Find the current flowing out of the battery.​

The weight of a male African elephant on the moon would be approximately 11,470.2 Newtons.

To calculate the weight of a male African elephant on the moon, we need to use the formula:

Weight = mass × acceleration due to gravity

The mass of the elephant is given as 7,000 kg. However, the acceleration due to gravity on the moon is approximately 1/6th (0.1667) of the acceleration due to gravity on Earth, which is approximately 9.8 m/s².

Weight on the Moon = 7,000 kg × 0.1667 × 9.8 m/s²

Weight on the Moon = 7,000 kg × 1.6386 m/s²

Weight on the Moon ≈ 11,470.2 Newtons

Therefore, the weight of a male African elephant on the moon would be approximately 11,470.2 Newtons.

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If the cup is perfectly insulated, no heat can flow in or out of the system. Therefore, the temperature of the system will remain constant at 0°C until all the ice has melted.

What amount of heat is needed to melt the ice?

To calculate the amount of heat required to melt the ice, we can use the formula:

Q = m * L

Where Q is the amount of heat required, m is the mass of ice, and L is the latent heat of fusion of water, which is 334 J/g.

The mass of ice is 50 grams, so:

Q = 50 g * 334 J/g = 16,700 J

Therefore, it will require 16,700 Joules of heat to melt the ice.

Once all the ice has melted, the temperature of the water will start to increase. We can use the specific heat capacity of water to calculate the amount of heat required to raise the temperature of the water.

The specific heat capacity of water is 4.18 J/g°C.

The temperature of the water needs to increase by 0°C to reach the melting point of ice (0°C), and then it needs to increase by another 1000°C to reach boiling point of water (100°C). So, the total temperature change is 100°C.

The mass of the water is 50 grams, so:

Q = m * c * ΔT

Q = 50 g * 4.18 J/g°C * 100°C

Q = 20,900 J

Therefore, it will require 20,900 Joules of heat to raise the temperature of the water from 0°C to 100°C.

In total, the amount of heat required to melt the ice and raise the temperature of the water to boiling point is:

Total Q = Q_ice + Q_water

Total Q = 16,700 J + 20,900 J

Total Q = 37,600 J

Therefore, it will require 37,600 Joules of heat to melt the ice and raise the temperature of the water to boiling point in the perfectly insulated cup.

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The radii of the first five Fresnel zones is 3.6 mm.

Distance from the light source to the wave surface, d₁ = 1 m

Distance from the wave surface to the observation point, d₂ = 1 m.

Wavelength of the light used, λ = 5 x 10⁻⁶m = 5 μm

The expression for the radius of the Fresnel zones is given by,

rₙ = √[nλd₁d₂/(d₁ + d₂)]

Therefore, the radii of the first five Fresnel zones is,

r₅ = √[5 x 5 x 10⁻⁶x 1 x 1/(1 + 1)]

r₅ = √(25 x 10⁻⁶/2)

r₅ = 3.6 x 10⁻³m

r₅ = 3.6 mm

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(a) The amplitude of the wave is 0.2 m.

(b) The period of the wave is  4 s.

(c) The wavelength of the wave is 100 m.

(a) The amplitude of the wave is the maximum displacement of the wave.

amplitude of the wave = 0.2 m

(b) The period of the wave is the time taken for the wave to make one complete cycle.

period of the wave = 5.5 s - 1.5 s = 4 s

(c) The wavelength of the wave is calculated as follows;

λ = v / f

where;

f = 1/t = 1 / 4s = 0.25 Hz

λ = ( 25 m/s ) / 0.25 Hz

λ = 100 m

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Answer:

They both have the same (avogadro's number) of atoms in one mole

Explanation:

Answer: 3

Explanation:

khan academy

When the teacher asked the student to make words out of the jumbled word gzeysktqix, the student was being tested on his ability to unscramble words. Unscrambling words is the process of taking a word or series of letters that are out of order and rearranging them to form a word that makes sense.

When trying to unscramble a word, it is important to look for any patterns that can help identify smaller words within the jumbled letters. This can help make the process easier and quicker. For example, in the jumbled word gzeysktqix, one might notice that the letters "sktqix" appear together.

