Distance and ___ are really the same quantity. A. length b. direction c. speed d. displacement

A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and

Alexxandr [17]

Answer:

a

$F =326.7 \ N$

b

$M = 6$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 200 \ kg$

The length of the small object from the rock is $d = 2 \ m$

The length of the small object from the branch $l = 12 \ m$

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The force exerted by the weight of the rock is mathematically evaluated as

$W = m_r * g$

substituting values

$W = 200 * 9.8$

$W = 1960 \ N$

So at equilibrium the sum of the moment about the fulcrum is mathematically represented as

$\sum M_f = F * cos \theta * l - W cos\theta * d = 0$

Here $\theta$ is very small so $cos\theta * l = l$

and $cos\theta * d = d$

Hence

$F * l - W * d = 0$

=> $F = \frac{W * d}{l}$

substituting values

$F = \frac{1960 * 2}{12}$

$F =326.7 \ N$

The mechanical advantage is mathematically evaluated as

$M = \frac{W}{F}$

substituting values

$M = \frac{1960}{326.7}$

$M = 6$

6 0

3 years ago

Metals are good conductors of electric current for which of the following reasons?They possess high concentrations of protonsThe

KatRina [158]

The correct answer is: They possess high concentrations of free electrons

The main characteristic of good conductors such as metals is the presence of movable electrically charged particles, or electrons. So, when an electric current is applied to a metal, the electrons will move and allow electricity to pass through them. Materials opposite of metals, with low electron mobility are not good conductors, instead they are called insulators.

3 0

3 years ago

A rope of length L has circular cross-sectional area A and density rho = m/V , where m is the mass of the rope and V = A · L is

hram777 [196]

Answer: µ = ρ¹ * A¹

Where x=1 and y=1

Explanation: According to the question, the mass per unit length (µ) is related to the density (ρ) and area A are related by the formulae below

µ = ρ * A

The dimension for each of these quantities is given below

Since µ is mass per unit length, unit is Kg/m and the dimension is ML^-1

ρ is density with unit kg/m³ and the dimension is ML^3

A is area with unit m², thus the dimension is M^2

Note that using dimensional analysis means we will be using the 3 fundamental quantities (mass, length and time) in our analysis.

Their dimensions below

Mass = M

Length = L

Time = T

Since the mass per unit length is related to density and area, we have a mathematical equation to provide a solution as shown below

µ = ρ^x * A^y.

By getting the power of x and y we will be able to get the formula that relates the quantities.

This is done by slotting in the dimensions of the respective quantities.

ML^-1 = (ML^-3)^x * (L²) ^y

By using law of indices on the right hand side of the equation, we have that

ML^-1 = (M^x * L^-3x) * (L^2y)

Also applying law of indices on the right hand side, we have that

ML^-1 = (M^x) * (L^-3x +2y)

The next step is to relate equal variables on both sides

For the M variable

M¹ = M^x which results to

x = 1

For the L variable

L^-1 = L^-3x+2y which results to

-1 = - 3x +2y

But x = 1

We have that

-1 = - 3(1) +2y

-1 = - 3 + 2y

-1 +3= 2y

2 = 2y

y = 1

Thus x=1 and y=1 and the formulae that relates the quantities is

µ = ρ¹ * A¹

3 0

3 years ago

A passenger compartment of a rotating amusement park ride contains a bench on which a book of mass

Basile [38]

a) 120 s

b) v = 0.052R [m/s]

Explanation:

a)

The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).

The graph of the problem is missing, find it in attachment.

To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.

The first point we take is t = 0, when the position of the book is x = 0.

Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.

Therefore, the period is

T = 120 s - 0 s = 120 s

b)

The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.

The perimeter of the wheel is:

$L=2\pi R$