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1 month ago

A 10.0 cm3 sample of copper has a mass of 89.6

Physics

1 answer:

inna [3.1K]1 month ago

4 0

Density is calculated as mass divided by volume, thus 89.6g / 10cm^3 = 8.96g/cm^3. *cm^3 is a standard unit of volume.*

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Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta

ValentinkaMS [3465]

Answer:

The amount of heat that enters the gas throughout this two-step process totals 120 cal.

Explanation:

Given that,

Moles present = 3

Heat capacity at volume held constant = 4.9 cal/mol.K

Heat capacity at pressure held constant = 6.9 cal/mol.K

Starting temperature = 300 K

Ending temperature = 320 K

We are tasked with determining the heat absorbed by the gas at constant pressure

Employing the heat formula

$\Delta H_{1}=nC_{p}\times\Delta T$

Substituting the values into the equation

$\Delta H_{1}=3\times6.9\times(320-300)$

$\Delta H_{1}=414\ cal$

Next, we calculate the heat absorbed by the gas at constant volume

Using the corresponding heat formula

$\Delta H_{1}=nC_{v}\times\Delta T$

Insert the values into the formula

$\Delta H_{1}=3\times4.9\times(300-320)$

$\Delta H_{1}=-294\ cal$

Now, it's necessary to evaluate the total heat flow into the gas during both steps

Using the total heat formula

$\Delta H_{T}=\Delta H_{1}+\Delta H_{2}$

$\Delta H_{T}=414-294$

$\Delta H_{T}=120\ cal$

Thus, the heat that transfers into the gas throughout this two-step process amounts to 120 cal.

7 0

2 months ago

Compressed air is used to fire a 60 g ball vertically upward from a 0.70-m-tall tube. The air exerts an upward force of 3.0 N on

Yuliya22 [3333]

Answer:

2.87 m

Explanation:

Given parameters:

Mass of the ball (m) = 60 g = 0.06 kg

Height of the tube (h) = 0.70 m

Force applied on the ball by compressed air (F) = 3.0 N

Initial velocity of the ball (u) = 0 m/s (Assumed)

Final velocity of the ball at the tube's exit (v) =?

Acceleration of the ball (a) =?

The ball's weight is derived from multiplying mass and gravity. Therefore,

Weight (W) = $mg=0.06\times 9.8=0.588\ N$

Thus, the total force acting on the ball equals the net of upward force minus the weight.

Net force = Air force - Weight

$F_{net}=F-mg\\F_{net}=3.0-0.588 = 2.412\ N$

According to Newton's second law, net force equals the mass multiplied by acceleration.

$F_{net}=ma\\\\a=\frac{F_{net}}{m}=\frac{2.412\ N}{0.06\ kg}=40.2\ m/s^2$

Acceleration (a) is calculated as 40.2 m/s².

Using the motion equation, we find:

$v^2=u^2+2ah\\\\v^2=0+2\times 40.2\times 0.7\\\\v=\sqrt{56.28}=7.5\ m/s$

Let’s denote the maximum height achieved as 'H'.

Next, we apply the principle of energy conservation from the pipe's peak to the maximum height.

A decrease in kinetic energy equals an increase in potential energy.

$\frac{1}{2}mv^2=mgH\\\\H=\frac{v^2}{2g}$

Substituting the values, we solve for 'H', yielding:

$H=\frac{56.28}{2\times 9.8}\\\\H=\frac{56.28}{19.6}=2.87\ m$

Hence, the ball ascends to a height of 2.87 m above the top of the tube.

6 0

2 months ago

A large box of mass m sits on a horizontal floor. You attach a lightweight rope to this box, hold the rope at an angle θ above t

inna [3103]

Answer:

The answer to the specified question will be " $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$ ".

Explanation:

Referring to the question,

$\sum F_{x}$

⇒ $TCos \theta-F_{s}=0$

⇒ $T_{m}Cos \theta =F_{s}$ ...(equation 1)

$\sum F_{y}$

⇒ $TSin \theta+F_{N}=m_{g}$

⇒ $M_{g}-TSin \theta=F_{N}$ ...(equation 2)

Now,

From equation 1 and equation 2, we conclude

⇒ $T_{m} Cos \theta = \mu_{s}F_{N}$

By substituting the value of $F_{N}$ , we derive

⇒ $T_{m} Cos\theta = \mu_{s}(M_{g}-T_{m}Sin \theta)$

⇒ $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$

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2 months ago

While looking at bromine (Br) on the periodic table, a student needs to find another element with very similar chemical properti

Sav [3153]

Response: Numerous elements can be found, all situated within the same vertical column as bromine.

Explanation:

Elements are organized by their atomic numbers on the periodic table. Those in the same vertical column (known as groups) exhibit the same valence electron configurations, resulting in similar chemical characteristics. Consequently, there are numerous elements sharing analogous chemical properties grouped with Bromine.

7 0

3 months ago

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