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alexira [117]
3 years ago
7

Why could the beam of particles not have a neutral charge?

Physics
1 answer:
tino4ka555 [31]3 years ago
4 0

Answer:

The positive and negative charges balance each other. Overall, the atom is uncharged (neutral). However, if something happens to make an atom lose or gain an electron then the atom will no longer be neutral. An atom that gains or loses an electron becomes an ion.

Explanation:

I think this is the answer : D

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A 2.0-kg object moving 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost
densk [106]

Answer:

20J

Explanation:

In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;

m₁u₁ + m₂u₂ = (m₁ + m₂)v          ---------------(i)

Where;

m₁ and m₂ are the masses of first and second objects respectively

u₁ and u₂ are the initial velocities of the first and second objects respectively

v  is the final velocity of the two objects after collision;

From the question;

m₁ = 2.0kg

m₂ = 8.0kg

u₁ = 5.0m/s

u₂ = 0        (since the object is initially at rest)

<em>Substitute these values into equation (i) as follows;</em>

(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v

(10.0) + (0) = (10.0)v

10.0 = 10.0v

v = 1m/s

The two bodies stick together and move off with a velocity of 1m/s after collision.

The kinetic energy(KE₁) of the objects before collision is given by

KE₁ = \frac{1}{2}m₁u₁² +  \frac{1}{2}m₂u₂²       ---------------(ii)

Substitute the appropriate values into equation (ii)

KE₁ = (\frac{1}{2} x 2.0 x 5.0²) +  (\frac{1}{2} x 8.0 x 0²)

KE₁ = 25.0J

Also, the kinetic energy(KE₂) of the objects after collision is given by

KE₂ = \frac{1}{2}(m₁ + m₂)v²      ---------------(iii)

Substitute the appropriate values into equation (iii)

KE₂ = \frac{1}{2} ( 2.0 + 8.0) x 1²

KE₂ = 5J

The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision

K = KE₂ - KE₁

K = 5 - 25

K = -20J

The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J

3 0
4 years ago
If 36C ofcharge pass through a write in 4s current is it carrying?​
charle [14.2K]

Answer:

The current flowing through the wire is 9 A.

Explanation:

Given;

quantity of charge passing through the wire, Q = 36 C

duration of the charge flow, t = 4 s

The current flowing through the wire is calculated as;

Q = It

I = Q / t

where;

I is the current flowing in the wire

I = 36/4

I = 9 A

Therefore, the current flowing through the wire is 9 A.

8 0
3 years ago
Monochromatic electromagnetic radiation illuminates an area of a surface. The electric and magnetic fields of the waves are then
emmasim [6.3K]

Answer:

c) True. factor four

Explanation:

The energy density of an electromagnetic wave is given by

             u = ½ ε₀ E² = B² / 2μ₀

Where ε₀ is the dielectric constant and μ₀ the magnetic permittivity.

Let's apply this equation to the present case

 

If we double the electro field

               E ’= 2 E₀

               u’= ½  ε₀ (2E₀)²

               u’= ½ ε₀ Eo² 4

               u’= 4 u₀

Therefore the energy is multiplied by four

Let's check the answers

a) False

b) False

c) True

d) False

e) False

6 0
3 years ago
Three polarizing filters are stacked, with the polarizing axis ofthe second and third filters at angles of 22.2^\circ and 68.0^\
andreev551 [17]

Answer:

I₂ = 25.4 W

Explanation:

Polarization problems can be solved with the malus law

     I = I₀ cos² θ

Let's apply this formula to find the intendant intensity (Gone)

Second and third polarizer, at an angle between them is

    θ₂ = 68.0-22.2 = 45.8º

    I = I₂ cos² θ₂

    I₂ = I / cos₂ θ₂

    I₂ = 75.5 / cos² 45.8

    I₂ = 155.3 W

We repeat for First and second polarizer

   I₂ = I₁ cos² θ₁

   I₁ = I₂ / cos² θ₁

   I₁ = 155.3 / cos² 22.2

   I₁ = 181.2 W

Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized

    I₁ = I₀ / 2

   I₀ = 2 I₁

   I₀ = 2 181.2

   I₀ = 362.4 W

Now we remove the second polarizer the intensity that reaches the third polarizer is

    I₁ = 181.2 W

The intensity at the exit is

    I₂ = I₁ cos² θ₂

    I₂ = 181.2 cos² 68.0

   I₂ = 25.4 W

8 0
4 years ago
A copper cylinder has a mass of 76.8 g and a specific heat of 0.092 cal/g.C. It is heated to 86.5 degrees C from 19.5 degrees C.
Sladkaya [172]

Given:

The mass of the copper cylinder is: m = 76.8 g = 0.0768 kg

The change in the temperature is: T = 86.5 deg C - 19.5 deg C = 67 deg C

The specific heat is: c = 0.092 cal/g.C

To find:

Heat energy needed to heat the copper cylinder.

Explanation:

The specific heat is defined as the amount of heat energy needed to raise the temperature of a substance by one degree celcius.

The expression relating heat Q, mass m, specific heat c and temperature difference T is:

Q=mcT

Substitute the values in the above equation, we get:

\begin{gathered} Q=76.8\text{ g}\times0.092\text{ Cal/g.C}\times67\text{ deg C} \\  \\ Q=473.40\text{ Cal} \end{gathered}

Final answer:

473.40 calories of heat is required to heat the copper cylinder.

6 0
1 year ago
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