A planetâs moon revolves around the planet with a period of 39 Earth days in an approximately circular orbit of radius of 4.8Ã10^8 m. How fast does the moon move?

Using the principle of continuity for incompressible fluids, if the diameter of a blood vessel is reduced by 59.0% due to plaque, the speed of the blood just as it enters this section will be approximately 2.44 times its initial speed.

The subject of the question falls under the topic of fluid flow in physics, specifically concerning the principle of continuity for incompressible fluids. This principle, often applied in fluid dynamics, suggests that in an area of steadily flowing fluid, the mass passing through one cross-section in a unit of time equals the mass passing through other sections.

Given this principle, if the cross-sectional area of the blood vessel decreases due to plaque buildup, the speed of the blood flow must increase accordingly to maintain a steady flow rate. If the diameter of the vessel decreases by 59.0%, the cross-sectional area A, which is proportional to the square of the diameter (A ~ D²), will be reduced to 0.41 of its original value (because (1 - 59/100)² = 0.41). Therefore, the speed V would be 1/0.41, or approximately 2.44 times the original speed V0.

So, if the blood vessel's diameter is reduced by 59.0%, then just as the blood enters the blocked portion of the vessel, its speed V will be 2.44 times the initial speed V0.

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Answer:

[tex]p_2 = 664081 N/m^{2}[/tex]

Explanation:

from the ideal gas law we have

PV = mRT

[tex]P = \rho RT[/tex]

[tex]\rho = \frac{P}{RT}[/tex]

HERE  R is gas constant for dry air  =  287  J K^{-1} kg^{-1}

[tex]\rho = \frac{7.00 10^{5}}{287(18+273)}[/tex]

[tex]\rho = 8.38 kg/m^{3}[/tex]

We know by ideal gas law

[tex]\rho = \frac{m_1}{V_1}[/tex]

[tex]m_1 = \rho V_1 = 8.38 *2*10^{-3}[/tex]

[tex]m_1 = 0.0167 kg[/tex]

for m_2

[tex]m_2 = \rho V_i - V_removed[/tex]

[tex]m_2 = 8.38*(.002 - 10^{-4})[/tex]

[tex]m_2 = 0.0159 kg[/tex]

WE KNOW

PV = mRT

V, R and T are constant therefore we have

P is directly proportional to mass

[tex]\frac{p_2}{p_1}=\frac{m_2}{m_1}[/tex]

[tex]p_2 = p_1 * \frac{m_2}{m_1}[/tex]

[tex]p_2 =7*10^{5} * \frac {.0159}{0.0167}[/tex]

[tex]p_2 = 664081 N/m^{2}[/tex]

This problem can be solved using Boyle's Law, which relates the pressure and volume of a gas at a constant temperature. The question asks for the new pressure of a bicycle tire after letting out a certain volume of air. The answer is approximately 7.37 x 10⁵ Pa.

The subject of this question is gas laws, specifically Boyle's Law which states that the pressure and volume of a gas have an inverse relationship when the temperature is kept constant. Assuming the temperature and volume of the tire remain constant before and after you let out the air, when a volume of 100 cm³ (which we will convert to 0.1 L for consistency) of air is let out, the new total volume of the gas is 1.9 L.

According to Boyle's Law, P1*V1 = P2*V2, where P1 and V1 represent the initial pressure and volume, and P2 and V2 represent the final pressure and volume. Plugging the values into this equation, we get:

(7.00 x 10⁵ Pa)(2.00 L) = P2 * (1.9 L)

Which gives us:

P2 = (7.00 x 10⁵ Pa * 2.00 L) / 1.9 L

Therefore, the pressure in the bike tire after letting out 100 cm³ of gas is approximately 7.37 x 10⁵ Pa.

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Answer:

2.07 m/s

Explanation:

m = 1.07 x 10^5 kg, u1 = 3.41 m/s, u2 = 1.4 m/s

Let the speed of three coupled car after collision is v

Use conservation of momentum

m x u1 + 2 m x u2 = 3 m x v

u1 + 2 u2 = 3 v

3.41 + 2 x 1.4 = 3 v

v = 2.07 m/s

Answer:

Option D is the correct answer.

Explanation:

Stress is the force per unit area that tend to change the shape of body.

Stress is defined as internal resistive force per unit area.

         [tex]\texttt{Stress}=\frac{\texttt{Internal resistive force}}{\texttt{Area}}[/tex]

         [tex]\sigma =\frac{F}{A}[/tex]

So, so stress distributed over an area is best described as internal resistive force.

