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2 years ago

8

Two wheels are identical but wheel b is spinning with twice the angular speed of wheel

1 answer:

Murljashka [212]2 years ago

6 0

<span>If two wheels are exactly the same but spin at different speeds, wheel b is twice te speed of wheel a, it is possible to find the ratio of the magnitude of radial acceleration at a singular point of the rim on wheel be to the spot is four.</span>

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If you have to apply 40n of force on a crowbar to lift a rock that weights 400n, what is the actual mechanical advantage of the

Mrrafil [7]

The mechanical advantage is defined as the ratio between the force produced by a machine and the force applied in input:
$MA= \frac{F_{out}}{F_{in}}$
For the crowbar of the problem, the force applied in input is 40 N, while the force produced in output is equal to the weight of the rock that is lifted, so 400 N. Therefore, the mechanical advantage is
$MA= \frac{400 N}{40 N}=10$

3 0

2 years ago

Based on the free-body diagram, the net force acting on this wheelbarrow is {blank} N.

Semenov [28]

Answer:

diagram?

Explanation:

5 0

2 years ago

Read 2 more answers

Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley

kenny6666 [7]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : x ( acceleration of mass 4m ) = $\frac{g}{7}$

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

Explanation:

Given data:

mass suspended = 4 meters

mass suspended at other end = 3 meters

first we have to express the kinetic and potential energy equations

The general kinetic energy of the system can be written as

T = $\frac{4m}{2} x^2 + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2$

T = $4mx^2 + 2my^2 -2mxy$

also the general potential energy can be expressed as

U = $-4mgx-3mg(-x+y)-mg(-x-y)+constant=-2mgy +constant$

The Lagrangian of the problem can now be setup as

$L =4mx^2 +2my^2 -2mxy +2mgy + constant$

next we will take the Euler-Lagrange equation for the generalized equations :

Euler-Lagrange equation = $4x-y =0\\-2y+x +g = 0$

solving the equations simultaneously

x ( acceleration of mass 4m ) = $\frac{g}{7}$

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

8 0

2 years ago

A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin

Dovator [93]

Answer:

fcosθ + Fbcosθ =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)

Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)

sum of x-direction forces = 0

fx+ Fbx=Wx

fcosθ + Fbcosθ =Wtanθ

7 0

2 years ago

What is the frequency of radiation whose wavelength is 11.5 a0 ?

irakobra [83]

Answer:

The frequency of radiation is $2.61 \times 10^{17} s^{-1}$

Explanation:

Given:

Wavelength $\lambda = 11.5 \times 10^{-10}$ m

Speed of light $c = 3 \times 10^{8}$ $\frac{m}{s}$

For finding the frequency of radiation,

$c = f \lambda$

$f = \frac{c}{\lambda}$

$f = \frac{3 \times 10^{8} }{11.5 \times 10^{-10} }$

$f = 2.61 \times 10^{17}$ $s^{-1}$

Therefore, the frequency of radiation is $2.61 \times 10^{17} s^{-1}$

4 0

2 years ago