathematically,

$P=(GMm)/(r^2)$

M and m are the masses

r is the distance between the masses

If the force between objects that are 10 meters apart, hence;

$P=(GMm)/(10^2)\nP=(GMm)/(100)\nGMm = 100P$

To find the force between objects that are any distance apart, we will use the same formula above to have;

$P'=(GMm)/(r^2)\n$

Substitute the result above into the expression to have:

$P'=(100P)/(r^2)$

Hence the force between objects that are any distance apart is expressed as $P'=(100P)/(r^2)$

Learn more on gravitational law here: brainly.com/question/11760568

Answer 2

Answer:

Answer:

F' = 100 F/r²

Explanation:

The gravitational force of attraction between two objects is given by the Newton's Gravitational Formula. The Newton's Gravitational Formula is as follows:

F = Gm₁m₂/r²

where,

F = Force between objects

G = Universal Gravitational Constant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects = 10 m

Therefore,

F = Gm₁m₂/10²

Gm₁m₂ = 100F --------------------- equation (1)

Now, we consider these objects at any distance r apart. So, the force becomes:

F' = Gm₁m₂/r²

using equation (1), we get:

F' = 100 F/r²

So, if the force (F) between objects 10 m apart is known, we can find it at any distance from the above formula.

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If the absolute pressure of gas is 550.280 kPa, its gauge pressure is

Answers

pressure absolute = pressure gage + pressure atmosphere

Answer:

650.280

Explanation: 100kpa + 550.280kpa

What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingrod of radius R1 inside of a thin-walled conducting shell of radius 2(both are infinite length). Suppose the inner rod hasradiusR1= 1.3 mm and outer shell has radiusR2= 10R1Ifthe net charge density on the center rod isq1= 3.4×10−12C/mand the outer shell isq2=−2q1,a.)What is the magnitude and direction of the electric field atradial distancer= 5R1from the center rod

Answers

Answer:

E = 9.4 10⁶ N / C, The field goes from the inner cylinder to the outside

Explanation:

The best way to work this problem is with Gauss's law

Ф = E. dA = qint / ε₀

We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.

The flow on the faces is zero, since the field goes in the radial direction of the cylinders.

The area of the cylinder is the length of the circle along the length of the cable

dA = 2π dr L

A = 2π r L

They indicate that the distance at which we must calculate the field is

r = 5 R₁

r = 5 1.3

r = 6.5 mm

The radius of the outer shell is

r₂ = 10 R₁

r₂ = 10 1.3

r₂ = 13 mm

r₂ > r

When comparing these two values we see that the field must be calculated between the two housings.

Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is

λ = q / L

Qint = λ L

Let's replace

E 2π r L = λ L /ε₀

E = 1 / 2piε₀ λ / r

Let's calculate

E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3

E = 9.4 10⁶ N / C

The field goes from the inner cylinder to the outside

The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail. When small prey, or even potential mates, swim within a few feet of the fish, they perturb the electric field. The change in the field is picked up by electric sensor cells in the skin of the elephant nose. These remarkable fish can detect changes in the electric field as small as 3.00 μN/C. How much charge, modeled as a point charge, in the fish would be needed to produce such a change in the electric field at a distance of 63.5 cm ?

Answers

Answer:

$1.34\cdot 10^(-16) C$

Explanation:

The strength of the electric field produced by a charge Q is given by

$E=k(Q)/(r^2)$

where

Q is the charge

r is the distance from the charge

k is the Coulomb's constant

In this problem, the electric field that can be detected by the fish is

$E=3.00 \mu N/C = 3.00\cdot 10^(-6)N/C$

and the fish can detect the electric field at a distance of

$r=63.5 cm = 0.635 m$

Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:

$Q=(Er^2)/(k)=((3.00\cdot 10^(-6) N/C)(0.635 m)^2)/(9\cdot 10^9 Nm^2 C^(-2))=1.34\cdot 10^(-16) C$

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the

Answers

Complete question:

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)$

k = 1.4

$T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)}\n\nT_2 = 1200((80)/(150))^{(1.4-1 )/(1.4)}\n\nT_2 = 1002.714K$

Work done is given as;

$W = (1)/(2) *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{(2W)/(m) } = √(2*C_p(T_1-T_2)) \n\nv_e = √(2*1004(1200-1002.714))\n\nv_e = √(396150.288) \n\nv_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

A popular physics lab involves a hand generator and an assortment of wires with different values of resistance. In the lab, the leads of the generator are connected across each wire in turn. For each wire, students attempt to turn the generator handle at the same constant rate. Students must push harder on the handle when the leads of the generator are connected__________. This is because turning the handle at a given constant rate produces__________ , regardless of what is connected to the leads. So, when turning the handle at a constant rate, lab students must push harder in cases where there is________

Answers

Answer:

Explanation:

Students must push harder on the handle when the leads of the generator are connected across the wire with the lowest resistance.

This is because turning the handle at a given constant rate produces a constant voltage across the leads, regardless of what is connected to the leads.

So, when turning the handle at a constant rate, lab students must push harder in case where there is a greater current through the connected wire.

You suspect that a power supply is faulty, but you use a power supply tester to measure its voltage output and find it to be acceptable. Why is it still possible that the power supply may be faulty?

Answers

While a power supply tester can be a useful tool for quickly checking voltage output, it might not reveal all the potential issues a faulty power supply can cause.

Even if a power supply tester shows that the voltage output of a power supply is within acceptable limits, it's still possible that the power supply may be faulty. Here's why:

1. Voltage Under Load: A power supply tester might only measure the voltage output under no load or very light load conditions.

A faulty power supply might provide the correct voltage at low loads but fail to deliver stable voltage under high loads, which could lead to system instability or crashes.

2. Voltage Ripple and Noise: Power supplies are expected to provide a stable and clean output voltage.

3. Short Circuits or Overloads: A power supply tester typically doesn't simulate the behavior of a real system.

4. Intermittent Issues: Faulty power supplies can exhibit intermittent issues. The power supply might work fine during the testing but fail when subjected to extended periods of operation or specific conditions.

5. Quality of Components: A power supply tester might not assess the quality of individual components within the power supply.

6. Compatibility Issues: Some power supplies might not be fully compatible with certain computer hardware. Even if the voltage seems fine, compatibility issues can still cause problems.

Learn more about Short Circuit here:

brainly.com/question/30778363

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