This could indicate that these letters could potentially form a word. By looking at the remaining letters, one could notice that the letters "g", "z", "e", and "y" could also form smaller words. After some rearranging, the letters can be unscrambled to form the words "sky", "zig", "sex", and "yet". These are just a few examples, as there are likely many other words that can be formed from this jumbled word.

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The potential difference across the battery is 4.2 volts.

What is  potential difference?

Potential difference is described as the amount of work energy required to move an electric charge from one point to another.

The unit of potential difference is the volt.

The potential difference across the battery is  calculated using the equation:

V = W / Q

Workdone  = Q * Vbattery = 2.3 C * 4.2 J/C = 9.66 J

Therefore, the voltage across the battery can be calculated as:

V = W / Q = 9.66 J / 2.3 C = 4.2 V

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a. The force should be applied 5 m from the fulcrum

b. The issue is that deformation of the rod will occur.

In order to avoid this deformation, with the help of friends, the distance from the can be reduced and more force applied.

The principle of moments states that for a system in equilibrium, the sum of the clockwise moments about a point of rotation is equal to the sum of the anticlockwise moments about that point.

Mathematically;

The moment of a force is the product of the force and the perpendicular distance from its point of action.

From the data provided:

Let the woolly mammoth move in a clockwise direction and the applied force in an anticlockwise direction.

Let d be the perpendicular distance of the applied force from the fulcrum

5000 * 2 m = 2000 * d

d = 5 m

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The molecules of O2 that are  present in 3.90 L flask at  a temperature of 273 K and a pressure of 1.00 atm is 1.047 x 10^23 molecules  of O2

Step  1:  used the ideal gas equation to calculate the moles of O2

that is Pv=n RT  where;

P(pressure)= 1.00 atm

V(volume) =3.90 L

n(number of moles)=?

R(gas constant) = 0.0821 L.atm/mol.K

T(temperature) = 273 k

by  making n the subject of the formula by  dividing  both side by RT

n= Pv/RT

n=[( 1.00 atm x 3.90 L)  /(0.0821 L.atm/mol.k  x273)]=0.174  moles

Step 2: use the Avogadro's  law constant  to calculate  the number of molecules

that  is  according to Avogadro's law

                          1  mole =  6.02 x10^23  molecules

                            0.174 moles=? molecules

by  cross  multiplication

the number of  molecules

= (0.174  moles x  6.02 x10^23  molecules)/ 1 mole  =1.047 x 10^23 molecules of O2

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Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time \(t\), the horizontal position \(x\) and vertical position \(y\) of the ball are given respectively by

\(x = v_i \cos(\theta_i) t\)

\(y = v_i \sin(\theta_i) t - \dfrac g2 t^2\)

and the horizontal velocity \(v_x\) and vertical velocity \(v_y\) are

\(v_x = v_i \cos(\theta_i)\)

\(v_y = v_i \sin(\theta_i) - gt\)

The ball reaches its maximum height with \(v_y=0\). At this point, the ball has zero vertical velocity. This happens when

\(v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g\)

which means

\(y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g\)

At the same time, the ball will have traveled half its horizontal range, so

\(x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g\)

Solve for \(v_i\) and \(\theta_i\) :

\(\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0\)

Since \(0^\circ<\theta_i<90^\circ\), we cannot have \(\sin(\theta_i)=0\), so we're left with (e)

\(3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}\)

Now,

\(\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}\)

\(\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}\)

so it follows that (d)

\(R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}\)

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

\(v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}\)

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

\(v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}\)

Then with \(v_y=0\), the ball's speed \(v\) is

\(v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}\)

Finally, in the work leading up to part (e), we showed the time to maximum height is

\(t = \dfrac{v_i \sin(\theta_i)}g\)

but this is just half the total time the ball spends in the air. The total airtime is then

\(2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}\)

and the ball is in the air over the interval (a)

\(\boxed{0 < t < 2\sqrt{\frac R{3g}}}\)

A. The weight of the mass on the earth is 117.6 N

A. The weight of the mass on the moon is 19.56 N

Weight is defined as follow:

Weight (W) = mass (m) × Acceleration due to gravity (g)

W = mg

Now we shall determine the weight. Details below:

A. Weight on earth

Weight (W) = mass (m) × Acceleration due to gravity (g)

