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3 years ago

In which application is a magnet used to store information?

Physics

2 answers:

galina1969 [7]3 years ago

6 0

Answer:

d the disk in a computer hard drive

Explanation:

because hard drive use magnetic memory tostore giga- and terabytes of data in computer.

tatiyna3 years ago

3 0

Answer:

D. The disk in a computer hard drive

The disk has a magnetic coating in which the information is stored on the magnetic strips.

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Green plants convert light energy from the sun into ______. * a. gravitational potential energy b. chemical potential energy c.

katrin [286]

Answer:

chemical potiential energy

6 0

3 years ago

A 61.0-kg person jumps from rest off a 10.0-m-high tower straight down into the water. Neglect air resistance. She comes to rest

9966 [12]

Answer:

Explanation:

In this case, law of conservation of energy will be implemented. It states that "the energy of the system remains conserved until or unless some external force act on it. Energy of the system may went through the conversion process like kinetic energy into potential and potential into kinetic energy.But their total always remain the same in conserved systems."

Given data:

Height of tower = 10.0 m

Depth of the pool = 3.00 cm

Mass of person = 61.0 kg

Solution:

Initial energy = Final energy

$U_{i} = (K.E) + U_{f}$

As the person was at height initially so it has the potential energy only.

$mg(h_{1} +h_{2}) = K.E + mgh_{2}$

$K.E = mgh_{1}$

$K.E = (61.0)(9.8)(10)\\K.E = 5978 J$

Lets find out the magnitude of the force that the water is exerting on the diver.

W =ΔK.E

$F.h_{2} = 5978\\$

$F = \frac{5978}{3}$

F = 1992.67 N

7 0

3 years ago

When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the

KonstantinChe [14]

Answer:

The ratio is KE : TM = 0.75

Explanation:

from the question we are told that

The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical energy of the mass is mathematically represented as

$TM = \frac{1}{2} * k * A^2$

Here k is the spring constant , A is the total displacement of the the mass from maximum compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

$TM = KE + PE$

=> $KE = TM - PE$

Here the potential energy of the mass is mathematically represented as

$PE = \frac{1}{ 2} * k * [ x ]^2$

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is

$x = \frac{A}{2}$

$PE = \frac{1}{ 2} * k * [ \frac{A}{2} ]^2$

$KE = \frac{1}{2} * k * A^2 - \frac{1}{2} * k * [\frac{A}{2} ]^2$

=> $KE = \frac{1}{2} * k * A^2 - \frac{1}{8} * k * A ^2$

=> $KE = 0.375 * k * A^2$

So the ratio of $KE : TM$ is mathematically represented as

$\frac{KE}{TM} = \frac{0.375 k A^2 }{0.5 k A^2}$

=> $\frac{KE}{TM} = 0.75$

3 0

3 years ago

Consider eight,eight-cubic centimeter (8 cm3) sugar cubes stacked so that they form a single 2 x 2 x 2 cube. How does the surfac

In-s [12.5K]

To find the surface area of a single cube we first nees to take the cube root of 8cm3 which is 2.

Now we know that the length of each side is 2 and we can find the area of one side by doing 2x2 which is 4.

To find the total surface area of one cube we do 4 times 6 side giving us a total of 24cm2.

To find the total surface area of the 8 individual cubes, we multiply 24cm2 by 8 to give us a total of 192cm2.

Now to find the total surface area of the one large cube, we know that each side of one of the small cubes is 4cm2 and the large cube is set up so that there are two levels of four cubes right on top of each other. So, the total area of each side of the large cube is 4cm2 times 4 which gives us 16cm2.

Then we multiply 16cm2 by 6 sides to give us a total surface area of 96cm2.

The ratio of the surface area of the single large cube comapred to the total surface area of the single cubes is 96:192

We can further simplify this ratio:

96:192

48:96

24:48

12:24

6:12

3:6

1:2

5 0

3 years ago

A spring with a force constant of 5.3 n/m has a relaxed length of 2.60 m. when a mass is attached to the end of the spring and a

olganol [36]

This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant.
 U = (1/2)kx^2
 U = (1/2)(5.3)(3.62-2.60)^2
 U = 2.75706 J

6 0

3 years ago