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2 years ago

How are some types of collisions different from others?

1 answer:

3 0

There are two general types of collisions, inelastic and elastic.
Inelastic collisions occur when two objects collide but neither of them bounce away from each other.
Collisions in which the objects do not touch each other are elastic. (Ex: Rutherford Scattering)

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(a) The electric potential due to a point charge is given by V = kq⁄r where q is the charge, r is the distance from q and k = 8.

LiRa [457]

Answer:

a) [volts] = [N m / C],

b) The lines or surface that has the same potential are called equipotential

c) the equipotential lines must also be perpendicular to the electric field lines

Explanation:

a) find the units of the volt

the electric potential energy is

V = k q / r

V = [N m² / C²] C / m

V = [N m / C]

The electric potential is defined as

V = E .s

V = [N / C] [m]

V = [N m / C] = [volt]

we see that in the two expressions the same result is obtained therefore the volt is

[volts] = [N m / C]

b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.

c) The electric potential is defined as the gradient of the electric field

v = $- \frac{dE}{dx} i^$

therefore the equipotential lines must also be perpendicular to the electric field lines

4 0

3 years ago

The capitary action of water is caused the cohesion of water molecules to different types of molecules and the adhesion of water

s344n2d4d5 [400]

Answer:

This is True

Explanation:

I just did this exact unit in bio last week I hope this helps ;)

5 0

2 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

Satuarated solution changes into unsatuarated by heating give reason

Crazy boy [7]

Answer:

Due to application of heat on a saturated solution, the interparticle space increases due to increase in the kinetic energy of the particles which allows more solutes to dissolve in the solution thereby making it unsaturated.

4 0

2 years ago

A 174 pound Jimmy Cheek is riding on a 54 ft diameter Ferris Wheel. The normal force on Jimmy Cheek is 146 pounds when Jimmy is

Veronika [31]

To solve this problem we will apply the concepts related to the balance of Forces, the centripetal Force and Newton's second law.

I will also attach a free body diagram that allows a better understanding of the problem.

For there to be a balance between weight and normal strength, these two must be equivalent to the centripetal Force, therefore

$F_c = W-N$