Weight (W) = 12 × 9.8

Weight on earth = 117.6 N

B. Weight on moon

Weight (W) = mass (m) × Acceleration due to gravity (g)

Weight (W) = 12 × 1.63

Weight on moon = 19.56 N

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Complete question:

The force of gravity on the Moon is said to be one-sixth of that on the Earth. What would a mass of 12 kg weigh; (a) on the Earth (b) on the moon

Answer: \(53\ rev/min\)

Explanation:

Given

angular speed of wheel is \(\omega_1 =530\ rev/min\)

Another wheel of 9 times the rotational inertia is coupled with initial wheel

Suppose the initial wheel has moment of inertia as I

Coupled disc has \(9I\) as rotational inertia

Conserving angular momentum,

\(\Rightarrow I\omega_1=(I+9I)\omega_2\\\\\Rightarrow \omega_2=\dfrac{I}{10I}\times 530\\\\\Rightarrow \omega_2=53\ rev/min\)

Graphical analysis of motion gives the graphical the relationships between position, velocity, and acceleration, versus time

2. Please find attached the required position versus time and acceleration versus time graphs

3. Please find attached the required velocity versus time and position versus time graphs

The reasons the attached graphs are correct are given as follows:

2. The given coordinates of the vertex points obtainable from the graph is first written as follows:

\(\begin{array}{|c|c|c|c|c|}\underline {Time, \ t}&\underline {Velocity, \ v}&Acceleration, a&\underline {Distance , \ s =v\cdot \Delta t+\dfrac{1}{2} \cdot a \cdot (\Delta t)^2}&Position\\0&-1&0&0&0\\2&1&1&0&0\\4&-1&-1&0&0\\6&1&1&0&0\\10&1&0&4&4\\12&0&-0.5&1&5\end{array}\)The distance covered between time intervals of time, s, is given as follows;

\(s = v \cdot \Delta t + (1/2) \cdot a \cdot ( \Delta t)^2\)

The position, Pₙ = s₁ + s₂ + s₃ + ... + sₙ

Other values on the graph obtained by calculation on a spreadsheet are;

\(\begin{array}{|l|cl|}Time \ (s)&&Position \ (m)\\0.5&&-0.375\\1&&-0.5\\1.5&&-0.375\\2.5&&0.375\\3&&0.5\\3.5&&0.375\\4.5&&-0.375\\5&&-0.5\\5.5&&-0.375\\6.5&&0.5\\7&&1\\7.5&&1.5\\8&&2\\8.5&&2.5\\9&&3\\10&&4\\10.5&&4.4375\\11&&4.75\\11.5&&4.9375\\12&&5\end{array}\right]\)

Please find attached the acceleration versus time graph

3. The coordinates of the points are presented as follows:

\(\begin{array}{|c|cc|}\underline{Time\ (s)}&&\underline{Acceleration\ (m/s^2)}\\0&&0.5\\2&&0.5\\2&&0\\4&&0\\4&&-0.5\\6&&-0.5\\6&&0\\8&&0\\8&&0.5\\10&&0.5\\10&&-0.5\\12&&-0.5\end{array}\right]\)

The distance covered between time intervals, s, is given as follows;

\(s = v \cdot \Delta t + (1/2) \cdot a \cdot ( \Delta t)^2\)

The position, Pₙ = s₁ + s₂ + s₃ + ... + sₙ

Using a spreadsheet application, more detailed values of the position can be found as shown in the graph created with MS Excel

0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3, 3.5, 4, 4.5, 5, 5.5, 6, 6.5, 7, 7.5, 8, 8.5, 9, 9.5, 10, 10.5, 11, 11.5, 12

0, 0.06, 0.25, 0.56, 1.00, 1.50, 2, 2.5, 3, 3.5, 3.81, 4.00, 4.06, 4.06, 4.06, 4.06, 4.06, 4.13, 4.31, 4.63, 5.06, 5.50, 5.81, 6, 6.06

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Answer:

Atomic mass is defined as the number of protons and neutrons in an atom, where each proton and neutron has a mass of approximately 1 amu (1.0073 and 1.0087, respectively). The electrons within an atom are so miniscule compared to protons and neutrons that their mass is negligible.

Explanation:

The mass will be moving at 14.3 m/s when it reaches the equilibrium position.

