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3 years ago

7

Westinghouse and Edison fought what was known as the war of the currents. Eventually, Westinghouse triumped using Alternating Cu

rrent for power distribution. Alternating current is better for power generation because A) direct current is unsafe. B) alternating current produces flutuating magnetic fields. C) alternating current provides a more constant current flow. D) direct current cannot be produce electricity without a battery.

2 answers:

yawa3891 [41]3 years ago

7 0

It is B! I know because i just took a USA test prep on it.

DerKrebs [107]3 years ago

6 0

The answer is B! :) Hope it helped

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A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.

SVETLANKA909090 [29]

A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

$W=mg$ : weight of the ladder, with m = 20 kg (mass) and $g=9.8 m/s^2$ (acceleration of gravity)

$W_M=Mg$ : weight of the person, with M = 54 kg (mass)

$N_1$ : normal reaction exerted by the wall on the ladder

$N_2$ : normal reaction exerted by the floor on the ladder

$F_f = \mu N_2$ : force of friction between the floor and the ladder, with $\mu$ (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

$W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}$

where

$Wsin 21^{\circ}$ is the component of the weight of the ladder perpendicular to the ladder

$W_M sin 21^{\circ}$ is the component of the weight of the man perpendicular to the ladder

$N_1 sin 69^{\circ}$ is the component of the normal force perpendicular to the ladder

And solving for $N_1$ , we find the force exerted by the wall on the ladder:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N$

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of $N_2$ .

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

$\sum F_y = 0\\N_2 - W - W_M =0$

And substituting and solving for N2, we find:

$N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N$

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

$\sum F_x = 0\\F_f - N_1 = 0$

And re-writing the equation,

$\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257$

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

$\mu = \frac{N_1}{N_2}$

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}$

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)$

And substituting, we get

$N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N$

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

$\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332$

Learn more about torques and equilibrium:

brainly.com/question/5352966

#LearnwithBrainly

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3 years ago

What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne

irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance $h= 1790 km = 1.790\times10^{6}\ m$

Magnetic field $B=4\times10^{-8}\ T$

Mass of proton $m=1.673\times10^{-21}\ Kg$

Radius of earth $R =6.38\times10^{6}\ m$

Radius of orbit $r=R+h$

$r=6.38\times10^{6}+1.790\times10^{6}$

$r=8170000\ m$

We need to calculate the speed

Using formula of magnetic field

$Bvq=\dfrac{mv^2}{r}$

$v=\dfrac{Bqr}{m}$

Put the value into the formula

$v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}$

$v=31.25\ m/s$

Hence, The velocity is 31.25 m/s and direction is toward west.

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3 years ago

In a ballistics test at the police station an officer

posledela

You times the 6 by the 350 duvided 1.8

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3 years ago

In some recent studies it has been shown that women are men when competing in similar sports (most notably in soccer and basketb

Elza [17]

Answer: The correct option is B (The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain)

Explanation:

Anterior cruciate ligament (ACL) is one of the important ligaments found at the knee joint which helps to stabilise the joint. It connects the femur to the tibia bone at the knee joint.

Anterior cruciate ligament tear is one of the common knee joint injury which is seen in individuals( especially females) involved in sports( example soccer and basketball which involves sudden change in direction causing the knee to rotate inwards)

ACL tear occurs through both contact and non contact mechanisms. The contact mechanism of ACL injury occurs when force is directly applied at the lateral part of the knee while in non contact mechanism,tear occurs when the tibia is externally rotated on the planted foot.

Research has proven that women are prone to have ACL tear than men when competing in similar sports. This disparity exists due to structural differences that pose as risk factors. These includes

- the female ACL size is smaller than the male.

- the ACL of female has a lower modulus if elasticity( that is, less stiff) than in males leading to greater joint mobility than in the male.. therefore the option, (The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain) is correct.

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3 years ago

Rank the four fundamental forces from strongest to weakest

kvasek [131]

1) the strong nuclear force, 2) the electromagnetic force, 3) the weak nuclear force, and 4) gravity

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