Answer:
Like charges repel each other; unlike charges attract. Thus, two negative charges repel one another, while a positive charge attracts a negative charge. The attraction or repulsion acts along the line between the two charges. The size of the force varies inversely as the square of the distance between the two charges.
Answer:
positive and negative ??
Explanation:
I dont really understand the question if im being honest
One end of a string is attached to an object of mass M, and the other end of the string is secured so that the object is at rest as it hangs from the string. When the object is raised to a position X that is a height H above its lowest point and released from rest, the object undergoes simple harmonic motion. When the object passes through the equilibrium position Y, it has a speed v0.
What methods could a student use to determine the total mechanical energy E at position Y, and why?
Answer:
v₀ = [tex]\sqrt{2gH}[/tex]
to determine this speed experimentally, the student must measure the height of the body as a function of time and with equation (1) he can find the speed for each point of interest
Explanation:
In this internal exercise the student must use the conservation of mechanical energy,
Starting point. Highest point of the trajectory
Em₀ = U = m g H
Point of interest. Point at height Y
[tex]Em_{f}[/tex] = K + U = ½ m v² + m g Y
energy is conserved
Em₀ = Em_{f}
m g H = ½ m v² + m g Y
v² = 2 g (H -Y) (1)
in this case they indicate that Y is the equilibrium position whereby Y = 0 and the velocity is v = v₀
v₀ = [tex]\sqrt{2gH}[/tex]
Therefore, to determine this speed experimentally, the student must measure the height of the body as a function of time and with equation (1) he can find the speed for each point of interest
You push a manual lawn mower across the lawn at constant speed. What is the value of the coefficient of friction between the mower and the grass
Answer:
0.27Explanation:
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
g A 240-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s
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
Velocity ratio of single movable pulley is 2 when Velocity ratio of single fixed pulley is 1. why
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