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3 years ago

ANswer the question and give example: Can friction ever perform positive work?

2 answers:

4 0

Yes. Think of block sitting on top of a bigger block. If the bottom block moves, it will drag the top block with it. Since the force of friction on the small block and its displacement are in the same direction, the "work" is positive. The static friction is a passive force, It is not a source of energy; it transmits the force placed on the bottom block. (And the "work" done by the friction on the bottom block is exactly the negative of the work done on the top block.)

quester [9]3 years ago

3 0

It do both Positive and Negative work.
If we have to stop some object then friction does a positive work.
For Example: to stop a car , necassary friction is required between the tyres of car and road to stop a car .

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Fruits and vegetables should take up half you plate at a meal.<br> A. <br> True<br> B. <br> False

umka21 [38]

Answer:

A. true

Explanation:

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3 years ago

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A baseball has a mass of 145 g. A bat exerts a force of 18,400 N on the ball. What is the acceleration of the ball?

amid [387]

The correct formula to use for the situation given above is: F = MA, where F is the applied force, M is the mass of the object and A is the acceleration.

From the details given in the question, we are told that:

F = 18, 400N

M = 145 g = 145 / 1000 = 0.145 kg

A = ?

From the equation F = MA

A = F / M

A = 18,400 / 0.145 = 126,896.55 = 1.27 *10^5.

Therefore, the correct option is C.

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3 years ago

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ben walks 2 m from his desk to the teachers desk. From the teachers desk he then walks 3 m in the same direction to the classroo

vaieri [72.5K]

Distance is the total length covered = 2m + 3m = 5m

Displacement is his distance from original position.

Displacement = 2m + (-3)m. Representing the 3m walked back as -3.

Displacement = 2m - 3m = -1m.

So his displacement is 1m behind his original starting point.

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3 years ago

The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ

LenaWriter [7]

The orbital period of the satellite[T] is given as $2*10^{6} S$ .

The radius of the satellite is given [R] $2.5*10^{12} m$ .

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V] written as -

V $=\frac{2\pi R}{T}$

where T is the time period of the satellite and R is the radius of the satellite.

$V=\frac{2*3.14*10^{12} }{2*10^{6} }$

$= 7.85*10^{6} m/s$

There is also another way through which we can get the solution as explained below-

We know that the tangential speed of a satellite V $=\sqrt{\frac{GM}{R^{2} } }$

where G is the gravitational constant and M is the mas of central object.

But we know that $g=\frac{GM}{R^{2} }$

⇒ $GM=gR^{2}$ where g is the acceleration due to gravity of that central object.

Hence $V=\sqrt{\frac{gR^{2} }{R} }$

⇒ $V=\sqrt{gR}$

By knowing the value of g due to that central object we can also calculate its tangential speed.

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3 years ago

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Why does a heavier object fall faster

Salsk061 [2.6K]

Explanation:

In a vacuum (no air resistance), it doesn't. All falling objects, regardless of mass, accelerate at the same rate.

However, when air resistance is taken into account, heavier objects indeed fall faster than lighter objects, provided they have the same shape and size. For example, a lead ball falls faster than a styrofoam ball.

To understand why, first look at what factors affect air resistance:

D = ½ρv²CA

where ρ is air density,

v is velocity,

C is drag coefficient,

and A is cross sectional area.

As falling objects accelerate, they eventually reach a maximum velocity where air resistance equals weight. This is called terminal velocity.

D = W

½ρv²CA = mg

v = √(2mg/(ρCA))

If we increase m while holding everything else constant, v increases. So two objects with the same size and shape but different masses will have different terminal velocities, with the heavier object falling faster.

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3 years ago

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