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Tpy6a [65]
3 years ago
8

Consider lifting a box of mass m to a height h using two different methods: lifting the box directly or lifting the box using a

pulley (as in the previous part). What is Wd/Wp, the ratio of the work done lifting the box directly to the work done lifting the box with a pulley?
Physics
1 answer:
cricket20 [7]3 years ago
8 0

Answer:

Wd/wp =1

Explanation:

The work done in lifting the box directly, is the same as lifting the box using pulley.

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A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that
Katyanochek1 [597]

To find the magnitude and direction of the electric field, let us find the horizontal and vertical components of the field separately, then we will use those values to calculate the total magnitude and direction.

The tension in the thread is 6.57×10⁻²N and the thread is aligned horizontally, so the tension force is directed entirely horizontally. The sphere is in static equilibrium, therefore the horizontal component of the electrostatic force acting on the sphere, Fx, must act in the opposite direction of the tension and have a magnitude of 6.57×10⁻²N. We know this equation relating a charge, an electric field, and the force that the field exerts on the charge:

F = Eq

F is the electric force, E is the electric field, and q is the charge

Let us adjust the equation for only the horizontal components of the above quantities:

Fx = (Ex)(q)

Fx is the horizontal component of the electric force and Ex is the horizontal component of the electric field.

Given values:

F = 6.57×10⁻²N

q = 6.80×10³C

Plug in these values and solve for Ex:

6.57x10⁻² = Ex(6.80×10³)

<u>Ex = 9.66×10⁻⁶N/C</u>

<u />

Since the sphere is in static equilibrium, the vertical component of the electrostatic force acting on the sphere, Fy, must have the same magnitude and act in the opposite direction of the sphere's weight. If we assume the weight to act downwards, then Fy must act upward.

We know the weight of the sphere is given by:

W = mg

W is the weight, m is the mass, and g is the acceleration of objects due to earth's gravity field near its surface.

We also know this equation:

F = Eq

Let us adjust for the vertical components:

Fy = (Ey)(q)

Set Fy equal to W and we get:

(Ey)(q) = mg

Given values:

q = 6.80×10³C

m = 0.018kg

g = 9.81m/s²

Plug in the values and solve for Ey:

(Ey)(6.80×10³) = 0.018(9.81)

<u>Ey = 2.60×10⁻⁵N/C</u>

<u />

Let's now use the Pythagorean theorem to find the total magnitude of the electric field:

E = \sqrt{Ex^{2}+Ey^{2}}

E = 2.77×10⁻⁵N/C

The direction of the electric field is given by:

θ = tan⁻¹(Ey/Ex)

θ = 20.4° off the horizontal

4 0
3 years ago
In a given chemical reaction the energy of the products is less than the energy of the reactants which statement is true for thi
GREYUIT [131]

Group of answer choices.

A. Energy is absorbed in the reaction.

B. Energy is released in the reaction.

C. There is no transfer of energy in the reaction.

D. Energy is lost in the reaction.

Answer:

B. Energy is released in the reaction.

Explanation:

A chemical reaction can be defined as a chemical process which typically involves the transformation or rearrangement of the atomic, ionic or molecular structure of an element through the breakdown and formation of chemical bonds to produce a new compound or substance.

Basically, there are two (2) main types of chemical reaction and these include;

I. Endothermic reaction: it's a chemical reaction in which heat is absorbed

II. Exothermic reaction: it's a chemical reaction in which heat is liberated into the environment.

In Chemistry, all chemical equation must follow or be in accordance with the Law of Conservation of Mass, which states that mass can neither be created nor destroyed by either a physical transformation or a chemical reaction but transformed from one form to another in an isolated (closed) system.

Generally, energy is released in a chemical reaction when the energy of the products is less than the energy of the reactants and it is referred to as an exothermic reaction.

However, when the energy of the products is greater than the energy of the reactants, energy is absorbed and it is referred to as an endothermic reaction.

5 0
3 years ago
Give me three examples of fossil fuel-based energy sources pls
svet-max [94.6K]
Gasoline, kerosene, and coal. 

4 0
3 years ago
Help with these questions please , 15 PTS &amp; brainliest
myrzilka [38]

For these two questions, first you need to know that the voltage across each branch of a parallel circuit is the same.

So, for Q5, we can first find out the voltage across R₂ by V=IR.

Voltage across R₂ = 2.5 × 8 = 20V

Since R₂ and R₃ are in parallel circuit, their voltage should be the same. Thus, voltage across R₃ is 20V.

So, by V=IR,

current of R₃ = \frac{20}{4} = 5A

Q6. voltage across R₁ = 2 × 4 = 8V

∴voltage across R₂ = 8V

current of R₂ = \frac{8}{8} = 1A

<h3><u>Alternative method</u></h3>

From these two examples, you can find out that the current of each branch of the parallel circuit is inversely proportional to the resistance of the branch.

ie. for Q5,

\frac{R2}{R3} = \frac{I3}{I2}

\frac{8}{4} = \frac{I3}{2.5}

I₃ = 5A

Q6. \frac{R1}{R2} = \frac{I2}{I1}

\frac{4}{8} = \frac{I2}{2}

I₂ = 1A

6 0
2 years ago
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
DiKsa [7]

Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

7 0
3 years ago
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