The erg is a unit of work in units of centimeters (cm), grams (g), and seconds (s), and 1 erg=1 g⋅cm^2/s^2 . Recall that the SI unit of force is the newton (N) and is equal to kg⋅m/s^2 . You push an object 0.032 m by exerting 0.010 N of force. The work is the force times the distance.How much work have you done, expressed in ergs?

A: 3,200 ergs

B: 32 ergs

C: 0.32 ergs

D: 0.00032 ergs

Answers

Answer 1

Answer:

For the given problem, the amount of work done expressed in ergs is 3200 ergs.

Answer: Option A

Explanation:

The work done on an objects are the force acting on it to move the object to a particular distance. So, work done on the object will be directly proportional to the force acting on it and the displacement.

Here, the force acting on the object is given as 0.010 N and the displacement of the object is 0.032 m. So, the work done on the object is

$\text { Work done }=\text { Force } * \text { displacement }$

$\text { Work done }=0.010 \mathrm{N} * 0.032 \mathrm{m}=0.00032 \mathrm{Nm}$

It is known that $1 N=1 \mathrm{kg} \mathrm{ms}^(-2)$

So, the work done can be expressed in $k g m s^(-2)$ as,

$\text { Work done }=0.00032 \mathrm{kgm}^(2) \mathrm{s}^(-2)$

It is known that $1 \mathrm{erg}=1 \mathrm{g} \mathrm{cm}^(2) / \mathrm{s}^(2)$ , so the conversion of units from Nm to erg will be done as follows:

$\text { Work done }=0.00032 \mathrm{kgm}^(2) \mathrm{s}^(-2) * \frac{1000 \mathrm{g}}{1 \mathrm{kg}} * \frac{100 * 100 \mathrm{cm}^(2)}{m^(2)}=3200 \mathrm{g} \mathrm{cm}^(2) \mathrm{s}^(-2)$

Thus, work done in ergs is 3200 ergs.

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Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. For the water, determine the heat transfer, in kJ per kg of water. Kinetic and potential energy effects can be ignored.

Answers

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

$u_(1) = u_(g)$ = 2553.6 kJ/kg

$v_(1) = v_(g) = 0.4625 m^(3)/kg$

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, $v_(2) = v_(g) = 0.4625 m^(3)/kg$

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at $v_(2) = v_(g) = 0.4625 m^(3)/kg$ and temperature $T_(2) = 360^(o)C$ so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

$u_(2) = u_{\text{at 5 bar, 400^(o)C}} + (\frac{v_(2) - v_{\text{at 5 bar, 400^(o)C}}}{v_{\text{at 7 bar, 400^(o)C - v_(at 5 bar, 400^(o)C)}}})(u_{at 7 bar, 400^(o)C - u_(at 5 bar, 400^(o)C)})$

$u_(2) = 2963.2 + ((0.4625 - 0.6173)/(0.4397 - 0.6173))(2960.9 - 2963.2)$

= 2963.2 - 2.005

= 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$ ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

$\Delta K.E = \Delta P.E$ = 0

Now, equation will be as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$

Q - 0 = $\Delta U + 0 + 0$

Q = $\Delta U$

Now, we will obtain the heat transfer per unit mass as follows.

$(Q)/(m) = \Delta u$

$(Q)/(m) = u_(2) - u_(1)$

= (2961.195 - 2553.6)

= 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

Final answer:

The heat transfer is 227.4 kJ per kg of water.

Explanation:

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. To determine the heat transfer in kJ per kg of water, we need to calculate the heat absorbed by the water as it reaches 360°C.

Using the specific heat capacity of water (4,186 J/kg°C) and the change in temperature (360°C - 100°C), we can calculate the heat transfer:

Qw = mw * cw * AT = (1 kg) * (4186 J/kg°C) * (360°C - 100°C) = 227,440 J = 227.4 kJ

Therefore, the heat transfer is 227.4 kJ per kg of water.

Heat transfer is the process by which thermal energy moves from one object or substance to another due to a difference in temperature. This fundamental phenomenon occurs through three main mechanisms: conduction, convection, and radiation. Conduction involves the direct transfer of heat through a material, such as metal. Convection is the transfer of heat through the movement of fluids (liquids or gases). Radiation is the emission of electromagnetic waves carrying heat energy. Understanding heat transfer is essential in various fields, including physics, engineering, and environmental science, as it governs temperature regulation, climate dynamics, and the functioning of countless technological devices.

Learn more about Heat transfer here:

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Every few years, winds in Boulder, Colorado, attain sustained speeds of 45.0 m/s (about 100 mi/h) when the jet stream descends during early spring. show answer No Attempt Approximately what is the force due to the Bernoulli effect on a roof having an area of 205 m2? Typical air density in Boulder is 1.14 kg/m3 , and the corresponding atmospheric pressure is 8.89 × 104 N/m2 . (Bernoulli’s principle assumes a laminar flow. Using the principle here produces only an approximate result, because there is significant turbulence.)

Answers

Answer:

The force exerted on the roof is $F =2.37*10^(5)N$

Explanation:

From the question we are told that

The speed of the wind is $v = 45.0 m/s$

The area of the roof is $A = 205 m^2$

The air density of Boulder is $\rho = 1.14 kg / m^3$

The atmospheric pressure is $P_(atm) = 8.89 * 10^(4) N/ m^2$

For a laminar flow the Bernoulli’s principle is mathematically represented as

$P_1 + (1)/(2) \rho v_a ^2 + \rho g h_a = P_2 + (1)/(2) \rho v_b ^2 + \rho h_b$

Where $v_1$ is the speed of air in the building

$v_b$ is the speed of air outside the building

$P_1 \ and \ P_2$ are the pressure of inside and outside the house

$h_a \ and \ h_b$ are the height above and below the roof

Now for $h_a = h_b$

The above equation becomes

$P_1 + (1)/(2) \rho v_a ^2 = P_2 + (1)/(2) \rho v_b ^2$

$P_1 - P_2 = (1)/(2) \rho (v_b^2 - v_a^2)$

Since pressure is mathematically represented as

$P = (F)/(A )$

The above equation can be written as

$F = (1)/(2) \rho ( v_b^2 - v_a ^2 ) A$

The initial velocity is 0

Substituting value

$F = (1)/(2) (1.14) [(45^2 - 0^2 ) ](205)$

$F =2.37*10^(5)N$

What is matter? explain and give example

Answers

Matter is generally any physical substance, and it's all around us... a form of matter would be liquid, and water is a liquid.

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the magnitude of the tension in the string is F. The boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. What happens to the tension in the sting?

Answers

Answer

given,

Tension of string is F

velocity is increased and the radius is not changed.

the string makes two complete revolutions every second

consider the centrifugal force acting on the stone

= $(mv^2)/(r)$

now centrifugal force is balanced by tension

T = $(mv^2)/(r)$

From the above expression we can clearly see that tension is directly proportional to velocity and inversely proportional to radius.

When radius is not changing velocity is increasing means tension will also increase in the string.

A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 35.0 m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?

Answers

Answer:

v = 14.35 m/s

Explanation:

As we know that crate is placed on rough bed

so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate

So here we have

$\mu mg = (mv^2)/(R)$