While looking at bromine (Br) on the periodic table, a student needs to find another element with very similar chemical properties but with a different atomic mass. Which best describes the student’s options?

There are two possible elements, and they are directly to the left and right of bromine.
There are two possible elements, and they are directly above and below bromine.
There are many possible elements, and they are all in the same vertical column as bromine.
There are many possible elements, and they are all in the same horizontal row as bromine.

Answer:

339.3 N

Explanation:

First, we start by converting the units.

1 rev/s = 2π rad/s, so

0.6 rev/s = 2π * 0.6 rad/s

0.6 rev/s = 1.2π rad/s

0.6 rev/s = 3.77 rad/s

Now we apply the equation of motion,

W(f) = w(o) + αt

3.77 = 0 + α * 2

3.77 = 2α

α = 3.77/2

α = 1.885 rad/s²

Torque = I * α

Torque = F * r

This means that

I * α = F * r, where I = 1/2mr²

Substituting for I, we have

1/2mr²α = F * r, making F the subject of formula, we have

F = 1/2mrα, then we substitute for the values

F = 1/2 * 240 * 1.5 * 1.885

F = 678.6 / 2

F = 339.3 N

Answer:

Explanation:

The question is incomplete. Here is the complete question:

You are pushing a 13.3 kg lawn mower across the lawn with a force of 200 N. What is the value of the coefficient of friction between the mower and the grass if the mower moves with a constant velocity? The force is applied downward at an angle of 65° with the horizontal.

According to Newton's second law of motion:

[tex]\sum F_x= ma_x\\F_{app} - F_f = ma_x\\[/tex]

[tex]F_f = \mu R\\[/tex]

[tex]F_{app} - \mu Rcos \theta = ma_x[/tex]

Fapp is the applied force = 200N

Ff is the frictional force

[tex]\mu[/tex] is the coefficient of friction between the mower and the grass

R is the reaction

m is the mass of the object

ax is the acceleration

Given

R = mg = 13.3*9.8

R = 130.34N

m = 13.3kg

ax  = 0m/s² (constant velocity)

Fapp = 200N

[tex]\theta = 65^0[/tex]

Substitute the given parameters into the formula and get the coefficient of friction as shown;

Recall that: [tex]F_f = \mu R\\[/tex]

[tex]\mu = \frac{F_f}{R}\\\mu = \frac{F_{x}cos65}{F_y+W} \\\mu =\frac{ 200cos65}{200sin65+13.3(9.8)}\\\mu = \frac{84.52}{181.26+130.34}\\\mu = \frac{84.52}{311.6}\\\mu = 0.27[/tex]

Hence the coefficient of friction between the mower and the grass is 0.27

Answer:

Velocity ratio of single pulley = 2

Explanation:

Given:

Tension T

Distance = T + T = 2T

Find:

Velocity ratio of single pulley

Computation:

Velocity ratio of single pulley = Total distance / Distance effort

Velocity ratio of single pulley = 2T / T

Velocity ratio of single pulley = 2