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3 years ago

How much work does an elephant do while moving a circus wagon 20 meters with a pulling force of 200n

Physics

1 answer:

olya-2409 [2.1K]3 years ago

8 0

Given:

s(distance)= 20 meters

F(force)=200N

Now we know that

work done= Force applied x distance

Substituting the given values in the above formula we get

Work done= 20 x 200= 4000J

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Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu

Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

$F_y: T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N$

$T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N$

$T_x = T*sin(50) = 0.0234 N$

The electric force is given by the expression:

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

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

$r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m$

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

$F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}$

$q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C$

O 0.247 μC

8 0

3 years ago

What is work? how do we solve for work?

slava [35]

Answer: Work can be calculated with the equation: Work = Force × Distance. The SI unit for work is the joule (J), or Newton • meter (N • m). One joule equals the amount of work that is done when 1 N of force moves an object over a distance of 1 m.

Explanation:

5 0

2 years ago

Read 2 more answers

When a hot and cold object are placed in contact, the hot one loses energy. Does this violate energy conservation? Why or why no

blsea [12.9K]

Answer:

This does not violate the conservation of energy.

Explanation:

This does not violate the conservation of energy because the hot body gives energy in the form of heat to the colder body, this second absorbs energy. This will be the case until both bodies reach the same temperature, reaching thermal equilibrium and reducing the transfer of thermal energy. In this way the energy was only transferred from one body to another but the total energy of the system (body 1 plus body 2) will be the same as in the beginning, respecting the principle of conservation of energy or also called the first principle of thermodynamics .

The part of physics that studies these processes is in turn called heat transfer or heat transfer or thermal transfer. Heat transfer occurs whenever there is a thermal gradient or when two systems with different temperatures come into contact. The process persists until thermal equilibrium is reached, that is, until temperatures are equalized. When there is a temperature difference between two objects or regions close enough, the heat transfer cannot be stopped, it can only be slowed down.

8 0

3 years ago

PLEASE HELP ME ASAP!!!!!<br> WHAT AM I SUPPOSED TO DO??

igomit [66]

Write what you can improve in your fitness test

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2 years ago

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Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a

butalik [34]

Answer:

(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N

(b) E = -42846.7 N/C

Explanation:

The diagram attached below explains some parameters.

Parameters given:

Charge Q1 = +50 nC at point (0, 0)

Charge Q2 = -50 nC at point (0.1, 0)

Charge Q3 = +150 nC at point (0.1, 0.08)

* The distances are in meters.

(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,

F = F1 + F2

FORCE DUE TO Q1 i.e. F(Q1, Q3)

We have to find the x and y components.

From the diagram, we can find θ using SOHCAHTOA:

θ = tan⁻¹ (0.08/0.1)

θ = 38.66⁰

The distance between Q1 and Q3 can be found using Pythagoras theorem:

x² = 0.08² + 0.1²

x = 0.128 m

F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j

F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ

F(Q1, Q3) = (k * Q1 * Q3) / r²

k = Coulombs constant

F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²

F(Q1, Q3) = 0.00412N

F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66

F1 = 0.00322i + 0.00257j N

FORCE DUE TO Q2 i.e. F(Q2, Q3)

We have to find the x and y components.

F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j

F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0

F(Q2, Q3) = (k * Q2 * Q3) / r²

F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²

F(Q2, Q3) = -0.0105N

F2 = -i0.0105 * cos90 - j0.0105 * cos0

F2 = - 0.0105j N

Hence, the total force will be

F = F1 + F2

F = 0.00322i + 0.00257j - 0.0105j

F = 0.00322i - 0.00793j N

The magnitude of this force is:

|F| = √(0.00322² + (-0.00793²)

|F| = 0.00856N

(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E1, electric field due to Q1 = kQ1/r²

E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)

E1 = 27465.8 N/C

E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)

E1 = -70312.5N/C

The total electric field:

E = E1 + E2

E = 27465.8 - 70312.5

E = -42846.7 N/C

3 0

3 years ago