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dezoksy [38]
3 years ago
15

When using a calorimeter, the initial temperature of a metal is 70.4C. The initial temperature of the water is 23.6C. At the end

of the experiment, the final equilibrium temperature of the water is 29.8C.
What is the final temperature of the metal?



C



What is the temperature change of the water?



C



What is the temperature change of the metal?



C
Physics
2 answers:
Sunny_sXe [5.5K]3 years ago
5 0

1) 29.8 C

At the beginning, the metal is at higher temperature (70.4 C) while the water is at lower temperature (23.6 C). When they are put in contact, the metal transfers heat to the water, until they reach thermal equilibrium: at thermal equilibrium the two objects (the metal and the water have same temperature). Therefore, since the temperature of the water at thermal equilibrium is 29.8 C, the final temperature of the metal must be the same (29.8 C).

2) 6.2 C

The temperature change of the water is given by the difference between its final temperature and its initial temperature:

\Delta T = T_f - T_i

where

T_f = 29.8 C\\T_i = 23.6 C

Substituting into the formula,

\Delta T=29.8 C-23.6 C=6.2 C

And the positive sign means that the temperature of the water has increased.

3) -40.6 C

The temperature change of the metal is given by the difference between its final temperature and its initial temperature:

\Delta T = T_f - T_i

where

T_f = 29.8 C\\T_i = 70.4 C

Substituting into the formula,

\Delta T=29.8 C-70.4 C=-40.6 C

And the negative sign means the temperature of the metal has decreased.

amm18123 years ago
4 0

Answer: 29.8, 6.2, -40.6

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<u>Explanation:</u>

As it is known that the kinetic energy is defined as the energy exhibited by the moving objects. So the kinetic energy is equal to the product of mass and square of the velocity attained by the car. Thus,

                  \text {Kinetic energy}=\frac{1}{2} m v^{2}

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As the work done is the energy required to start the car from zero velocity to 5 m/s velocity.  

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                         Work done = 7500 J - 0 J = 7500 J

Thus, the initial kinetic energy of the car is zero, the final kinetic energy is 7500 J and the work utilized by the car is also 7500 J.

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On what two properties of a material does its density depend on?
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Snail 3, trying to keep up with Snail 2, managed to get to
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Snail 2 changed from 7 cm/min. down to 1 cm/min. in 3 minutes

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NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant
Viktor [21]

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

so

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8  / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

3 0
3 years ago
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