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2 years ago

8

An iron ball and an aluminum ball of mass 100 g each are heated to the same temperature and then cooled to a temperature of 20°C

. The heat lost by the iron ball is 3.6 kJ. The heat lost by the aluminum ball is 7.2 kJ. What does this imply?

1 answer:

Mekhanik [1.2K]2 years ago

4 0

As the temperature changes and their masses are the same, heat lost by the balls is directly proportional to their specific heat values. The heat lost by the aluminum ball is higher implies aluminum has higher specific heat.

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Answer:240 meters

Explanation:60×4

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3 years ago

Which of the following correctly describes Albert Einstein’s interpretation of the photoelectric effect.

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There is no correct description on that list of choices.

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3 years ago

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A 35.8 kg box initially at rest is pushed 2.38 m along a rough, horizontal floor with a constant applied horizontal force of 108

tiny-mole [99]

Answer:

The work done by the applied force is 259.22 J.

Explanation:

The work done by the applied force is given by:

$W = F*d$

Where:

F: is the applied horizontal force = 108.915 N

d: is the distance = 2.38 m

Hence, the work is:

$W = F*d = 108.915 N*2.38 m = 259.22 J$

Therefore, the work done by the applied force is 259.22 J.

I hope it helps you!

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2 years ago

Drag the positive or negative feedback loop on the left to each process on the right. terms may be used once, more than once, or

slamgirl [31]

The order of the positive and negative feedback loops are positive, positive, negative, positive, positive, negative.

<h3>What is a feedback loop?</h3>

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Both negative and positive feedback loops are possible. Insofar as they stay within predetermined bounds, negative feedback loops are self-regulating and helpful for sustaining an ideal condition. One of the most well-known examples of a self-regulating negative feedback loop is an old-fashioned home thermostat that turns on or off a furnace using bang-bang control.

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2 years ago

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

$KE =\frac{1}{2}mv^2$

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$2M_2V_1^2=0.5\times M_2V_2^2$

or

$2V_1^2=0.5\times V_2^2$

or

$4V_1^2= V_2^2$

or

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

or

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

or

$2(V_1+9.0)^2=(2V_1+9.0)^2$

or

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

or

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

or

$(0.404V_1)=(3.72)$

or

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

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2 years ago

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