Ask question

Ask question

All categories

3 years ago

8

Claudia uses 100 N of force on a rope attached to a pulley to lift an anvil that weighs 400 N. What is the Mechanical Advantage

of the pulley?

1 answer:

Mashcka [7]3 years ago

7 0

Answer:

100 N/ 400 N= .25

Explanation:

You might be interested in

The national high magnetic field laboratory holds the world record for creating the strongest magnetic field. for brief periods

max2010maxim [7]

Newton’s 2nd law states that Force is equal to the product of mass (m) and acceleration (a):

F = m a ---> 1

While in magnetic forces, force can also be expressed as:

F = q v B ---> 2

where,

q = total charge

v = velocity = 45 cm / s = 0.45 m / s

B = the magnetic field = 85 T

First we solve for the total charge, q:

q = 3.8 × 10^-23 g (1 mol / 23 g) (6.022 × 10^23 electrons / mol) (1.602 × 10^-19 C / electron)

q = 1.594 × 10^-19 C

We equate equations 1 and 2 then solve for acceleration a:

m a = q v B

a = q v B / m

a = [1.594 × 10^-19 C * 0.45 m / s * 85 T] / 3.8 × 10-26 kg

a = 160,437,862.2 m/s^2

Therefore the maximum acceleration of Na ions is about 160 × 10^6 m/s^2.

5 0

3 years ago

Read 2 more answers

1. During the Middle Ages, armies often attacked castles using large siege engines such as the counterweight trebuchet at left.

fredd [130]

Answer: Some challenges that I could see would be the walls, possibly a moat, tar, and the towers.

Explanation: The wall is obliviously a main problem, trying to get over it or through it is a difficult challenge. The moat (if it has one) means that there is more than likely only one way to get in or out. If it does have tar it means that the attackers are going to be put in a "sticky situation" And finally the towers, they have people at the top shooting arrows down at you, or throwing things at you.

8 0

3 years ago

A uniform electric field is created by two parallel plates separated by a distance of 0.04 m. What is the magnitude of the elect

FromTheMoon [43]

Complete question:

A uniform electric field is created by two parallel plates separated by a

distance of 0.04 m. What is the magnitude of the electric field established

between the plates if the potential of the first plate is +40V and the second

one is -40V?

Answer:

The magnitude of the electric field established between the plates is 2,000 V/m

Explanation:

Given;

distance between two parallel plates, d = 0.04 m

potential between first and second plate, = +40V and -40V respectively

The magnitude of the electric field established between the plates is calculated as;

E = ΔV / d

where;

ΔV is change in potential between two parallel plates;

d is the distance between the plates

ΔV = V₁ -V₂

ΔV = 40 - (-40)

ΔV = 40 + 40

ΔV = 80 V

E = ΔV / d

E = 80 / 0.04

E = 2,000 V/m

Therefore, the magnitude of the electric field established between the plates is 2,000 V/m

7 0

2 years ago

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

2 years ago

Shawn uses 45 N of force to stop the cart 27 meter from running his foot over. How much work does he do?

Ede4ka [16]

Answer:

W=f×d

w=45×27

w=1215 j.............

3 0

3 years ago