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ratelena [41]
3 years ago
11

A man seeking to set a world record wants to tow a 101,000-kg airplane along a runway by pulling horizontally on a cable attache

d to the airplane. The mass of the man is 84 kg, and the coefficient of static friction between his shoes and the runway is 0.76. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.
Physics
1 answer:
NemiM [27]3 years ago
7 0

Answer:

6.19 x 10^{-3} m/s^{2}

Explanation:

For this exercise we need to sum the forces on the y-axis and x-axis as follows:

∑F_{y} = N - mg = m.a_{y} = 0

From the exercise, we deduce there is no motion in y-axis, so:

N = mg

Then for x-axis we have:

∑F_{x} = H - f^{s} = m.a_{x} = 0

Now, from the exercise we deduce that we are looking for the greatest static friction which means to have the maximun static friction to start moving, so at this point the acceleration is zero, so we can find horizontal force (H), which then will act in the airplane to move it. Therefore we have:

H = f^{s} = f^{sma_{x} } = u_{s}N = u_{s}mg

H = (0.76)(84Kg)(9.8m/s^{2})

H = 625.63 N

Now we apply this force to the weight of the plane to find the greatest acceleration the mann can give to start moving the plane.

a = \frac{F}{m} = \frac{H}{m}

a = \frac{6325.63N}{101000Kg}

a = 6.19 x 10^{-3} m/s^{2}

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a body of radius R and mass m is rolling horizontally without slipping with speed v. it then rolls us a hill to a maximum height
ki77a [65]

Answer:

mR²/2

Explanation:

Here is the complete question

An object of radius′

R′  and mass ′

M′  is rolling horizontally without slipping with speed ′

V′

. It then rolls up the hill to a maximum height h = 3v²/4g. The moment of inertia of the object is (g= acceleration due to gravity)

Solution

Since it rolls without slipping, there is no friction. So, its initial mechanical energy at the horizontal surface equals its final mechanical energy at the top of the hill.

Since the object is rolling initially, and on horizontal ground, it initial energy is kinetic and made up of rotational and translational kinetic energy.

So, E = K + K'

E = 1/2mv² + 1/2Iω² where m = mass of object, v = speed of object, I = moment of inertia of object and ω = angular speed of object = v/r where v = speed of object and R = radius of object.

Also, the final mechanical energy of the object, E' is its potential energy at the top of the hill. So, E' = mgh.

Since E = E',

1/2mv² + 1/2Iω² = mgh

substituting the values of ω and h into the equation, we have

1/2mv² + 1/2Iω² = mgh

1/2mv² + 1/2I(v/R)²= mg(3v²/4g)

Expanding the brackets, we have

1/2mv² + 1/2Iv²/R²= 3mv²/4

Dividing through by v², we have

1/2m + I/2R²= 3m/4

Subtracting m/2 from both sides, we have

I/2R² = 3m/4 - m/2

Simplifying, we have

I/2R² = m/4

Multiplying through by 2R², we have

I = m/4 × 2R²

I = mR²/2

6 0
3 years ago
A cassette recorder uses a plug-in transformer to convert 120 V to 12.0 V, with a maximum current output of 200 mA. What is the
goldfiish [28.3K]

Answer:

20 mA

Explanation:

We have given that the transformer convert 120 V to 12 V means it is a step down transformer

The turn ratio of the transformer is given as a=\frac{N_1}{N_2}=\frac{V_1}{V_2}=\frac{120}{12}=10 here V_! is primary voltage and V_2 is secondary voltage

As the voltage decreases from primary to secondary so current will increase from primary to secondary

We have given output current that is secondary current so primary that is input current will be less

So input current =\frac{200}{10}=20mA

4 0
3 years ago
The line that is drawn perpendicular to the point at which a wave intersects a boundary is know as the
fomenos
<span>The line that is drawn perpendicular to the point at which a wave intersects a boundary is know as the Normal .

When the normal is drawn, the incident ray makes an angle with it known as the angle of incidence and the reflected ray makes an angle with it known as the angle of incidence. These angles are always equal.
 
The refracted ray makes an angle with the normal known as angle of refraction. The sin of angle of incidence to the sin of angle of refraction is called the refractive index( </span>μ= <span>sin i / sin r) .

hope all of it helps you!</span>
5 0
3 years ago
The box resting on the inclined plane above has a mass of 20kg. The incline sits at a 30o angle. Find the friction force between
tekilochka [14]

The friction force between the box and the incline if the box does not slide down the incline will be 0.577

The force preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction. There are several kinds of friction: Two solid surfaces in touch are opposed to one another's relative lateral motion by dry friction.

Given the box resting on the inclined plane above has a mass of 20kg and the The incline sits at a 30 degree angle

We have to find the friction force between the box and the incline if the box does not slide down the incline

Since the frictional force F₁ must equal or exceed gravitational force F₂ down the incline:

F₁ = F₂

μmgcosΘ = mgsinΘ

μ = (mgsinΘ)/(mgcosΘ)

μ = tanΘ

μ = 0.577

Hence the friction force between the box and the incline if the box does not slide down the incline will be 0.577

Learn more about friction force here:

brainly.com/question/24386803

#SPJ4

3 0
2 years ago
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The human Skeleton ? Lol
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