What gases can CFC and HCFC refrigerants decompose into at high temperatures

A spring with a mass of 5 Kg has natural length 0.5m. A force of 35.6 N is required to maintain it stretched to a length of 0.5m

aleksandrvk [35]

Answer:

Explanation:

force constant of spring k = force / extension

= 35.6 / 0.5

k = 71.2 N / m

angular frequency ω of oscillation by spring mass system

$\omega = \sqrt{\frac{k}{m} }$

where m is mass of the body attached with spring

Putting the values

$\omega = \sqrt{\frac{71.2}{5} }$

ω = 3.77 radian / s

The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2

so the equation for displacement from equilibrium position that is middle point can be given as follows

x = .5 sin ( ω t + π / 2 )

= 0.5 cos ω t

= 0.5 cos 3.77 t .

x = 0.5 cos 3.77 t .

8 0

3 years ago

A proton, mass 1.67 × 10−27 kg and charge +1.6 × 10−19 c, moves in a circular orbit perpendicular to a uniform magnetic field of

stellarik [79]

Time taken by proton to complete one complete circular orbit= 7.28 x 10⁻⁸ s

Explanation:

For proton, the centripetal force required for circular motion is provided by the magnetic force,

so Fm= Fc

q v B = m v²/r

m= mass of charged particle

v= velocity

B =magnetic field

q= charge

r= radius of circular path

v= q B r/m

now v= r ω

ω= angular velocity

ω r = q B r /m

ω=q B /m

now ω= 2π/T where T =time period

so 2π/T=q B/m

T= 2 πm/q B

T= 2π (1.67 x 10⁻²⁷)/ [( 1.6 x 10⁻¹⁹)* (0.9)]

T= 7.28 x 10⁻⁸ s

6 0

2 years ago

Two point charges are 10.0cm apart and have charges of 2.0uC and -2.0uC, respectively. What is the magnitude of the electric fie

elena-14-01-66 [18.8K]

The electric field generated by a point charge is given by:
$E= k_e \frac{Q}{r^2}$
where
$k_e = 8.99 \cdot 10^9 Nm^2 C^{-2}$ is the Coulomb's constant
Q is the charge
r is the distance from the charge

We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them.

Let's calculate first the electric field generated by the positive charge at that point:
$E_1=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =+7.19 \cdot 10^6 N/C$
where the positive sign means its direction is away from the charge.

while the electric field generated by the negative charge is:
$E_2=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(-2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =-7.19 \cdot 10^6 N/C$
where the negative sign means its direction is toward the charge.

If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
$E_{tot} =E_1 + E_2 = 7.19 \cdot 10^6 N/C+7.19 \cdot 10^6 N/C=1.44 \cdot 10^7 N/C$

3 0

3 years ago

You're carrying a 3.2-m-long, 24kg pole to a construction site when you decide to stop for a rest. You place one end of the pole

Aleksandr [31]

Answer:

Tension of 132N

Explanation:

We need to apply Summatory of Force to find the tension in the hand.

We define te tensión in the hand as $F_2$ and the Tension in fence post as $F_1$ , then