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2 years ago

What type of energy is thermal energy?

Physics

1 answer:

BigorU [14]2 years ago

5 0

Thermal energy is an example of kinetic energy , due to motion of particles .

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The mass of a hot-air balloon and its cargo (not including the air inside) is 170 kg. The air outside is at 10.0°C and 101 kPa.

scoundrel [369]

Answer:

108.37°C

Explanation:

P₁ = Initial pressure = 101 kPa

V₁ = Initial volume = 530 m³

T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K

P₂ = Final pressure = 101 kPa (because it is open to atmosphere)

V₂ = Final volume = 530 m³

P₁V₁ = n₁RT₁

⇒101×530 = n₁RT₁

⇒53530 J = n₁RT₁

P₂V₂ = n₂RT₂

⇒53530 J = n₂RT₂

$\frac{m_1}{m_2}=\frac{\rho V_1}{\rho V_1-170}\\\Rightarrow \frac{m_1}{m_2}=\frac{1.244\times 530}{1.244\times 530-170}=1.347\\\Rightarrow \frac{m_1}{m_2}=1.347\\\Rightarrow \frac{n_1}{n_2}=1.347$

Dividing the first two equations we get

$1=\frac{n_1}{n_2}\frac{T_1}{T_2}\\\Rightarrow 1=1.347\frac{283.15}{T_2}\\\Rightarrow T_2=1.347\times 283.15= 381.52\ K$

∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C

8 0

2 years ago

If a car has a mass of 200 kilograms and produces a force of 500 N, how fast will the car accelerate?

Hitman42 [59]

Answer:

2.5m/s²

Explanation:

Given parameters:

Mass of car = 200kg

Force on car = 500N

Unknown:

Acceleration of the car = ?

Solution:

According to Newton's second law of motion:

Force = mass x acceleration

Insert the given parameters and find the acceleration;

500 = 200 x acceleration

acceleration = 2.5m/s²

3 0

2 years ago

Read 2 more answers

Define rectilinear propagation of light.??

Tju [1.3M]

Answer:

it relates to the light propensity to travel over one straight line without having any interference in its trajectory

Explanation:

5 0

2 years ago

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Which statement is true about two isotopes of the same element?

scZoUnD [109]

Answer:

atoms that have the same # of protons but a DIFFERENT # of NEUTRONS

Explanation:

3 0

2 years ago

Read 2 more answers

0.180-kg stone rets on a frictionless, horizontal surface. A bullet of mass 7.50 g, traveling horizontally at 390 m/s, strikes t

Musya8 [376]

Answer:

The magnitud of the velocity is

$8.46m/s$

and the direccion:

$-28.3$ degrees from the horizontal.

Explanation:

Fist we define our variables:

$m_{s}=0.18kg\\m_{b}=7.5g = 0.0075kg\\v_{1b}= 390m/s -i\\v_{1s}=0m/s\\v_{2b}=210m/s -j\\v_{2s}=?$

The letters i and j represent the direction of the movement, i in this case is the horizontal direction, and j is perpendicular to i.

velocities with sub-index 1 are the speeds before the crash, and with sub-index 2 are the velocities after the crash.

Using conservation of momentum:

$m_{s}v_{1s}+m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}\\v_{1s}=0, so\\m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}$

Clearing for the velocity of the stone after the crash:

$v_{2s}=\frac{m_{b}v_{1b}-m_{b}v_{2b}}{m_{s}}$

Substituting known values:

$v_{2s}=\frac{0.0075kg(((390m/s-i)-210m/s-j)}{0.18kg}\\v_{2s}=(16.25m/s-i) - (8.75m/s-j)$

The magnitud of the velocity is :

$|v_{2s}|=\sqrt{(16.25m/s-i)^2 + (8.75m/s-j)^2}\\|v_{2s}|=18.46m/s$

and the direction:

$tan^{-1}(-8.75/16.25)=-28.3$

this is -28.3 degrees from the +i direction or the horizontal direcction.

Note: i and j can also be seen as x and y axis.

3 0

3 years ago