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Fittoniya [83]
3 years ago
15

He specific heat of Aluminum is 0.9 J/g K. The specific heat of Copper is 0.39 J/g K. If samples of equal mass of both Aluminum

and Copper are heated up to 100°C and then dropped in a cold water bath. Compare the heat lost by the two samples.
A) There is not enough information to conclude anything.
B) The copper loses a little more than twice the heat of the Aluminum.
C) The Aluminum loses a little more than twice the heat of the Copper.
D) The heat lost is the same because the temperature change is the same.
Physics
2 answers:
Alexus [3.1K]3 years ago
7 0

Answer:

C) The Aluminum loses a little more than twice the heat of the Copper.

Explanation:

Since specific heat is part of the equation. A smaller specific heat will create a smaller heat gain or loss. The Aluminum loses a little more than twice the heat of the Copper.

iris [78.8K]3 years ago
3 0

the real answer is c

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Difference between Pascal’s law and law of flotation
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Hi, I didn’t understand too well your question, but I hope this helps!


Archimedes principle is based on the weight of the object to push the object upward. ​Law of floation is the priciple which tells us about the density of the object with the liquid in which it is placed.
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A kid pulls on a rope with 20 newtons of force. The block and tackle system pulls up a 160 newton box. What is the mechanical ad
Minchanka [31]
Answer:
The mechanical advantage of the system is 8

Explanation:
the mechanical advantage measures how much the system multiplies the input force to get the output.

In the given:
The input force (effort) is 20 Newton
The output force (load) is 160 Newton

This means that the mechanical advantage is:
mechanical advantage = load / effort = 160 / 20 = 8

Note that the mechanical advantage is unit-less (has no unit) since it is a ratio between two forces.

Hope this helps :)
3 0
3 years ago
During a cross-country flight you picked up rime icing which you estimate is 1/2" thick on the leading edge of the wings. You ar
igor_vitrenko [27]

Answer:

Use a faster than normal approach and landing speed.

Explanation

For pilots, it is one of the critical moments of the flight that concentrates 12% of fatal accidents. The main difficulty lies in reaching enough speed to take flight within the space of the runway. At present, it ceased to be a challenge for the aircraft, since the engine power improved, so the takeoff ceased to be the most dangerous moment of the flight.

One of the risks that aircraft face today is that some of the engines fail while the plane accelerates. In that case, the pilot must decide in an instant whether it is better to take flight and solve the problem in the air or if it is preferable not to take off.

Although for many staying on the ground might seem the most sensible option, it is not as simple as it seems: to suddenly decelerate an aircraft, with the weight it has and the speed it reaches can cause accidents. However, today a special cement was designed that runs around the runways of the airports, which when coming into contact with the wheels of the aircraft the ground breaks and helps to slow down.

6 0
3 years ago
What is the wavelength of a wave if the wave speed is 24 m/s and the frequency is 48 Hz?
Mice21 [21]

Answer:

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Explanation:

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4 0
3 years ago
Read 2 more answers
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
2 years ago
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