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Ierofanga [76]
3 years ago
11

A student must design an experiment to determine the gravitational mass of an object. Which of the following experiments could t

he student use? Select two answers.
Physics
1 answer:
Elena L [17]3 years ago
5 0

Answer: a. Place the object on one side of a lever at a known distance away from the fulcrum. Place known masses on the other side of the fulcrum so that they are also paced on the lever at known distance from the fulcrum. Move the known masses to a known distance such that the lever is in static equilibrium.

d. Place the object on the end of a vertically hanging spring with a known spring constant. Allow the spring to stretch to a new equilibrium position and measure the distance the spring is stretched from its original equilibrium position.

Explanation:

The options are:

a. Place the object on one side of a lever at a known distance away from the fulcrum. Place known masses on the other side of the fulcrum so that they are also paced on the lever at known distance from the fulcrum. Move the known masses to a known distance such that the lever is in static equilibrium.

b. Place the object on a surface of negligible friction and pull the object horizontally across the surface with a spring scale at a non constant speed such that a motion detector can measure how the objects speed as a function of time changes.

c. Place the object on a surface that provides friction between the object and the surface. Use a surface such that the coefficient of friction between the object and the surface is known. Pull the object horizontally across the surface with a spring scale at a nonconstant speed such that a motion detector can measure how the objects speed as a function of time changes.

d. Place the object on the end of a vertically hanging spring with a known spring constant. Allow the spring to stretch to a new equilibrium position and measure the distance the spring is stretched from its original equilibrium position.

Gravitational mass simply has to do with how the body responds to the force of gravity. From the options given, the correct options are A and D.

For option A, by balancing the torque, the mass can be calculated. Since the known mass and the distance has been given here, the unknown mass can be calculated.

For option D, here the gravitational force can be balanced by the spring force and hence the mass can be calculated.

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An asteroid of mass 5.7×104 kg carrying a negative charge of 14.0 μC is 190 m from a second asteroid of mass 5.5×104 kg carrying
Finger [1]

Answer:

repulsion force es  R = 5,224 10⁻³ N

Explanation:

We use Newton's second law to add forces, the elective force is repulsive because the asteroids have the same type of charge and gravitational force is always attractive.

Gravitational force

      Fg = G m1 m2 / r²

      Fg = 6,673 10⁻¹¹ 5.7 10⁴ 5.5 10⁴/190²

      Fg = 5,795 10⁻⁶ N

Electric force

     Fe = k q1 q2 / r²

     Fe = 8.99 10⁹ 14.0 10⁻⁶ 15.0 10⁻⁶ / 190²

     Fe = 5.2296 10⁻³ N

The resulting force is

     R = Fe - Fg

     R = 5.2296 10⁻³ - 5.795 10⁻⁶=  5229.6 10⁻⁶ - 5,795 10⁻⁶

     R = 5,224 10⁻³ N

this is a repulsive force and the asteroids move away from each other

7 0
3 years ago
Plasma is the most abundant state of matter in the universe
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You rubbed two identical balloons in your hair's friend. This causes the balloons to become charged negatively with a magnitude
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Answer:

a)  F = 2.25 10³ N, b) REPULSIVE.

Explanation:

a) The electric force is given by coulomb's law

           F = k \frac{q_1q_2}{r^2}k q1q2 / r2

in this case nso indicate that the two charges have the same value

           q₁ = q₂ = 2.5 10⁻⁶ C

Let's reduce the magnitudes to the SI system

           r = 0.5 cm (1m / 100cm) = 5 10⁻³ m

let's calculate

           F = 9 \ 10^9 \frac{(2.5 \ 10^{-6})^2 }{(5 \ 10^{-3})^2}

           F = 2.25 10³ N

b) In electricity, electric charges of the same sign repel and those of the opposite sign attract

In this exercise, the balls are equal and are rubbed with the same material, for which it acquires charges of the same type, consequently, as the charges are of the same type, they indicate that the negative force is REPULSIVE.

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3 years ago
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