What is effort arm
don't say the answer of gogle

Answers

Answer 1

Answer:

Answer:

effort arm mean the use of any work by using your hand force motion or by hand power

Related Questions

Where would the normal force exerted on the rover when it rests on the surface of the planet be greater
As in problem 80, an 76-kg man plans to tow a 128000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that he instead attempts the feat by pulling the cable at an angle of 6.7° above the horizontal. The coefficient of static friction between his shoes and the runway is 0.87. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.
The position of a particle changes linearly with time, i.e. as one power of t, as given by the following: h(t) = ( 4.1 t + 5.5 ) meters. Find the speed of the particle, in meters per second.
A conveyor belt is used to move sand from one place to another in a factory. The conveyor is tilted at an angle of 18° above the horizontal and the sand is moved without slipping at the rate of 2 m/s. The sand is collected in a big drum 5 m below the end of the conveyor belt. Determine the horizontal distance between the end of the conveyor belt and the middle of the collecting drum.
Find the average speed of the electrons in a 1.0 cm diameter, copper power line, when it carries a current of 20 A.

A farmer is using a rope and pulley to lift a bucket of water from the bottom of a well. the farmer uses a force f1=57.5 n to pull the bucket of water upwards. the total mass of the bucket of water is f2= 3.9kg. -Calculate how much work Wg in J gravity does on the bucket filled with water as the farmer lifts it up the well.
-Calculate the net work Wnet in J done on the bucket of water by the two forces F1 and Fg.

Answers

Final answer:

To calculate the work done by gravity on the bucket of water as it is lifted up the well, multiply the weight of the bucket by the lifting distance. The net work done on the bucket by the force applied by the farmer and gravity is the sum of the work done by both forces. The net work is represented by the equation Wnet = W1 + Wg.

Explanation:

To calculate how much work gravity does on the bucket filled with water as the farmer lifts it up the well, we need to multiply the force of gravity (weight) by the vertical distance the bucket is lifted. The equation for work is W = Fd, where W is the work done, F is the force, and d is the distance. In this case, the force of gravity is the weight of the bucket, which can be calculated by multiplying the mass by the acceleration due to gravity (9.8 m/s^2).

So, the work done by gravity (Wg) on the bucket is Wg = Fg * d = (m * g) * d = (3.9 kg * 9.8 m/s^2) * d = 38.22 d Joules.

To calculate the net work done on the bucket by the two forces, we can use the equation Wnet = W1 + Wg, where W1 is the work done by force F1 and Wg is the work done by gravity. Since force F1 and the displacement (lifting distance) are both vertical, the work done by F1 is given by W1 = F1 * d.

Therefore, the net work done on the bucket by forces F1 and gravity is Wnet = F1 * d + Fg * d = (57.5 N) * d + (3.9 kg * 9.8 m/s^2) * d = (57.5 N + 38.22 d) Joules.

Learn more about Work done by forces here:

brainly.com/question/31428590

#SPJ12

Why a switch is connected in phase wire and never is neutral wire?

Answers

If you had the wire connected to a neutral wire, you would never get a charge. You could receive a charge pulse from the phase wire.

Answer:

ifyou have a wre connect you should not have to connected

i think that is the answer

A laser (electromagnetic wave) has the maximum electric field strength of 1.0x1011 V/m. What is the force the laser applies on a mirror (totally reflective) of 5.0 mm2 area? A. 2.76 x105N B. 1.21 x106N C. 1.94 x106N D.4.43 x105 N E. 7.82 x104N

Answers

Answer:

The correct option is D

Explanation:

From the question we are told that

The maximum electric field strength is $E = 1.0 *10^(11) \ V/m$

The area is $A = 5.0 \ mm^2 = 5.0 *10^(-6) \ m^2$

Generally the force the laser applies is mathematically represented as

$F = \epsilon_o * E ^2 * A$

Here $\epsilon_o = 8.85*10^(-12) C/(V \cdot m)$

$F = 8.85*10^(-12) * (1.0 *10^(11)) ^2 * 5.00*10^(-6 )$

=> $F = 4.43 *10^(5) \ N$

why did thomson's from experermenting with cathode rays cause a big change in scientific thought about atoms

Answers

Answer:

His results gave the first evidence that atoms were made up of smaller particles.

1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of charge +4.00 μC and mass 10.0 mg moving along the y-axis. If Q3 starts from rest at y = 2.00 cm, what is its speed when it reaches y = 4.00 cm?

Answers

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = $\sqrt{3^(2) +2^(2) }$ = $√(13)$ = 3.61cm = 0.036m

r₂ = $\sqrt{4^(2) + 3^(2) }$ = $√(25)$ = 5cm = 0.05m

electric potential V = $(kq)/(r)$

change in potential ΔV = $V_(1) - V_(2)$

ΔV = $(2kq_(1) )/(r_(1)) - (2kq_(2) )/(r_(2) )$ , where $q_(1) = q_(2)=$ 2.00μC

ΔV = $2kq((1)/(r_(1)) - (1)/(r_(2) ))$

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × $((1)/(0.036) - (1)/(0.05) )$

ΔV= 2.789×10⁵

$(1)/(2)mv^(2)$ = ΔV × q₃

$(1)/(2)$ ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s

We are given;

Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C

Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C

Distance of charge 1 along x = 3 cm = 3 × 10⁻² m

Distance of charge 2 along x = -3 cm = -3 × 10⁻² m

Charge 3; q₃ = +4.00 μC = 4 × 10⁻⁶ C

mass; m = 0.01 g

distance of charge 3 along y = 4 cm = 4 × 10⁻² m

q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.

Thus;

Distance of charge 1 from the initial position of q₃;

r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)

r₁ = 0.0361 m

Distance of charge 2 from the final position of q₃;

r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)

r₂ = 0.05 m

Now, formula for electric potential is;

V = kq/r

Where k = 9 × 10⁹ N.m²/s²

Thus,change in potential is;

ΔV = V₁ - V₂

Now, Net V₁ = 2kq₁/r₁

Net V₂ = 2kq₂/r₂

Thus;

ΔV = 2kq₁/r₁ - 2kq₂/r₂

ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]

ΔV = 277229.92 V

Now, from conservation of energy;

½mv² = q₃ΔV

Thus;

½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92

v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01

v = √(221.783936)

v = 14.89 m/s

Read more about point charges at;brainly.com/question/13914561

7. If the impact of the golf club on the ball in the previous question occurs over a time of 2 x 10 seconds, whatforce does the ball experience to accelerate from rest to 73 m/s?

Answers

Answer:

3.65 x mass

Explanation:

Given parameters:

Time = 20s

Initial velocity = 0m/s

Final velocity = 73m/s

Unknown:

Force the ball experience = ?

Solution:

To solve this problem, we apply the equation from newton's second law of motion:

F = m $(v - u)/(t)$

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

So;

F = m ( $(73 - 0)/(20)$ ) = 3.65 x mass

Final answer:

To calculate the force experienced by the ball to accelerate from rest to 73 m/s, use Newton's second law of motion.

Explanation:

To calculate the force experienced by the ball to accelerate from rest to 73 m/s, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = m * a).

Since the ball starts from rest, its initial velocity (vi) is 0 m/s. The final velocity (vf) is 73 m/s. The time (t) taken for the impact is given as 2 x 10 seconds. So, the acceleration (a) can be calculated using the formula a = (vf - vi) / t.

Substituting the given values into the equation, we have a = (73 - 0) / (2 x 10) = 3.65 m/s^2.

Now, we can find the force (F) using the formula F = m * a. If the mass of the ball is known, we can substitute it into the equation to find the force experienced by the ball.