Option D is the correct answer.

Final answer:

Stress distributed over an area refers to the internal resistive forces that develop within a material in response to applied external forces. It is best described as an internal force, specifically termed internal resistive force, and is measured as the force per unit area.

Explanation:

Stress distributed over an area is best described as d) Internal resistive force. Stress is a physical quantity that represents the internal forces per unit area within a material that develop as a response to applied external forces or changes in temperature. It is calculated by the ratio of force to area and measured in Newtons per square meter (N/m²). Stress caused by forces perpendicular to the cross-section of the material is called normal stress, which can be tensile or compressive. Similarly, stress caused by forces parallel to the area, such as shear stress, represents deformation through sliding layers.

For example, when a metal rod is pulled from both ends, the internal resistive forces that develop within the material to oppose elongation are a manifestation of tensile stress. In contrast, when a book is pushed down upon by a hand, the internal resistive forces that prevent the book from compressing are an example of compressive stress.

Explanation:

It is given that,

Mass of the shot, m = 7.27 kg

Time taken to accelerate, t = 1.3 s

It is shot from rest to 16 m/s and it raises to a height of 0.9 m. We need to find the power output of the shot-putter. It is given by :

[tex]P=\dfrac{energy}{time}[/tex]

Energy = kinetic energy + potential energy

[tex]E=\dfrac{1}{2}\times 7.27\ kg\times (16\ m/s)^2+7.27\ kg\times 9.8\ m/s^2\times 0.9\ m[/tex]

E = 994.68 J

Power, [tex]P=\dfrac{994.68\ J}{1.3\ s}[/tex]

P = 765.13 Watts

We know that, 1 horse power = 745.7 watts

Or P = 1.02 horse power

Hence, this is the required solution.

Answer: The average atomic mass of the given element is 20.169 amu.

Explanation:

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]     .....(1)

We are given:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882  

Putting values in equation 1, we get:

[tex]\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)][/tex]

[tex]\text{Average atomic mass}=20.169amu[/tex]

Hence, the average atomic mass of the given element is 20.169 amu.

Answer:

Well..

Explanation:

That's impossible. I know because I once weighed 11.1 kN, and I was temporarily immobile. It's probably the same for a car, and therefore it can not be "traveling" anywhere at all.. unless you put the car on an airplane or a boat or something.

Answer:

[tex]\rho = 12580.7 kg/m^3[/tex]

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

[tex]F = \frac{GMm}{(r + h)^2}[/tex]

here we have

[tex]F =\frac {mv^2}{(r+ h)}[/tex]

[tex]\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}[/tex]

here we have

[tex]v = \sqrt{\frac{GM}{(r + h)}}[/tex]

now we can find time period as

[tex]T = \frac{2\pi (r + h)}{v}[/tex]

[tex]T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}[/tex]

[tex]1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}[/tex]

[tex]M = 4.54 \times 10^{23} kg[/tex]

Now the density is given as

[tex]\rho = \frac{M}{\frac{4}{3}\pi r^3}[/tex]

[tex]\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}[/tex]

[tex]\rho = 12580.7 kg/m^3[/tex]

Answer:

d

Explanation:

HOPE THIS HELPS!!

The owner's velocity, with the same momentum as the combined momentum of the dog and cat, is 1.28 m/s directed 9.46 degrees east of the north.

To solve this problem, we need to calculate the dog's momentum, the cat's momentum, and then use these two results to find the owner's velocity and direction.

First, let's calculate the momentum for each pet. Momentum (p) is defined as mass (m) times velocity (v). For the dog, p = mv = 26.2 kg * 3.21 m/s = 84.042 kg*m/s northward. For the cat, p = mv = 5.30 kg * 2.64 m/s = 13.992 kg*m/s eastward.

To find the combined momentum vector of the two animals, we will use Pythagorean theorem because the vectors are perpendicular to each other. So, resultant momentum = sqrt[(84.042^2) + (13.992^2)] = 85.87 kg*m/s.

The owner's momentum equals the total momentum of the dog and cat, so that's 85.87 kg*m/s. The magnitude of the owner's velocity (v) is therefore the momentum divided by his mass: v = p / m = 85.87 kg*m/s / 67.2 kg = 1.28 m/s. The direction of the owner's velocity can be found using trigonometry. The angle is arctan (cat's momentum / dog's momentum) = arctan (13.992 / 84.042) = 9.46° east from north.

