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3 years ago

In the example given below, Aaron applies a force of 300N and Bob applies a force of 450N :

Physics

1 answer:

garri49 [273]3 years ago

3 0

Answer:

Explanation:

This problem is all about torque. The "rules" are that in order for a system to be in rotational equilibrium, the sum of the torques on the system have to equal 0 (in other words, they have to equal each other {cancel each other out}). The equation for torque is

τ = F⊥r where τ is torque, F⊥ is the perpendicular force, and r is the lever arm length in meters. We also have to understand that in general Forces moving clockwise are negative and Forces moving counterclockwise are positive. Now we're ready for the problem:

A. The counterclockwise torque:

τ = 300(3) so

τ = 900N*m

B. The clockwise torque:

τ = -450(2.5) so

τ = -1100N*m

C. Obviously the system is not in roational equilibrium because one side is experiencing a greater torque than the other. This system will move clockwise as it currently exists.

D. In order for the system to be in rotational equilibrium, we have to move Bob's location from the fulcrum. Let's see to where.

The torques have to be the same on both sides of the fulcrum; mathematically, that looks like this:

F⊥r = F⊥r Filling in:

300(3) = 450r and

900 = 450r so

2 = r. This means that Bob will have to move closer to the fulcrum by a half of a meter to 2 meters from the fulcrum in order for the system to be in balance.

Isn't this so much fun?!

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A 15.0-m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.08 angle with the horizontal. (a

grigory [225]

Answer:

a) fr = 266.92 N, fy = 1300 N, b) μ = 0.36

Explanation:

a) This is a balancing act.

Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive

-w x - W x₂ + R y = 0 (1)

usemso trigonometry to find distances

cos 60.08 = x / 7.5

x = 7.5 cos 60.08

x = 3.74 m

fireman

cos 60.08 = x₂ / 4

x2 = 4 cos 60

x2 = 2 m

wall support

sin 60.08 = y / 15

y = 15 are 60.08

y = 13 m

we substitute in equation 1

R y = w x + W x2

R = (w x + W x2) / y

R = (500 3.74 +800 2) / 13

R = 266.92 N

now let's write the expressions for the translational equilibrium

X axis

R -fr = 0

R = fr

fr = 266.92 N

Y Axis

Fy - w-W = 0

fy = 500 + 800

fy = 1300 N

b) ask the friction coefficient

the firefighter's distance is

cos 60.08 = x₃ / 9.00

x₃ = 9 cos 60

x₃ = 5.28 m

from equation 1

R = (w x + W x₃) / y

R = 500 3.74 + 800 5.28) / 13

R = 468.769 N

we saw that

fr = R = 468.769

The expression for the friction force is

fr = μ N

in this case the normal is the ratio to pesos

N = Fy

N = 1300 N

μ N = fr

μ = fr / N

μ = 468,769 / 1300

μ = 0.36

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3 years ago

How does the entropy of steam compare to the entropy of ice?

Dmitriy789 [7]

The answer is; The entropy of steam is larger than because it is more disordered than ice

Entropy is synonymous to the degree of disorder or randomness of molecules in a system. The molecules of steam are far apart from each other and move randomly in the system colliding with each other. Those of ice has less kinetic energy, vibrate more or less in a fixed position in the structiure, and are arranged in a orderd fashion

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3 years ago

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A light spring stretches 0.13 m when a 0.35 kg mass is hung from it. The mass is pulled down from this equilibrium position an a

Alenkasestr [34]

Answer:

v = 1.30 m/s

Explanation:

given,

mass hung = 0.35 Kg

spring stretched when load is hanged (x)= 0.13 m

now,

weight of the mass attached = Kx

m g = k x

0.35 x 9.8 = k x 0.13

k = 26.38 N/m

now, using conservation of energy

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx'^2$

$v = \sqrt{\dfrac{kx'^2}{m}}$

$v = \sqrt{\dfrac{26.38 \times 0.15^2}{0.35}}$

$v = \sqrt{1.6958}$

v = 1.30 m/s

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3 years ago

Why are the layers of the earth in the order that they are in?

REY [17]

Answer:

Each layer has its own properties, composition, and characteristics that affects many of the key processes of our planet. They are, in order from the exterior to the interior – the crust, the mantle, the outer core, and the inner core.

Explanation:

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3 years ago

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A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif

Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c) w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

μ = N*I*A

Where,

N: The number of turns

I : Current in coil

A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

vec ( τi ) = vec ( μi ) x vec ( B )

= 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

= ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

$\left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right] = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\$

- The initial torque ( τi ) is written as follows:

vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

Ui = - vec ( μi ) . vec ( B )

Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

Ui = - [(-0.000605556 + 0.00011)]

Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

Uf = - vec ( μf ) . vec ( B )

Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

ΔU = Uf - Ui

ΔU = -0.000252315 - 0.000495556

ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

Ui + Eki = Uf + Ekf

Where,

Eki : The initial kinetic energy ( initially at rest ) = 0

Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.

-ΔU = Ekf

0.5*T*w^2 = -ΔU

w^2 = -ΔU*2 / T

Where,

w: The angular speed at second position

w = √(0.000747871*2 / 6.50×10^−7)

w = 47.97 rad / s

6 0

3 years ago