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Helga [31]
4 years ago
15

How much work is required to lift an object with a mass of 5.0 kilograms to a height of 3.5 meters?

Physics
2 answers:
KATRIN_1 [288]4 years ago
7 0
Work is Force multiplied by distance: work = Fd

work = mgh

work = (5.0 kg)(9.8 m/s)(3.5 m)

work = 171.5 joules or for two sigfigs, 170 joules
Veronika [31]4 years ago
6 0
I believe it would be 172 J (or 1.72 · 10² J)
You might be interested in
Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree
kvv77 [185]

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is    

                    I_f_3 = 27.57 W/cm^2

2 When third  polarizer is removed the intensity after it passes through the stack is    

                I_f_2 = 102.24 W/cm^2

Explanation:

  From the question we are told that

       The angle of the second polarizing to the first is  \theta_2 = 21^o  

        The angle of the third  polarizing to the first is     \theta_3 = 61^o

        The unpolarized light after it pass through the polarizing stack   I_u = 60 W/cm^2

Let the initial intensity of the beam of light before polarization be I_p

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

                     I_1 = \frac{I_p}{2}

Now according to Malus’ law the  intensity of light that would emerge from the second polarizing filter is mathematically represented as

                    I_2 = I_1 cos^2 \theta_1

                       = \frac{I_p}{2} cos ^2 \theta_1

The intensity of light that will emerge from the third filter is mathematically represented as

                  I_3 = I_2 cos^2(\theta_2 - \theta_1 )

                          I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]

making I_p the subject of the formula

                  I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}

    Note that I_u = I_3 as I_3 is the last emerging intensity of light after it has pass through the polarizing stack

         Substituting values

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}

                           =234.622W/cm^2

When the second    is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

                      I_f_3 = \frac{I_p}{2} cos ^2 \theta_2

I_f_3 is the intensity of the light emerging from the stack

                     

substituting values

                     I_f_3 = \frac{234.622}{2} * cos^2(61)

                       I_f_3 = 27.57 W/cm^2

  When the third polarizer is removed  the  second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be  

                  I_f_2 = \frac{I_p}{2} cos ^2 \theta_1

I_f_2 is the intensity of the light emerging from the stack

Substituting values

                  I_f_2 =  \frac{234.622}{2} cos^2 (21)

                     I_f_2 = 102.24 W/cm^2

   

7 0
3 years ago
For a freely falling object dropped from rest, what is the acceleration at the
Zinaida [17]

The acceleration of the object after 3 seconds of fall is -9.8 m/s².

The given parameters;

  • initial velocity of the object, u = 0
  • time of motion of the object, t = 3 seconds

Acceleration is the change in velocity per change in time of motion.

The acceleration of the object after 3 seconds of fall is calculated as follows;

  • Since the object is in free fall, the object experiences only acceleration due to gravity.
  • the magnitude of this acceleration due to gravity is 9.8 m/s²
  • the direction of this acceleration is downwards

 

Thus, the acceleration of the object after 3 seconds of fall is -9.8 m/s².

Learn more here: brainly.com/question/13197713

7 0
3 years ago
Two pith balls hang side by side close to each other without touching as shown in the figure below. They are both neutral to beg
Mila [183]

a)

When a plastic rod is rubbed with wool, charging by friction occurs.

This is one of the three mechanisms of charging an object: in this particular case, as the rod is rubbed with the wool, electrons jump from the wool to the rod.

As a result, the plastic rod will remain with an excess of electrons (negative charge), while the wool will remain with a lack of electrons (positive charge).

However, the number of electrons gained by the rod is equal to the number of electrons given off by the wool: therefore, the total net charge of the system rod + cloth has not changed.

b)

When the rod touches sphere A, some electrons are transferred from the rod to the sphere (this method of charging is called charging by conduction).

Then, the sphere A will also remain negatively charged, as well as the plastic rod.

Later, the rod is brought close to sphere A without touching it: here, the electrons in the rod repel the electrons in the sphere (like charges repel each other), so the rod and the sphere will repel each other.

c)

Before part b), the negative charge on the plastic rod was equal to the positive charge on the wool.

After part b), the plastic rod has given off part of its electrons to sphere A. As a result, the negative charge on the plastic rod will be now less than the positive charge on the wool.

Therefore, the net charge of the rod + wool system will now be positive.

d)

The electrostatic force between two charged object is:

- Attractive if the charges on the objects have opposite sign

- Repulsive if the charges on the objects have same sign

Since sphere A is charged negatively, when it is brought close to sphere B, the electrons in sphere B will be attracted on the opposite side relative to sphere A, while the positive charges will migrate close to sphere A. Therefore, the two spheres will attract each other, since opposite charges attract each other.

e)

When the plastic rod (which is negatively charged) touches sphere B, some electrons are transferred to sphere B (charging by induction). As a result, charge B will also be negatively charged now, as it has an excess of electrons.

At the same time, sphere A is also negatively charged. Therefore, as like charges repel each other, this means that the two spheres will now repel each other.

f)

The net charge of the rod and the two spheres is negative.

In fact, the electrons given off by the rod to the two spheres were initially on the rod itself: this means that the total net charge of the rod + 2 spheres is equal to the net charge on the rod before the process.

Since the net charge on the rod before the process was negative, then the net charge of the system consisting of the rod + 2 spheres must be negative as well.

g)

When a glass rod is rubbed with silk, charging by friction occurs, similar to part a). However in this case, the glass rod gives off electrons to the silk: therefore, the glass rod remains positively charged.

Later, the glass rod touches sphere B: as a result, positive charge is transferred from the rod to sphere B, so sphere B will remain positively charged.

In the meantime, sphere A is negatively charged: since opposite charges attract each other, this means that now sphere A and sphere B will attract each other.

h)

When the two spheres are moved a bit further apart, the attractive force between them decreases a bit. In fact, the electrostatic force between two charged objects is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the  charges on the two objects

r is the separation between the objects

As we can see, the force is inversely proportional to the square of the distance: therefore, if the separation between the spheres increases, the force will decrease.

4 0
3 years ago
What is the displacement of a spring if it has a spring constant of 10 N/m, and a force of 2.5 N is applied?
pochemuha
O.25 m is the displacement
5 0
3 years ago
Read 2 more answers
How many butter, which has a usable energy content of 6.0 Cal/g (= 6000 cal/g), would be equivalent to the change in gravitation
Mazyrski [523]

Answer:

175.96 g

Explanation:

Potential energy required for the man to climb 7.07 km = m g h.

= 64 x 9.8 x 7070

= 4.434 x 10⁶ J

= 4.434 X 10⁶ / 4.2 cals

= 1.0557 x 10⁶ cals

= 1.0557 x 10⁶ / 6000 g of butter

= 175.96 g of butter.

3 0
3 years ago
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