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kherson [118]
3 years ago
5

If two boxes, one with a mass of 4 kg and one with a mass of 10 kg are sitting on a shelf. Explain the differences in their amou

nt of potential energy.
Physics
1 answer:
vovikov84 [41]3 years ago
4 0

Answer:

The box with a mass of 10 kilograms would have a greater amount of potential energy than the box with a mass of 4 kilograms.

Explanation:

Potential energy can be defined as an energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

P.E = mgh

Where,

P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Let's assume the height is 5 meters.

Acceleration due to gravity, g = 10 m/s²

For box with mass of 10 kg

P.E = 10*10*5

P.E = 500 Joules.

For box with mass of 4 kg;

P.E = 4*10*5

P.E = 200 Joules.

Hence, the box with a mass of 10 kilograms would have a greater amount of potential energy than the box with a mass of 4 kilograms.

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You are setting up a physics experiment to measure the electric field of a dipole along its axis. At what distance would you exp
snow_lady [41]

Answer:

y = 1.73 √s

Explanation:

For this question, let's look for the complete formula for the elective field of a dipole and then compare with the approximate formula.

A dipole is two charges of equal magnitude and different sign separated a distance 2, the field on axes at the midpoint is

               E₁ = E₂ = k q² / r²

For distance we use Pythagoras' theorem

              r² = y² + s²

The total electric field is

              E = 2 E₁ cos θ

The field perpendicular to the dipole axis is canceled, let's use trigonometry

            cos θ = s / r

Let's replace

              E = 2 k q² / (y² + s²) a / √(y² + s²)

            E = 2 q s / (y² + s²)^{3/2}

This is the exact formula.

The approximate formula is

            E’= 2 q s / y³

If we relate these two formulas

          E’/ E = (y² + s²).^{3/2}/y³

We see that the error in the distance propagates in an error for the electric field, they ask us that the uncertainty be 5% (er = 0.05)

The approximate formula is the measured value and the exact formula is the actual calculated value, so the relative uncertainty is

       E’= E  (y² + s²).^{3/2} / y³

      ΔE ’= dE’ /dy Δy + dE’/ds Δs + dE’ /dE ΔE

The last term is considered zero since the value is exact

      dE’/ dy =  (y² + s²).^{3/2} (-3y⁻⁴) + y⁻³ 3/2 (y² + s²).^{1/2}  2y

      dE ’/ dy = -3 (y² + s²).^{3/2}/y⁴ - 3y  (y2 + s2).^{1/2} /y³

      dE ’/ ds = 3/2 (y² + s²).^{1/2} 2s/y³

       dE'/ds = 3s (y²+s²).^{1/2} /y³

      ΔE’= E [+3 (y² + s²).^{3/2}/y⁴ + 3y  (y2 + s2).^{1/2} /y³] Δy

                  + [3s (y²+s²).^{1/2} /y³] Δs

     ΔE’/E’ = Δy [3y - 3 / (y² + s²)] + Δs [3s / (y² + s²)]

     ΔE’/E' =  3Δy [(1- / (y² + s²)] + 3Δs  s / (y² + s²)

In general the distance and is measured with a tape measure, large value with an uncertainty of Δy = 0.1 cm and the distance between the charge is measured with a caliper Δs = 0.05 cm

Let's replace the values

           0.05 = 0.1  3[1 – 1/ (y² + s²)] + 0.05 3s /(y² + s²)

 

This is the formula of the error between the approximate field and the exact field, so that the error is at 0.05, the first term must be eliminated by which  y >> s

          0.05 = 0.05 3s / y²

          1 = 3s / y ²

          y = √3s

          y = 1.73 √s

4 0
3 years ago
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FrozenT [24]
The most common injuries from sports could be ankle sprains, concussion, bone fractures, kneetears, hamstring strains, and groin strain.
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3 years ago
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A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu
Vesnalui [34]

Answer:

The magnitude of the resultant acceleration is 2.2 m/s^2

Explanation:

Mass (m) of the sailboat =  2000 kg

Force acting on the sailboat due to ocean  tide is F_1 = 3000N

Eastwards means takes place along the positive x direction

ThenF_{1x} = 3000N and F_{1y}= 0

Wind Force acting on the Sailboat isF_2  = 6000N directed towards the northwest that means at an angle  45 degree above the negative x axis

Then  

F_{2x} = -(6000N) cos 45 degree = -4242.6 N

F_{2y}  = (6000N) cos 45 degree = 4242.6 N

Hence  , the net force acting on the sailboat in x direction is  

F_x = F_{1x}+ F_{2x}

=  - 3000 N + 4242.6 N

=  - 3000 N +4242.6 N

= 1242.6N

Net Force acting on the sailboat in y direction is  

F_y = F_{1y}+ F_{2y}

= 0+ 4242.6N

= 4242.6N

The magnitude of the resultant force =

Using pythagorean theorm of 1243 N and 4243 N

\sqrt{(1242.6)^2 + (4242.6)^2

\sqrt{(1544054.76) + (17999654.8)}

\sqrt{(19543709.5)^2}

4420.8 N

F = ma

a = \frac{F}{m}

a =\frac{4420.8}{ 2000}

=2.2 m/s^2

4 0
4 years ago
A hot-air balloon is floating above a straight road. To estimate their height above the ground, the balloonists simultaneously m
Karolina [17]

Answer:

3.1 miles

Explanation:

To solve this question it is important to remember that the distance between two mile markers is approximately 1 mile

Once this is known, the question becomes very easy to solve. We make two triangle, which have the following three points

Triangle 1: Hot-Air-Balloon, Ground, Milepost 1 - With angle of depression 20

Triangle 2: Hot-Air-Balloon, Ground, Milepost 2 - With angle of depression 18

As a reminder, the angle of depression is simply the angle the balloonist's head makes with the horizontal plane to be able to see the milepost.

From this we can simply drive two formulas using the Tan function

Equation 1 - Tan(90-18) = \frac{b+1}{h}

Equation 2 - Tan(90-20) = \frac{b}{h}

Solving them simultaneously we get the value of height (h) to be 3.0852 miles or 3.1 miles

8 0
4 years ago
The sun produces 3.826 x 1026 Joules of energy every second as it combines smaller hydrogen atoms into larger helium atoms. What
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I think the correct answer from the choices listed above is option B.  The name of the process which allows for the production of so much energy in the sun would be fusion reaction. It <span>is a nuclear </span>reaction<span> in which two or more atomic nuclei come close enough to form one or more different atomic nuclei and subatomic particles</span>
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