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serg [7]
3 years ago
5

A system releases 20 J of heat into its surroundings while the surroundings

Physics
1 answer:
Allushta [10]3 years ago
7 0

Answer:

The change of energy is 60J

Explanation:

We have a closed system, which receives 80 J of energy. It means this energy is entering the system. And also we have 20 J that is dissipating in the shape of heat.

Therefore the amount remaining in the system will be 80 - 20 = 60 J

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Explain how motion and energy are related to roller coasters, please be specific!
Fittoniya [83]
Because when an object is in motion it has kinetic energy
4 0
3 years ago
Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected b
zmey [24]

Answer:

The maximum amount of work is  W = 1563.289 \ J

Explanation:

From  the question we are told that

   The temperature of the environment is  T = 280\ K

    The volume of container A is  V_A = 2 m^3

    Initially the number of moles  is  n = 1.2 \ moles

     The volume of container B is V_B = 3.5 \ m^3

     

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

             W =  P_A V_A  ln[ \frac{V_B}{V_A} ]

Now from the Ideal gas law

          P_A V_A =  nRT

So substituting for P_A V_A in the equation above

          W =  nRT ln [\frac{V_B}{V_A} ]

Where R is the gas constant with a values of  R =  8.314 \  J/mol

Substituting values we have that

            W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]

          W = 1563.289 \ J

8 0
3 years ago
Joe Burrow of the Cincinnati Bengals ate a turkey sandwich and an apple for lunch. Later that day, he spent 2 hours on the footb
Nostrana [21]

Answer:

A Thermal energy was converted to kinetic energy

7 0
2 years ago
A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?
Olenka [21]
Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m

Formula:
F = k*x

Solving: 
F = k*x
F = 50*0.15
\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark

Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
E = \frac{k*x^2}{2}

Solving:(Energy associated with this stretching)
E = \frac{k*x^2}{2}
E =  \frac{50*0.15^2}{2}
E =  \frac{50*0.0225}{2}
E =  \frac{1.125}{2}
\boxed{\boxed{E = 0.5625\:J}}\end{array}}\qquad\quad\checkmark

7 0
3 years ago
A hockey puck moving at 0.4600 m/s collides with another puck that was at rest. The pucks have equal mass. The first puck is def
Sladkaya [172]

Answer:

Speed =0.283m/ s

Direction = 47.86°

Explanation:

Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane

MU1 =MU2cos38 + MV2cos y ...x plane

0 = MU2sin38 - MV2sin y .....y plane

Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2

Substitute into equation above

.46 = .34cos38 + V2cos y ...equ1

.34sin38 = V2sin y...equ2

.19=V2cos Y...x

.21=V2sin Y ...y

From x

V2 =0.19/cost

Sub V2 into y

0.21 = 0.19(Sin y/cos y)

1.1052 = tan y

y = 47.86°

Sub Y in to x plane equ

.19 = V2 cos 47.86°

V2=0.283m/s

7 0
3 years ago
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