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Dmitry_Shevchenko [17]
3 years ago
12

What are the relationships of the different types of matter, mass, and the change in the average kinetic energy of particles?

Physics
1 answer:
Evgesh-ka [11]3 years ago
6 0

matter = solid, liquid gas.

energy typyes, kinetc in a gas, kinetc and electrostatic in solids and liquids

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On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0
2 years ago
Joey drives along the 110 South freeway and notices a mile marker that reads 260 miles. He drives until he reaches the 150 mile
Nostrana [21]

Answer:

85 miles .

Explanation:

Displacement along the 110 South freeway = 260 - 150  =  110 miles

Displacement along the 110 North freeway = 150 - 175   = - 25 miles

Net displacement = 110 - 25 = 85 miles

So Joey's displacement from the 260 mile marker is 85 miles .

8 0
3 years ago
In order to be accepted, a scientific theory must be
bija089 [108]

Answer:

The answer is the 1st one

6 0
3 years ago
Water enters a typical garden hose of diameter 0.016 m with a velocity of 3 m/s. Calculate the exit velocity of water from the g
Vladimir79 [104]

Answer:

v₂ = 306.12 m/s

Explanation:

We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:

A₁v₁ = A₂v₂

where,

A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²

v₁ = entrance velocity = 3 m/s

A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²

v₂ = exit velocity = ?

Therefore,

(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂

v₂ = (0.006 m³/s)/(0.0000196 m²)

<u>v₂ = 306.12 m/s</u>

5 0
3 years ago
A running back with a mass of 70 kg travels down the field with a velocity of 5.0 m s . Calculate the kinetic energy of the foot
jarptica [38.1K]

K.E = 1/2 mv²

    =  1/2 (70kg) (5.0ms)²

    =  1 ˣ ( 1750) / 2

    = 875 kgms²

     

3 0
3 years ago
Read 2 more answers
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