To determine the speed of the mass when it reaches the equilibrium position, we need to consider the conservation of mechanical energy, which includes both the spring potential energy and the gravitational potential energy.

The total mechanical energy (E) of the system is the sum of the potential energy and the kinetic energy:

E = PE (potential energy) + KE (kinetic energy)

At the equilibrium position, all the potential energy is converted into kinetic energy, so the total mechanical energy is entirely in the form of kinetic energy.

The potential energy stored in the spring (PE_spring) is given by Hooke's Law:

PE_spring = (1/2) * k * \(x^{2}\)

Where k is the spring constant (250 N/m) and x is the displacement from the equilibrium position (0.32 m).

PE_spring = (1/2) * 250 N/m *\((0.32 m)^2\)

PE_spring = 12.8 J

The change in gravitational potential energy (ΔPE_gravity) is given by:

ΔPE_gravity = m * g * h

Where m is the mass (0.125 kg), g is the acceleration due to gravity (9.8 m/\(s^2\)), and h is the change in height (which is zero in this case since the height doesn't change).

ΔPE_gravity = 0 J

Therefore, the total mechanical energy (E) is equal to the potential energy stored in the spring:

E = PE_spring

E = 12.8 J

Since the total mechanical energy is entirely in the form of kinetic energy at the equilibrium position, we can calculate the speed (v) using the equation:

E = (1/2) * m * \(v^2\)

Rearranging the equation, we get:

\(v^2\) = (2 * E) / m

\(v^2\) = (2 * 12.8 J) / 0.125 kg

\(v^2\) = 204.8 \(m^2\)/\(s^2\)

Taking the square root of both sides:

v = √ 204.8 \(m^2\)/\(s^2\)

v ≈ 14.3 m/s

Therefore, the mass will be moving at approximately 14.3 m/s when it reaches the equilibrium position.

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The distance between the central bright fringe and the first bright fringe on the screen is approximately 8.52E-3 meters.

In Young's double-slit experiment, the distance between the central bright fringe (m = 0) and the first bright fringe (m = 1) can be calculated using the following formula:

y = (m * λ * L) / d

where:

y is the distance between the fringes on the screen,

m is the order of the fringe (0 for the central bright fringe, 1 for the first bright fringe),

λ is the wavelength of the light,

L is the distance from the slits to the screen, and

d is the distance between the slits.

Given the values:

λ = 4.57E-7 m (blue light wavelength)

d = 2.42E-4 m (distance between the slits)

L = 4.5 m (distance from the slits to the screen)

For the central bright fringe (m = 0), the distance (y) is:

y = (0 * 4.57E-7 m * 4.5 m) / 2.42E-4 m

y = 0

Therefore, the central bright fringe coincides with the point where the two beams of light overlap.

For the first bright fringe (m = 1), the distance (y) is:

y = (1 * 4.57E-7 m * 4.5 m) / 2.42E-4 m

y ≈ 8.52E-3 m

This calculation demonstrates how the interference pattern in Young's experiment is formed, with bright and dark fringes being produced based on the constructive and destructive interference of the light waves from the two slits. The distance between these fringes depends on the wavelength of light, the separation of the slits, and the distance between the slits and the screen.

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As the problem tells us, the field at point P is 0, thus, the field exerted by charges q1 and q2 have the same magnitude, and exactly the opposite direction, as it can be seen on the following drawing:

Thus, we know that charge q2 will have to be a source of field (as opposed to a sink), and thus, a positive charge. Now all we have to do is find out what charge could produce a field with the same magnitude of the one from q1. As the electric field can be written as:

We'll have:

Thus, our answer is q2=1.43775*10^(-8)C

Note: Your lesson requests the answer to be inserted as: +1.44

Answer: \(3.816\ m/s\)

Explanation:

Given

Mass of Henry is 95 kg

Normal weight of Henry is \(mg=95\times 9.8=931\ N\)

The scale reads the weight as 830 N for first 3.6 s i.e. less than the normal weight i.e. Elevator is moving downwards

Apparent weight is given by

\(\Rightarrow 830=m(g-a)\quad [a=\text{acceleration of elevator}]\\\Rightarrow 830=95(9.8-a)\\\Rightarrow 8.736=9.8-a\\\Rightarrow a=1.06\ m/s^2\)

After 3.6 s weight becomes 930 N which is approximately equal to normal weight. It implies elevator starts moving with constant velocity i.e. no acceleration.

