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lisov135 [29]
3 years ago
6

What happens to the kinetic energy of the particles in a substance when it is heated?

Physics
1 answer:
raketka [301]3 years ago
3 0
When kinetic energy of the particles get heated the motion of the particles increases as the particles become more energetic.


When particles are heated up, space is being created. The atoms started to get "overly excited" and started to move faster than they usually do. When this happens, energy is released in the form of heat, light or etc. Because of this, kinetic energy increases and atoms colliding with each other happens more often.
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What is a net force on an object that has a mass of 20.0 kg, an applied force of 100 n moving on a surface with a friction coeff
sergiy2304 [10]

The net force on the object as described is; 58.84N

Two forces acting on the object are;

  • The <em>applied force and the frictional force.</em>

In essence; the frictional force can be evaluated as;

  • Frictional force; = coefficient × Weight of object.

  • Frictional force = 0.21 × 20 × 9.8.

  • Frictional force = 41.16N

  • The Net force = Applied force - frictional force

  • Net force = 100 - 41.16N

Net Force = 58.84 N.

Read more:

brainly.com/question/94428

5 0
3 years ago
What does the term wavelength refer to?
Andru [333]
Are you asking what it means?

4 0
3 years ago
I will pick the branliest if answered these questions.
Bond [772]

1.

n₁ = index of refraction of diamond = 2.4

n₂ = index of refraction of water = 1.33

θ₁ = angle of incidence = 24 deg

θ₂ = angle of refraction = ?

using snell's law

n₁ Sinθ₁ = n₂ Sinθ₂

inserting the values

(2.4) Sin24 = (1.33) Sinθ₂

θ₂ = 47.22 deg


2.

c = specific heat = 4.18

m = mass of water = 3.5 kg

ΔT = change in temperature = 55 - 25 = 30 C

Q = heat taken

heat taken is given as

Q = m c ΔT

inserting the values

Q = (3.5) (4.18) (30)

Q = 439 J


3.

n = number of moles = 1

m = molar mass = 0.0399 kg

T = temperature = 27 K

v = average velocity

average velocity is given as

v = sqrt(3RT/m)

v = sqrt(3 x 8.314 x 275/0.0399)

v = 414.6m/s


4.

n = number of moles = 2

T = temperature = 35 C  = 35 + 273 K = 308 K

U = internal energy

internal energy is given as

U = 2.5 n RT

U = 2.5 (2) (8.314) (308

U = 1.28 × 10⁴ J


5.

F₁ = 3300 N

F₂ = ?

A₁ = 0.060 m²

A₂ = 0.18 m²

Using pascal's law

F₁/A₁ = F₂/A₂

3300/0.060 = F₂/0.18

F₂ = 9900 N


6.

R = resistance of each resistance in series = 10 ohm

R' = equivalent resistance in series

equivalent resistance is given as

R' = 3 R

R' = 3 x 10

R' = 30 ohm


7.

R = resistance of each resistance in parallel = 10 ohm

R'' = equivalent resistance in parallel

equivalent resistance is given as

R'' = R/3

R'' = 10/3

R'' = 3.3 ohm


8.

for series circuit

R' = equivalent resistance in series = 30 ohm

V = Voltage applied = 60 Volts

i' = current in each resistor in series

current in each resistor in series is given as

I' = V/R'

i' = 60/30

i' = 2 A


for parallel circuit :

i = current in each resistor = V/R = 60/10 = 6 A








6 0
4 years ago
Read 2 more answers
3. An engine’s fuel is heated to 2,000 K and the surrounding air is 300 K. Calculate the ideal efficiency of the engine. Hint: T
maksim [4K]

Answer: E = 0.85

Therefore the efficiency is: E = 0.85 or 85%

Explanation:

The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input heat QH

E = W/QH.

W=QH – QC,

Where Qc is the output heat.

That is,

E=1 - Qc/QH

E =1 - Tc/TH

where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir.

Note: The unit of temperature must be in Kelvin.

Tc = 300K

TH = 2000K

Substituting the values of E, we have;

E = 1 - 300K/2000K

E = 1 - 0.15

E = 0.85

Therefore the efficiency is: E = 0.85 or 85%

4 0
4 years ago
Suppose you design an apparatus in which a uniformly charged disk of radius R is to produce an electric field. The field magnitu
ANEK [815]

Answer:

The electric field will be decreased by 29%

Explanation:

The distance between point P from the distance z = 2.0 R

Inner radius = R/2

Outer raidus = R

Thus;

The electrical field due to disk is:

\hat {K_a} = \dfrac{\sigma}{2 \varepsilon _o} \Big( 1 - \dfrac{z}{\sqrt{z^2+R_i^2}} \Big))

\implies \dfrac{\sigma}{2 \vaepsilon _o} \Big ( 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0\ R)^2+(R)^2}} \Big)

Similarly;

\hat {K_b} = \hat {k_a} - \dfrac{\sigma}{2 \varepsilon_o} \Big( 1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ r)^2 + (\dfrac{R}{2}^2)}}\Big)

However; the relative difference is: \dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{E_a -E_a + \dfrac{\sigma}{2 \varepsilon_o  \Big[1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ R)^2 + (\dfrac{R}{2})^2}} \Big] } } { \dfrac{\sigma}{2 \varepsilon_o \Big [ 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0 \ R)^2 + (R)^2}} \Big] }}

\dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{1 - \dfrac{2.0}{\sqrt{(2.0)^2 + \dfrac{1}{4}}} }{1 - \dfrac{2.0 }{\sqrt{(2.0)^2 + 1}}}

= 0.2828 \\ \\ \mathbf{\simeq  29\%}

3 0
3 years ago
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