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kondaur [170]
3 years ago
15

A metal ball of mass 100 g is heated to 90°C and then cooled to 25°C. The heat lost in the process is 2.5 kJ. Another metal ball

of mass 200 g is heated to 90°C and then cooled to 25°C. The heat lost in the process is 5.0 kJ. What can be concluded from the data?
Physics
2 answers:
Masteriza [31]3 years ago
8 0

Answer:

( B )  The first ball is made of a material with a higher specific heat.

Explanation:

kobusy [5.1K]3 years ago
3 0

Explanation:

q = mCΔT

where q is heat,

m is mass,

C is specific heat capacity,

and ΔT is temperature change.

For the first ball:

2500 J = (100 g) C (90°C − 25°C)

C = 0.385 J/g/°C

For the second ball:

5000 J = (200 g) C (90°C − 25°C)

C = 0.385 J/g/°C

The two metals have the same specific heat, and are likely the same metal (possibly copper or zinc).

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The best and most correct answer among the choices provided by the question is the first choice "warm, dry air"


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I hope my answer has come to your help. God bless and have a nice day ahead!
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Which explains how time of winds creates high waves?
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The smaller waves created by the constant winds gradually add up to form larger ones.
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Jason applies a force of 4.00 newtons to a sled at an angle 62.0 degrees from the ground. What is the component of force effecti
Andreas93 [3]
Answer: 1.88 N

Explanation:

Data:

Force = 4.00N
angle = 62°
horizontal force = ?

Solution:

The trigonometric ratio that relates horizontal - leg to hypotenuse is the cosine.

That ratio is:

                        horizontal - leg
cos(angle) = -------------------------
                          hypotenuse

So, applied to the force, that is:

                             horizontal force
cos (angle) = -----------------------------------
                                 total force

So, clearing the horizontal component you get:
                                         
horizontal force = force * cos (angle)

Substitute the data given:

horizontal force = 4.00N * cos(62°) = 4.00N * 0.4695 = 1.88 N

Answer: 1.88N


7 0
3 years ago
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147.5 ml of ethyl alcohol (coefficient of volume expansion = 1120 x 10-6K-1) at a temperature of 273.1 K is measured into a 150.
marishachu [46]

Answer:

288.2 K

Explanation:

\beta = Volumetric expansion coefficient = 1120\times 10^{-6}\ /K

T_i = Initial temperature = 273.1 K

T_f = Final temperature

v_0 = Original volume = 150 mL

Change in volume is given by

\Delta v=\beta v_0(T_f-T_i)\\\Rightarrow T_f=\frac{\Delta v}{\beta v_0}+T_i\\\Rightarrow T_f=\frac{150-147.5}{1120\times 10^{-6}\times 147.5}+273.1\\\Rightarrow T_f=288.2\ K

The temperature of the ethyl alcohol should be 288.2 K to reach 150 mL

8 0
3 years ago
An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up. Now a slab of dielectric material is
natita [175]

Answer:

The charge on the capacitor had increased

Explanation:

The expression for the capacitance of an air-filled parallel-plate capacitor is as follows as;

C=\frac{\epsilon _{0}A}{d}  

Here, C is the capacitance, \epsilon _{0} is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.

When a slab of dielectric material is placed between the plates of the capacitor then the expression for the capacitance is as follows;

C=\frac{K\epsilon _{0}A}{d}

Here, K is the dielectric constant.

In the given problem, a slab is inserted between the plates of the capacitor then the capacitance of the capacitor will increase in this case. Therefore, the option (a) is true.

The expression for the charge stored in the capacitor is as;

Q= CV

Here, Q is the charge and V is the potential.

The charge will also increase in this case as the charge stored in the capacitor is directly proportional to the capacitance. Therefore, the option (d) is not true.

The expression for the energy stored in the capacitor is as follows;

E=\frac{CV^{2}}{2}

The voltage is constant in the given problem but the capacitance increases then the energy stored in the capacitor will increase. Therefore, the option (b) is not true.

The voltage across the capacitor will remain same as the capacitor is still connected to the battery. Therefore, the option (c) is not true.

Therefore, only option (a) is true.

3 0
2 years ago
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