If elevator starts from rest, it velocity after 3.6 s is

\(v=u+at\\\Rightarrow v=0+1.06(3.6)\\\Rightarrow v=3.816\ m/s\)

This velocity will remain continues as after 3.6 s, elevator starts moving with constant velocity.

Thermal expansion is defined as the property of metals to expand when they are heated.

The metal expands when heated because the atoms move apart more.

An excellent method to demonstrate the concept of thermal expansion is using the bar and gauge experiment.

Both the diameter and the length are precisely comparable when both are at room temperature.

The bar will not fit within the gauge once it has been heated and cooled. Similar to the last instance; the bar will no longer fit flush inside the gauge if the gauge is heated and the bar is cooled.

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Answer:

Explanation:

Speed  = Distance/Time

Given

Speed = 6.12m/s

Distance = 0.750m

Substitute

Time - Distance//Speed

Time = 0.75/6.12

Time = 0.122secs

Hence it takes the pulse 0.122secs

The work done in stretching the spring from 50 cm to 80 cm is 67.5 J.

Hooke's law states that the force applied to an elastic material is directly proportional to its extension, provided its elastic limit is not exceeded.

To calculate the amount of work done by Hooke's law, first, we need to find the force constant of the spring.

Formula:

Where:

make k the subject of the equation

From the question,

Given:

Substitute these values into equation 2

Finally, To find the work done in stretching the spring from 50 cm to 80 cm, we use the formula below.

Where:

Also, From the question,

Given:

Substitute these values into equation 3

Hence, The work done in stretching the spring from 50 cm to 80 cm is 67.5 J.

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Answer:

85

Explanation:

soln

given that;

frequency=17Hz

wavelength=5m

speed?

formula for wavelength is;

wavelength= speed/frequency

then ; making v the subject formula

we have that v=wavelength*frequency

v=17*5=>85ms

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Answer: 4.8N

Explanation: Using Force (N) = Mass (Kg) * Acceleration (m/s2)
We know the Mass, but its in grams. convert it into kg.
40g / 1000 = 0.04 KG

Then we need to find the acceleration. We can find it by the given values
acceleration = Delta S / Delta T. Or Change in speed / Change in time
So, 60m\s / 0.5s = 120m\s2

now we can use f=ma
X= 0.04 * 120
X= 4.8N

Answer:

49.33333m/s²

Explanation:

The formula for acceleration is acceleration=change in velocity/change in time

So:

Iv= 0m/s

Fv= 267m/s

a=267m/s divided by 4.5s

a=49.3333m/s²

Answer:

The change in momentum is:  \(13.5\,\frac{kg\,m}{s}\)

Explanation:

Let's define that vectors pointing up are positive, and vectors pointing down negative.

Then we express the initial momentum of the ball as the product of its mass times the velocity, and include the negative sign since this momentum is pointing down (velocity vector is pointing down):

\(P_i=-0.5\,(15) \frac{kg\,m}{s} =-7.5\, \frac{kg\,m}{s}\)

the final momentum is positive (pointing up) and given by the product:

\(P_f=0.5\,(12) \frac{kg\,m}{s} =6\, \frac{kg\,m}{s}\)

Therefore, since the change in momentum is defined as the difference between the final momentum minus the initial one, we get:

\(P_f-P_i=6-(-7.5)\,\frac{kg\,m}{s} =13.5\,\frac{kg\,m}{s}\)

It is not recommended to fire a gun straight up into the air.

When a bullet is fired into the air, it will eventually come down and can pose a danger to people and property below. The bullet can still be lethal when it reaches the ground, especially if it lands on a hard surface or hits someone directly.

Additionally, firing a gun in a residential area can be illegal and can result in legal consequences. In general, guns should only be fired in designated shooting ranges or in self-defense situations where there is an immediate threat to life. It is important to handle firearms responsibly and follow all safety guidelines to prevent accidents and injuries.

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Answer: 12 ounce steak on venus

Explanation: venus = closer to the sun. The sun = more potential gravity, which would make it heavier than a quarter pound hamburger on jupiter.