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2 years ago

A/An _____ is described as a device used to store electrical energy, usually two conductors separated by an insulator.

Physics

1 answer:

Daniel [21]2 years ago

8 0

Answer: A capacitor.

Explanation:

The capacitor is a passive element that is used in electronics to store electrical energy maintaining an electrical field. The simpler case of a capacitor is the parallel plates capacitor.

It consists of two parallel metal plates separated by a distance D, in this case, the air between the plates works as a dielectric, as the plates do not touch each other and are separated by a dielectric, the charge is stored in the surface plates.

There are a lot of other types of capacitors, the most used in actuality may be the cylindrical one, where instead of parallel plates, it uses two concentric cylinders, and the space between the cylinders is filled with a dielectric/insulator.

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A basketball player throws a chall -1 kg up with an initial speed of his hand at shoulder height = 2.15 m Le gravitational poten

Talja [164]

Complete Question:

A basketball player tosses a basketball m=1kg straight up with an initial speed of v=7.5 m/s. He releases the ball at shoulder height h= 2.15m. Let gravitational potential energy be zero at ground level

a) Give the total mechanical energy of the ball E in terms of maximum height hn it reaches, the mass m, and the gravitational acceleration g.

b) What is the height, hn in meters?

Answer:

a) Energy = mghₙ

b) Height, hₙ = 5.02 m

Explanation:

a) Total energy in terms of maximum height

Let maximum height be hₙ

At maximum height, velocity, V=0

Total mechanical energy , E = mgh + 1/2 mV^2

Since V=0 at maximum height, the total energy in terms of maximum height becomes

Energy = mghₙ

b) Height, hₙ in meters

mghₙ = mgh + 1/2 mV^2

mghₙ = m(gh + 1/2 V^2)

Divide both sides by mg

hₙ = h + 0.5 (V^2)/g

h = 2.15m

g = 9.8 m/s^2

V = 7.5 m/s

hₙ = 2.15 + 0.5(7.5^2)/9.8

hₙ = 2.15 + 2.87

hₙ = 5.02 m

6 0

3 years ago

Light shines through a single slit whose width is 5.7 x 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a

lilavasa [31]

Answer:

$\lambda = 570\ nm$

Explanation:

Given,

Width of slit, W = 5.7 x 10⁻⁴ m

Distance between central bright fringe, L = 4 m

distance between central bright fringe and first dark fringe, y = 4 mm

Diffraction angle

$tan \theta = \dfrac{y}{L}$

$tan \theta = \dfrac{4}{4\times 10^3}$

$\theta = 0.0572$

Now.

$W sin \theta = m \lambda$

m = 1

$5.7 \times 10^{-4} \times sin (0.0572) = 1 \times \lambda$

$\lambda = 569.99 \times 10^{-9}\ m$

$\lambda = 570\ nm$

4 0

3 years ago

Which of these rotational quantities is analogous to mass in a linear system?

Rom4ik [11]

Answer:

Explanation:

i think it is rotational inertia

because analogue of mass in rotational motion is moment of inertia. It plays the same role as mass plays in transnational motion.

hope it's right & helps !!!!!!!!!

5 0

2 years ago

Determine the mass of Ar in 1.00 liter of a gas mixture at 25oC which contains 0.300 atm of Ne and has a total pressure of 4.00

sergejj [24]

Answer:

The mass of Ar is 36.91g

Explanation:

The gas mixture consist of Neon(Ne) and Argon(Ar)

Partial pressure of Ar = total pressure of mixture - partial pressure = 4 - 0.3 = 3.7 atm

Mole fraction of Ar = partial pressure of Ar ÷ total pressure of mixture = 3.7/4 = 0.925

Mass of Ar = 0.925 × molecular weight of Ar = 0.925 × 39.9 = 36.91g

4 0

3 years ago

An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be

lord [1]

Answer:

the quarterback will be moving back at speed of 0.080625 m/s

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0

2 years ago