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Wittaler [7]
3 years ago
8

In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres w

ith masses of 1.40 kg and 14.0 g whose centers are separated by about 2.30 cm. Calculate the gravitational force between these spheres, treating each as a particle located at the center of the sphere.
Physics
2 answers:
Pepsi [2]3 years ago
8 0

Answer:

The gravitational force F =2.47*10⁻⁹N

Explanation:

Given Data;

First mass m1 = 1.40kg

Second mass m2 = 14.0g = 14/1000 = 0.014kg

distance (r) = 2.30cm = 0.023m

Gravitational constant = 6.67* 10⁻¹¹N/m²kg²

gravitational force (F)?

For calculating the gravitational force, we use the formula;

F = (Gm₁m₂)/r²

Substituting into the formula, we have

F = (6.67 * 10⁻¹¹ * 1.40 * 0.014)/0.023²

   = 1.30732*10⁻¹²/5.29*10⁻⁴

   = 2.47*10⁻⁹N

NeTakaya3 years ago
6 0

Answer:

F=2.47*10^{-10} N

Explanation:

The gravitational force is calculated by using

F=G\frac{M_1M_2}{r^2}

G: Cavendish constant = 6.67*10^{-12}Nm^2/kg^2

r=2.30cm=0.023m

M1=1.4kg

M2=14.0g=0.014kg

By replacing we have

F=2.47*10^{-10} N

hope this helps!!

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a ball is thrown straight up into the air with a speed of 13 m/s. if the ball has a mass of 0.25 kg, how high does the ball go?
evablogger [386]
<h2>Hello!</h2>

The answer is: 8.62m

<h2>Why?</h2>

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

E=PE+KE

Where:

PE=m.g.h

KE=\frac{1*m*v^{2}}{2}

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

<em>h </em>is the reached height of the object

<em>v</em> is the velocity of the object

Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:

KE_{1}+PE_{1}=KE_{2}+PE_{2}

Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.

\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}

So,

h_{2}=\frac{1*m*v_{1} ^{2}}{2*m*g}\\h_{2}=\frac{1*v_{1} ^{2}}{2g}=\frac{(\frac{13m}{s})^{2} }{2*\frac{9.8m}{s^{2}}}\\h_{2}=8.62m}

Hence,

The height at the second moment (maximum height) is 8.62m

Have a nice day!

5 0
3 years ago
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