A certain part of a flat screen TV has a thickness of 150 nanometers. How<br> many meters is this?

A container of volume 0.6 m^3 contains 5.3 mol of argon gas at 24°C. Assuming argon behaves as an ideal gas, find the total inte

Vitek1552 [10]

Answer:

the internal energy of the gas is 433089.52 J

Explanation:

let n be the number of moles, R be the gas constant and T be the temperature in Kelvins.

the internal energy of an ideal gas is given by:

Ein = 3/2×n×R×T

= 3/2×(5.3)×(8.31451)×(24 + 273)

= 433089.52 J

Therefore, the internal energy of this gas is 433089.52 J.

5 0

2 years ago

A circular wire loop lies inside a region of space containing a magnetic field. The direction of the magnetic field is out of th

jekas [21]

Answer:

clockwise

Explanation:

According to the law given by Lenz, known as the Lenz law, it is said that a current induced in the circuit which is due to the change in the magnetic field and is so directed so as to oppose the change in the flux and to apply a force in the opposite direction if the force.

Here, as the magnetic field is directed out of the screen, the current flows in the direction which is clockwise in the loop and it opposes the increasing magnetic field.

The clockwise induced current will produce magnetic field in to the screen.

6 0

2 years ago

A particle moving in the x direction is being acted upon by a net force F(x)=Cx2, for some constant C. The particle moves from x

elixir [45]

Answer:

Change in kinetic energy is ( 26CL³)/3

Explanation:

Given :

Net force applied, F(x) = Cx² ....(1)

Displacement of the particle from xi = L to xf = 3L.

The work-energy theorem states that change in kinetic energy of the particle is equal to the net amount of work is done to displace the particle.

That is,

ΔK = W = ∫F·dx

Substitute equation (1) in the above equation.

ΔK = ∫Cx²dx

The limit of integration from xi = L to xf = 3L, so

$\Delta K=\frac{C}{3}(x_{f} ^{3} - x_{i} ^{3})$

Substitute the values of xi and xf in the above equation.

$\Delta K=\frac{C}{3}((3L) ^{3} - L ^{3})$

$\Delta K=\frac{C}{3}\times26L^{3}$

5 0

2 years ago

You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w

Anit [1.1K]

Answer:

v₀ =3.8 m/s

Explanation:

Newton's second law of the box:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg mass of the box

d= 5.4m length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof and the y-axis in the direction perpendicular to it.

W: Weight of the box : In vertical direction

N : Normal force : perpendicular to the direction the roof

fk : Friction force: parallel to the direction to the roof

Calculated of the weight of the box

W= m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N

Calculated of the Normal force

∑Fy = m*ay ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86 = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34 m/s²

Kinematics of the box

Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:

d:displacement = 5.4 m

v₀: initial speed

vf: final speed = 0

a : acceleration of the box = -1.34 m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

$v_{o} =\sqrt{14.472}$

v₀ = 3.8 m/s

7 0

2 years ago

One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra

sveticcg [70]

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration ( $a_{r}$ ), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

$a_{r} = 4\pi^{2}\cdot f^{2}\cdot R$ (1)

Where:

$f$ - Frequency, measured in hertz.

$R$ - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension ( $T$ ), measured in newtons, through the string, then we derive the following model:

$\Sigma F = T = m\cdot a_{r}$ (2)

Where $m$ is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

$T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R$ (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

$\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right)$ (4)

$\frac{T_{2}}{T_{1}} = 4\cdot 2$

$\frac{T_{2}}{T_{1}} = 8$

If both the radius and frequency are doubled, then the tension is increased 8 times.

5 0

1 year ago

A certain part of a flat screen TV has a thickness of 150 nanometers. How many meters is this?