How do you calculate energy lost due to friction in an experiment?

a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe

Elden [556K]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

$Mg(r - L/2) = mg(L/2 - 8)$

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

$400,000(r - 10) = 19,600 (10 - 8)$

$r - 10 = 0.098$

$r = 10.098 m$

so pillar is at distance 10.098 m from one end of the slab

3 0

3 years ago

A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout

nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
So, we can write the following equation:

$E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)$

As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
So, on the outer surface of the shell there must be a charge that be the difference between them:

$Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C$

Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

$E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C$

As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.

6 0

3 years ago

A 50 kg astronaut floating in space throws her 2 kg wrench to the left at 10

Hoochie [10]

The astronaut will move at 0.4 m/s in the opposite direction to the wrench

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the astronaut-wrench system must be conserved before and after the launch.

Before the launch, the total momentum is zero, since the astronaut is at rest:

p = 0 (1)

After the launch, the total momentum is:

$p=mv+MV$ (2)

where :

m = 2 kg is the mass of the wrench

v = 10 m/s is the velocity of the wrench

M = 50 kg is the mass of the astronaut

V is the recoil velocity of the astronaut

Since momentum is conserved, we can write (1) = (2), and so we can solve for V:

$0=mv+MV\\V=-\frac{mv}{M}=-\frac{(2)(10)}{50}=-0.4 m/s$

And the negative sign means that the astronaut will move in the opposite direction to the wrench.

Learn more about conservation of momentum:

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8 0

3 years ago

A satellite is revolving the earth 4km above the surface.find the orbital velocity of the satellite (R =6400km,g=9.8m/s^2)

Karo-lina-s [1.5K]

Answer:

v ≈ 7900 m/s

Explanation:

centripetal force will equal gravity force

mv²/R = mg

v²/R = g

v² = Rg

v = √(Rg)

v = √(6.4e6(9.8))

v = 7.91959...e+3

v ≈ 7900 m/s

of course, at those velocities and that deep into the atmosphere, the satellite would quickly burn up, slow down, and cause tremendous damage to buildings etc. with the sonic boom shock wave. It would also have to avoid a lot of mountains as 4000 m is not that high.

3 0

3 years ago

Assuming the diode is ideal but with a forward voltage drop of. 65 volts, what is the current in ma if v1=0v, and r1=490ω?

weeeeeb [17]

The current in the ideal diode with forward biased voltage drop of 65V is 132.6 mA.

To find the answer, we have to know more about the ideal diode.

<h3>What is an ideal diode?</h3>

A type of electronic component known as an ideal diode has two terminals, only permits the flow of current in one direction, and has less zero resistance in one direction and infinite resistance in another.
A semiconductor diode is the kind of diode that is used the most commonly.
It is a PN junction-containing crystalline semiconductor component that is wired to two electrical terminals.

<h3>How to find the current in ideal diode?</h3>

Here we have given with the values,

$V_2=65V\\V_1=0V\\R_1=490Ohm.$

We have the expression for current in mA of the ideal diode with forward biased voltage drop as,

$I=\frac{V_2-V_1}{R_1} =\frac{65}{490} =132.6mA$

Thus, we can conclude that, the current in mA of the ideal diode with forward biased voltage drop of 65 V is 132.6.

Learn more about the ideal diode here:

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7 0

2 years ago