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weqwewe [10]
3 years ago
11

Consider the following setup with three identical springs, a ruler for length measurements and three known masses and three unkn

own masses. You previously explored the force applied by the spring on a mass by hanging each of the known masses on a separate identical spring to determine the spring constant kk. Here we want to determine the mass of some unknowns. You hang each of the unknown colored masses on the same springs you characterized previously. From the displacement of the springs from the original equilibrium position, what is the mass of the green mass
Physics
1 answer:
svetlana [45]3 years ago
4 0

Answer:

To find the value of the unknown weight, we previously placed the 3 known weights and made a graph of the force against displacement

When hanging the weight is known, we measure the displacement and from the graph we can find the value of the hanging masses

We can also use the equation and multiply the constant K by the displacement and this is the applied weight.

Explanation:

For this problem we will use the translational equilibrium relation

        F –W = 0

        F = W

        W = mg

The spring elastic force is

        F = - k x

We substitute

        k x = m g

Where we see that the force of the spring is equal to the weight of the body.

To find the value of the unknown weight, we previously placed the 3 known weights and made a graph of the force against displacement

When hanging the weight is known, we measure the displacement and from the graph we can find the value of the hanging masses

We can also use the equation and multiply the constant K by the displacement and this is the applied weight.

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How does the gravitational force between two objects change if the distance
Anestetic [448]

Answer:

The Gravitational Force is reduced 4 times

Explanation:

The equation of Gravitational force follows:

F = (G*m1*m2)/r^2

Assume that G*m1*m2 = 1 and r = 1:

F = 1/1^2 = 1 N

Multiply the radius by 2

F = 1/2^2 = 1/4 N

So doubling the distance reduces the force 4 times.

7 0
3 years ago
according to Newton's third law what is the equal and opposite force to the downward force of gravity pulling on a man standing
Fudgin [204]

When you're talking about gravity, it's easy to identify the equal
opposite forces.

Gravity ALWAYS produces an equal pair of opposite forces. 
They both act between the centers of the two objects, one in
each direction.

Consider the equal pair of opposite gravitational forces between
you and the Earth.  One force acts on you, and draws you toward
the center of the Earth.  We call that force "your weight". 
The other one acts on the Earth, and draws it toward the center
of you.  Hardly anybody ever talks about that one, but the two
forces are equal ... your weight on Earth is equal to the Earth's
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7 0
3 years ago
Read 2 more answers
A 3.1 kg ball is dropped from the top of a 38 m tall building. What is the speed of the ball when it is halfway from the buildin
Archy [21]

Answer:

19.3m/s

Explanation:

Use third equation of motion

v^2-u^2=2gh

where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity

insert values to get answer

v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s

4 0
2 years ago
A pure substance can be a ..................... (a) element (b) compound (c) either element or compound (d)none of these
stiks02 [169]

Answer:

C

Explanation:

An element is a pure substance that can not be broken down into anything simpler

A compound in also a pure substance held together in fixed proportion through chemical bonds

7 0
3 years ago
Read 2 more answers
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What
Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 \alpha = 6261 \  rev/minutes^2

b

 \theta  = 613 \ revolutions

Explanation:

From the question we are told that

   The initial  angular speed is w_i =  1120 \ rev/minutes

    The angular speed after t = 13.8 s = \frac{13.8}{60 }  = 0.23 \ minutes  is w_f = 2560 \ rev/minutes

    The time for revolution considered is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  

 Generally the angular acceleration is mathematically represented as

         \alpha = \frac{w_f - w_i }{t}

=>      \alpha = \frac{2560  - 1120 }{0.23}  

=>      \alpha = 6261 \  rev/minutes^2

Generally the number of revolution made is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  is mathematically represented as

           \theta  =  \frac{1}{2}  * (w_i + w_f)*  t

=>      \theta  =  \frac{1}{2}  * (1120+ 2560 )*  0.333

=>      \theta  = 613 \ revolutions

5 0
3 years ago
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