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4 years ago

The answer please!!!

Physics

1 answer:

Tomtit [17]4 years ago

8 0

Ahhh ! Tricky tricky question !

All of the books are just sitting there, "RESTING" on the table. They're not slipping, sliding, rising, falling, rolling, swinging, swooshing, wandering, strolling, or meandering. Just "RESTING" in their stack.

None of the books has a changing speed or direction of motion. So none of the books is experiencing any acceleration.

So the NET FORCE on any book <em>must be</em> <em>zero</em>.

Pick a book. It doesn't matter whether you picked the bottom, middle, or top book. And the book's mass, weight, color, age, author, cost, or topic also don't make any difference. If the book is not accelerating, then the NET FORCE on it is zero. Just ask Dr. Newton. He'll tell you "F = m A" .

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he membrane that surrounds a certain type of living cell has a surface area of 6.0 x 10-9 m2 and a thickness of 1.6 x 10-8 m. As

k0ka [10]

Answer:

$1.54481175\times 10^{-12}\ C$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area = $6\times 10^{-9}\ m^2$

d = Thickness = $1.6\times 10^{-8}\ m$

k = Dielectric constant = 5.4

V = Voltage = 86.2 mV

Charge is given by

$Q=CV\\\Rightarrow Q=k\epsilon\dfrac{A}{d}V\\\Rightarrow Q=5.4\times 8.85\times 10^{-12}\times \dfrac{6\times 10^{-9}}{1.6\times 10^{-8}}\times 86.2\times 10^{-3}\\\Rightarrow Q=1.54481175\times 10^{-12}\ C$

The charge on the outer surface is $1.54481175\times 10^{-12}\ C$

4 0

3 years ago

What is the equation for gravity and what does each part mean?

Andrew [12]

This equation description the force between any two object in the universe in the equation F is the force of the gravity Newton lawe

6 0

3 years ago

A 15.7 kg block is dragged over a rough, horizontal surface by a constant force of 83.1 N

Kay [80]

Answer:

The work done by the 83.1 N force is 4687.5 J.

The magnitude of the work done by the force of friction is 1187.5 J.

Explanation:

Given:

Mass of the block, $m=15.7$ kg

Force acting on it, $F=83.1$ N

Angle of application of force, $\theta = 25.7$ °

Displacement of the block, $d=62.6$ m

Coefficient of friction, $\mu =0.161$

Acceleration due to gravity, $g=9.8$ m/s²

Work done by a force is given as:

$Work,W=F\times d\times \cos\theta$

So, work done by the constant force is given as:

$W_{force}=83.1\times 62.6\times \cos(25.7)\\W_{force}=4687.5\textrm{ J}$

Now, in order to find work done by friction, we need to evaluate friction.

For the vertical direction, the net force is zero as there is no vertical motion.

Therefore,

$N + F\sin\theta = mg\\ N = mg-F\sin\theta$

Frictional force is given as:

$f=\mu N=\mu (mg-F \sin \theta)=0.161\times ((15.7\times 9.8)-(83.1\times \sin(25.7))=18.97\textrm{ N}$

Now, friction acts in the direction opposite to the displacement. Thus, angle between frictional force and displacement is 180°.

Therefore, work done by friction is:

$W_{fric}=f\times d\times \cos\theta_f\\W_{fric}=12.72\times 62.6\times \cos(-180)\\W_{fric}=18.97\times 62.6\times -1\\W_{fric}=-1187.5\textrm{ J}$

Negative sign indicates that frictional force is acting opposite to motion.

So, the magnitude of the work done by the force of friction is 1187.5 J.

8 0

4 years ago

How much GPE is stored when an 80kg astronaut climbs to the top of a 5m high lunar lander? The gravity strength on the moon is 1

8_murik_8 [283]

Answer:

The GPE, stored is 640 Joules

Explanation:

The given parameters are;

The given mass of the astronaut, m = 80 kg

The height of the top of the lunar lander to which the astronaut climbs, h = 5 m

The gravity strength on the moon, g = 1.6 N/kg

The Gravitational Potential Energy, GPE, stored is given according to the following equation;

GPE stored = m·g·h

Therefore, by substituting the known values, we have;

GPE Stored = 80 kg × 1.6 N/kg × 5 m = 640 Joules

The GPE, stored = 640 Joules.

6 0

3 years ago

A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m. The force of friction

boyakko [2]

Given :

A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m.

The force of friction between the refrigerator and the floor is 230 N.

To Find :

How much work has been performed by the mover on the refrigerator.

Solution :

Since, refrigerator is moving with constant velocity.

So, force applied by the mover is also 230 N ( equal to force of friction ).

Now, work done in order to move the refrigerator is :

$W = Force\times distance\\\\W = 230 \times 8.1\ N\ m\\\\W = 1863\ N\ m$

Hence, this is the required solution.

5 0

3 years ago