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3 years ago

When personal information stored digitally on a computer is accidentally

Physics

1 answer:

Schach [20]3 years ago

4 0

Answer:

C. The information is always easily restored."

Explanation:

When the personal information stored digitally on a computer is accidentally

deleted, the data is even stored in the hard disk in various fragments until it is not permanently deleted. Hence the correct option will be "C. The information is always easily restored."

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On a velocity vs. time graph, what does it mean when the line crosses over the x-axis?

Phantasy [73]

On a velocity - time graph, if the line crosses the x - axis it depicts that the object has started moving in the opposite direction.

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3 years ago

A see-saw is balanced on a pivot with two children on it. One child is sitting 1.5 m to the left of the pivot and has a mass of

Lorico [155]

Answer:

Explanation:

Remark

This is a second class lever. It is much more efficient than the fishing pole problem. All distances are measured from the pivot in these kinds of questions.

Givens

d1 = 1.5

d2 = ?

m1 = 50 kg

m2 = 30 kg

The lighter child will have to sit further away from the pivot to make the two conditions equal.

Formula

d1*m1 = d2*m2

1.5*50 = d2 * 30

75 = 30 * d2

75/30 = d2

d2 = 2.5

Remark

Notice that the distance is longer for the lighter child. The fact that these are masses and not forces does not matter, but you should take note of it. There is a difference between masses and forces. See the fishing pole problem.

Answer to the multiple Choice question. No motion on this kind of problem means equal moments. The answer is D

Problem 2

1) The wheels are further apart making B more stable. The wider the distance the wheels are apart, the harder it would be to tip the concrete mixer over

2) The center of gravity is lower. The higher the force is the more chance you have of exerting an external force to tip the mixer over.

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3 years ago

According to Boyle’s law, PV= K, what was the volume at the time of the first measurement given the following information? Round

alexandr402 [8]

Answer:

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6 0

3 years ago

A single circular loop of wire of radius 0.75 m carries a constant current of 3.0 A. The loop may be rotated about an axis that

NISA [10]

Answer:

B = 0.8 T

Explanation:

It is given that,

Radius of circular loop, r = 0.75 m

Current in the loop, I = 3 A

The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.

When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.

We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :

$\tau=NIAB\sin\theta$

B is magnetic field

$B=\dfrac{\tau}{NIA\sin\theta}\\\\B=\dfrac{1.8}{1\times \pi \times (0.75)^2\times 3\times \sin(25)}\\\\B=0.8\ T$

So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.

6 0

3 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

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5 0

3 years ago