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The magnitude of the owner's velocity is approximately [tex]\( 1.267 \, \text{m/s} \)[/tex], and the direction is [tex]\( 45^\circ \)[/tex] northeast.

To find the direction and magnitude of the owner's velocity, we need to calculate the total momentum of the dog and cat and then equate that to the owner's momentum.

 First, we calculate the momentum of the dog and cat separately using the formula p = mv , where p  is the momentum,  m  is the mass, a v is the velocity.

For the dog:

[tex]\[ p_{\text{dog}} = m_{\text{dog}} \times v_{\text{dog}} \][/tex]

[tex]\[ p_{\text{dog}} = 26.2 \, \text{kg} \times 3.21 \, \text{m/s} \][/tex]

[tex]\[ p_{\text{dog}} = 84.002 \, \text{kg} \cdot \text{m/s} \][/tex]

For the cat:

[tex]\[ p_{\text{cat}} = m_{\text{cat}} \times v_{\text{cat}} \][/tex]

[tex]\[ p_{\text{cat}} = 5.30 \, \text{kg} \times 2.64 \, \text{m/s} \][/tex]

[tex]\[ p_{\text{cat}} = 14.032 \, \text{kg} \cdot \text{m/s} \][/tex]

The total momentum of the dog and cat is the vector sum of their individual momenta. Since they are moving in perpendicular directions (northward and eastward), we can use the Pythagorean theorem to find the magnitude of the total momentum:

[tex]\[ p_{\text{total}} = \sqrt{p_{\text{dog}}^2 + p_{\text{cat}}^2} \][/tex]

[tex]\[ p_{\text{total}} = \sqrt{(84.002)^2 + (14.032)^2} \][/tex]

[tex]\[ p_{\text{total}} = \sqrt{7056.0624 + 196.82784} \][/tex]

[tex]\[ p_{\text{total}} = \sqrt{7252.89} \][/tex]

[tex]\[ p_{\text{total}} \ =85.136 \, \text{kg} \cdot \text{m/s} \][/tex]

The direction of the total momentum vector is northeast, which is [tex]45^\circ \)[/tex] from the northward direction (the direction of the dog's velocity).

Now, we equate the owner's momentum to the total momentum of the pets:

[tex]\[ p_{\text{owner}} = p_{\text{total}} \][/tex]

[tex]\[ m_{\text{owner}} \times v_{\text{owner}} = p_{\text{total}} \][/tex]

[tex]\[ 67.2 \, \text{kg} \times v_{\text{owner}} = 85.136 \, \text{kg} \cdot \text{m/s} \][/tex]

[tex]\[ v_{\text{owner}} = \frac{85.136 \, \text{kg} \cdot \text{m/s}}{67.2 \, \text{kg}} \][/tex]

[tex]\[ v_{\text{owner}} \ = 1.267 \, \text{m/s} \][/tex]

Answer:

rate of change in temperature of copper is more than the rate of change in temperature of aluminium.

so here copper will reach to our body temperature first

Explanation:

As we know that rate of energy absorb by the two sphere is same

so here we will have

[tex]\frac{dQ}{dt} = ms\frac{\Delta T}{\Delta t}[/tex]

now for copper sphere we will have

[tex]\frac{dQ}{dt} = 50(0.4)\frac{\Delta T}{\Delta t}[/tex]

[tex]\frac{\Delta T}{\Delta t} = \frac{1}{20}\frac{dQ}{dt}[/tex]

now for Aluminium sphere we will have

[tex]\frac{dQ}{dt} = 25(0.9)\frac{\Delta T}{\Delta t}[/tex]

[tex]\frac{\Delta T}{\Delta t} = \frac{1}{22.5}\frac{dQ}{dt}[/tex]

So rate of change in temperature of copper is more than the rate of change in temperature of aluminium.

so here copper will reach to our body temperature first

Answer:

Part a)

[tex]V = 7.2 \times 10^{11} Volts[/tex]

Part b)

this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.

Part C)

here we can assume the sphere is placed at vacuum so that there is no break down of air.

Explanation:

Part a)

As we know that the potential near the surface of metal sphere is given by the equation

[tex]V = \frac{kQ}{R}[/tex]

here we have

Q = 8 C

R = 10.0 cm

now we have

[tex]V = \frac{(9\times 10^9)(8 C)}{0.10}[/tex]

[tex]V = 7.2 \times 10^{11} Volts[/tex]

Part b)

this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.

Part C)

here we can assume the sphere is placed at vacuum so that there is no break down of air.

