Igneous rocks are formed from ____

Refer to Concept Simulation 4.4 for background relating to this problem. The drawing shows a large cube (mass = 28.9 kg) being a

Usimov [2.4K]

Answer:

smallest magnitud is P=33.3 N

Explanation:

We are analyze the situation as an external force is applied and there is a friction force. We have a problem with Newton's second law.

F = ma

As the two blocks go together they must have the same acceleration, so we can calculate this for the entire system

P = (m1 + m2) a

a = P / (m1 + m2)

In this case there is no friction force because the small block does not touch the ground.

In order to calculate the friction force, we must analyze each system component separately.

The large block on the X axis has an applied P force and as it moves feels a force from the small block. In the Y axis has the weight (W1) and the reaction to normal (N1)

For the small block on the X axis, the force it feels is the thrust of the large block, note that this is an action and reaction force between the two blocks, it is the same definition we have of the normal one, so we can call this force (N)

Y axis it has the weight (W2) down, the force of friction (fr) that opposes the movement, so it is directed upwards. we write these equations

N = m2 a

fr -W2 = 0

fr = W2

The definition of friction force is

fr = μ N

Let's replace and calculate

μ (m2 a) = m2 g

μ (P / (m1 + m2)) = g

P = g /μ (m1 + m2)

Let's calculate the value of this force

P = 9.8 / 0.710 (28.9 +4.4)

P = 13.80 (33.3)

P = 33.3 N

This is the minimum friction force that prevents the block from sliding down

6 0

3 years ago

A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavel

erastovalidia [21]

Answer:

D.

Explanation:

To solve the problem it is necessary to apply the concepts of Destructive and constructive interference. The constructive interference in tin film is given by

$2t = (m+\frac{1}{2})\frac{\lambda}{n}$

Where,

t = thickness

$\lambda=$ Wavelenght

m= is an integer

n= film/refractive index

We use this equaton because phase change is only present for gasoline air interface, but not at the gasoline-water interface. <em>The minimum t only would be when the value of m=0 then</em>

$2nt = \frac{\lambda}{2}$

$t = \frac{560nm}{4*1-4}$

$t = 100nm$

Therefore the correct answer is D. The minimum thickness of the film to see ab right reflection is 100nm

4 0

4 years ago

One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

3 years ago

What type of ion is formed by a non-metal atom?

Verizon [17]

Non metals produce something called protons they are ions

7 0

3 years ago

The cylinder with piston locked in place is immersed in a mixture of ice and water and allowed to come to thermal equilibrium wi

lukranit [14]

Answer:

a. volume of gas: (decreases)

b. temperature of gas: (same)

c. internal energy of gas: (same)

d. pressure of gas: (increases)

Explanation:

We have a gas (let's suppose that is ideal) in a piston with a fixed volume V.

Then we put in a reservoir at 0°C (the mixture of water and ice)

remember that the state equation for an ideal gas is:

P*V = n*R*T

and:

U = c*n*R*T

where:

P = pressure

V = volume

n = number of mols

R = constant

c = constant

T = temperature.

Now, we have equilibrium at T = 0°C, then we can assume that T is also a constant.

Then in the equation:

P*V = n*R*T

all the terms in the left side are constants.

P*V = constant

And knowing that:

U = c*n*R*T

then:

n*R*T = U/c

We can replace it in the other equation to get:

P*V = U/c = constant.

Now, the piston is (slowly) moving inwards, then:

a) Volume of the gas: as the piston moves inwards, the volume where the gas can be is smaller, then the volume of the gas decreases.

b) temperature of the gas: we know that the gas is a thermal equilibrium with the mixture (this happens because we are in a slow process) then the temperature of the gas does not change.

c) Internal energy of the gas:

we have:

P*V = n*R*T = constant

and:

P*V = U/c = constant.

Then:

U = c*Constant

This means that the internal energy does not change.

d) Pressure of the gas:

Here we can use the relation:

P*V = constant

then:

P = (constant)/V

Now, if V decreases, the denominator in that equation will be smaller. We know that if we decrease the value of the denominator, the value of the quotient increases.

And the quotient is equal to P.

Then if the volume decreases, we will see that the pressure increases.

4 0

3 years ago

Igneous rocks are formed from _____.