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3 years ago

Physics help me this is idek

Physics

1 answer:

blagie [28]3 years ago

5 0

Assuming B is a stoping point that means there is acceleration because acceleration is slowing down to a stop, turning around, or speeding up. Now we also know that its speed is decreasing due to the fact that the dots are getting closser to each other. Its only velocity is it tells a direction and speed so that rules out B and C. All you have left is A and D so, you need to figure out which one has acceleration as slowing to a stop and the speed decreasing. (They kind of walk hand in hand.)

I hope that helped you.

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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

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

Learn more about torques and equilibrium:

brainly.com/question/5352966

#LearnwithBrainly

7 0

3 years ago

Use Newton’s Universal Law of Gravitation to calculate the magnitude of the gravitational force between a 200 kg refrigerator an

expeople1 [14]

Answer:

3.735×10⁻⁶ N

Explanation:

From newton' s law of universal gravitation,

F = Gmm'/r² .............................. Equation 1

Where F = Gravitational force between the person and the refrigerator, m = mass of the person, m' = mass of the refrigerator, r = distance between the person and the refrigerator. G = gravitational universal constant.

Given: m = 70 kg, m' = 200 kg, r = 0.5 m

Constant: G = 6.67×10⁻¹¹ Nm²/kg².

F = (6.67×10⁻¹¹×70×200)/0.5²

F = 93380×10⁻¹¹/0.25

F = 373520×10⁻¹¹

F = 3.735×10⁻⁶ N

Hence the force between the person and the refrigerator = 3.735×10⁻⁶ N

6 0

4 years ago

A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in.-thick plate by welding along a helix that forms an angle of 25

Varvara68 [4.7K]

Answer:

Normal stress = 66/62.84 = 1.05kips/in²

shearing stress = T/2 = 0.952/2 = 0.476 kips/in²

Explanation:

A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in

4 -in.-thick

angle of 25°

Axial force P = 66 kip axial force

determine the normal and shearing stresses

Normal stress б = force/area = P/A

= 66/ (П* (d₂²-d₁²)/4

=66/ (3.142* (12²-8²)/4

= 66/62.84 = 1.05kips/in²

Tangential stress T = force* cos ∅/area = P/A

= 66* cos 25/ (П* (d₂²-d₁²)/4

=59.82/ (3.142* (12²-8²)/4

= 59.82/62.84 = 0.952kips/in²

shearing stress = tangential stress /2

= T/2 = 0.952/2 = 0.476 kips/in²

8 0

3 years ago

The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the wave shown?

belka [17]

Answer:

B) 0.3Hz

Explanation:

I just took the test i hope i helped and i hope you pass the test

7 0

3 years ago

Read 2 more answers

A 4 kg rock is dropped from 5 m. There is no friction. What kind of energy does is have before? What kind of energy does it have

denpristay [2]

1) The total mechanical energy of the rock is:
$E=U+K$
where U is the gravitational potential energy and K the kinetic energy.

Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
$E_i=U=mgh$
where $m=4 kg$ is the mass, $g=9.81 m/s^2$ is the gravitational acceleration and $h=5 m$ is the height.
Putting the numbers in, we find the potential energy
$U=mgh=(4 kg)(9.81 m/s^2)(5 m)=196.2 J$

2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
$E_f=K= \frac{1}{2}mv^2$
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
$K=U=196.2 J$

3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
$W=196.2 J-0 J=196.2 J$

6 0

3 years ago