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2 years ago

Look at the question on the pic :)

Physics

2 answers:

MAVERICK [17]2 years ago

7 0

A it has to be solid

Mrrafil [7]2 years ago

6 0

Answer:

Explanation:

Has to be a solid

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Jill applies a force of 250 N to a machine. The machine applies a force of 25 N to an object. What is the mechanical advantage o

Vinvika [58]

Mechanical advantage is defined as the ratio of output load to the input load. The mechanical advantage of the machine will be 0.1.

<h3>What is mechanical advantage?</h3>

Mechanical advantage is a measure of the ratio of output force to input force in a system,

It is used to obtain the efficiency of forces in levers and pulleys. It is an effective way of amplifying the force in simple machines like levers.

The theoretical mechanical advantage is defined as the ratio of the force responsible for the useful work in the system to the applied force.

Given

applied force = 250 N

Output force = 25

Mechanical advantage = work output / work input

$\rm{Mechanical advantage}=\frac{F_O}{F_I}$

$\rm{Mechanical advantage}=\frac{25}{250}$

$\rm{Mechanical advantage}=0.1$

Hence the mechanical advantage of the machine will be 0.1

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2 years ago

What is one way that early scientific practice differed from modern scientific practice?

mote1985 [20]

Early hypotheses were not based on observations.
Early hypotheses were not tested by experimentation.
Early hypotheses were formed from scientific questions.
Early hypotheses were influenced by creative thinking

6 0

3 years ago

Read 2 more answers

Is this right? plzz anwser soon

-Dominant- [34]

Answer:

yes

Explanation:

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2 years ago

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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:

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:

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

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:

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

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.

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

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

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3 years ago

Initially a car accelerates at 2 m/s2 for x seconds. The car then travels at a velocity of -6 m/s for x seconds. If the car disp

Luda [366]

Answer:

The time travel is

t=8 s

Explanation:

$a= 2 \frac{m}{s^{2} } \\v=-6 \frac{m}{s} \\x=16m$

$x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}$

$x_{f}=0-6*t+\frac{1}{2} *2*t^{2}$

$t^{2}-6*t-16=0\\ using :\\\frac{-b+/-\sqrt{b^{2}-4*c*a } }{2} \\\frac{-(-6)+/-\sqrt{(-6)^{2}-4*(-16)*(1) } }{2}=\frac{3}{2} +/- \frac{10}{2} \\t_{1} = 2s \\t_{2} = 8s$

Check

$t_{2}=8s$

$x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}$

$x_{f}=0-6*+\frac{1}{2} *2*8^{2}$

$x_{f}=-48+64\\x_{f}=16$

5 0

3 years ago