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3 years ago

Pertaining to simple machines and levers what changes when the fulcrum position is modified?

Physics

1 answer:

Inessa [10]3 years ago

7 0

Explanation :

There are many types of simple machines i.e. lever, wedge, pulley, wheel and axle etc.

Fulcrum is the pivot point about which the lever turns. When the position of the fulcrum is shifted, the position of effort changes.

There are three classes of the lever as class 1, class 2 and class 3.

Class 1 includes seesaws and scissors.

class 2 includes wheelbarrow

class 3 includes tongs and tweezers.

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A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0

3 years ago

An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields a

Lapatulllka [165]

Answer:

The speed of the electron is $2.55\times 10^3\ m/s$ .

Explanation:

Given that,

The magnitude of electric field, $E=1.32\ kV/m=1.32\times 10^3\ V/m$

The magnitude of magnetic field, B = 0.516 T

Both the magnetic and electric fields are acting on the moving electron. Then, the magnitude of electric field and magnetic field is balanced such that :

$evB=eE\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.32\times 10^3}{0.516}\\\\v=2558.13\ m/s$

$v=2.55\times 10^3\ m/s$

So, the speed of the electron is $2.55\times 10^3\ m/s$ . Hence, this is the required solution.

3 0

3 years ago

Two bicyclist, originally separated by a distance of 20 miles, are each traveling at a uniform speed of 10 miles per hour toward

Radda [10]

Answer:

D = 25 miles

Explanation:

To solve this problem, we just need to know how much time it took both bicyclists to collide and that will be the same amount of time that the bee flew at 25miles per hour. With those values we could calculate the distance it traveled.

Since both bicyclists collide, we know that Xa=Xb, so:

Xa = V*t = 10*t and Xb = 20 - V*t = 20 - 10*t

10*t = 20 - 10*t Solving for t:

t = 1 hour Now we can calculate the distance for the bee:

D = Vbee * t = 25 * 1 = 25 miles

6 0

3 years ago

MRU(movimiento rectilineo uniforme) un movil viaja en linea recta con una velocidad media de 12metros /segundos durante 9segundo

liubo4ka [24]

Answer:

a) 141.6m

b) 8.4m/s

Explanation:

a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:

$x_1=v_1t_1=(12m/s)(9s)=108m\\\\x_2=v_2t_2=(4.8m/s)(7s)=33.6m\\\\x_T=x_1+x_2=108m+33.6m=141.6m$

hence, the total distance is 141.6m

b) the mean velocity of the total trajectory is given by:

$v_m=\frac{v_1+v_2}{2}=\frac{12m/s+4.8m/s}{2}=8.4\frac{m}{s}$

hence, the mean velocity is 8.4 m/s

8 0

3 years ago

Hey what is you answering streak and what fact dose it say

Valentin [98]

Answer:

It’s 53:) Or - quit

Explanation:

That’s my strong streak

5 0

2 years ago

Read 2 more answers