Final answer:

The voltage near a 10.0 cm diameter metal sphere with 8.00 C of excess charge is calculated to be 1.438 x 10^12 V, which is unreasonable due to the high value leading to inevitable discharge. The assumption of an 8.00 C charge on such a small sphere is responsible for this unrealistic result.

Explanation:

Calculating the Voltage near a Charged Sphere

To find the voltage near a 10.0 cm diameter metal sphere with an excess positive charge of 8.00 C, we use the formula V = kQ/r, where V is the voltage, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), Q is the charge, and r is the radius of the sphere. For a diameter of 10.0 cm, the radius (r) is 0.05 m. Thus, V = (8.99 x 10^9 N m^2/C^2 * 8.00 C) / 0.05 m = 1.438 x 10^12 V.

Unreasonable Voltage

This voltage is extremely high and unreasonable because a metal sphere of that size could not sustain such a high voltage without discharging. The consequence of such a high voltage would include electric breakdown of the air around the sphere, leading to sparks or lightning-like discharges.

Erroneous Assumptions

The assumption responsible for this unreasonable result is the magnitude of charge being considered. An 8.00 C charge on a small metal sphere is significantly larger than what could realistically accumulate on the surface, given the limits of charge density and material breakdown thresholds.

Answer:

Answer to the question is: 1837.65 millimeters of mercury are equal to 245 kPa.

Explanation:

1 kPa are equal to 7.50062 millimeters of mercury.

Final answer:

To convert the pressure from 245 kPa to mmHg, first convert kPa to atm, then multiply by the conversion factor from atm to mmHg. The pressure is 1837.68 mmHg.

Explanation:

To convert the pressure in a container from kilopascals (kPa) to millimeters of mercury (mmHg), we use the conversion factor that 1 atmosphere (atm) is equivalent to 760 mmHg. First, we convert the given pressure in kilopascals to atmospheres:

1 atm = 101.325 kPa

So, to convert 245 kPa to atm, we divide 245 kPa by 101.325 kPa/atm:

245 kPa / 101.325 kPa/atm = 2.418 atm

Next, we convert atmospheres to millimeters of mercury (mmHg) using the conversion factor:

2.418 atm x760 mmHg/atm = 1837.68 mmHg

Therefore, the pressure in the container is 1837.68 mmHg.

Answer:

[tex]T_C=256.2^{\circ}C[/tex]

Explanation:

Given that,

Efficiency of heat engine, [tex]\eta=40\%=0.4[/tex]

Temperature of hot source, [tex]T_H=427^{\circ}C[/tex]

We need to find the temperature of cold sink i.e. [tex]T_C[/tex]. The efficiency of heat engine is given by :

[tex]\eta=1-\dfrac{T_C}{T_H}[/tex]

[tex]T_C=(1-\eta)T_H[/tex]

[tex]T_C=(1-0.4)\times 427[/tex]

[tex]T_C=256.2^{\circ}C[/tex]

So, the temperature of the cold sink is 256.2°C. Hence, this is the required solution.

Answer:

The wavelength of the light is 633 nm.

Explanation:

Given that,

Distance between the two slits d= 0.025 cm

Distance between the screen and slits D = 120 cm

Distance between the slits y= 1.52 cm

We need to calculate the angle

Using formula of double slit

[tex]\tan\theta=\dfrac{y}{D}[/tex]

Where, y = Distance between the slits

D = Distance between the screen and slits

Put the value into the formula

[tex]\tan\theta=\dfrac{1.52}{120}[/tex]

[tex]\theta=\tan^{-1}\dfrac{1.52}{120}[/tex]

[tex]\theta=0.725[/tex]

We need to calculate the wavelength

Using formula of wavelength

[tex]d\sin\theta=n\lambda[/tex]

Put the value into the formula

[tex]0.025\times\sin0.725=5\times\lambda[/tex]

[tex]\lambda=\dfrac{0.025\times10^{-2}\times\sin0.725}{5}[/tex]

[tex]\lambda=6.326\times10^{-7}\ m[/tex]

[tex]\lambda=633\ nm[/tex]

Hence, The wavelength of the light is 633 nm.

Answer:

49.85 V

Explanation:

u = 0, s = 5.62 cm, t = 1.15 x 10^-6 s

Let the electric field is E and voltage is V.

Use second equation of motion

s = ut + 1/2 a t^2

5.62 x 10^-2 = 0 + 0.5 a x (1.15 x 10^-6)^2

a = 8.5 x 10^10 m/s^2

m x a = q x E

E = m x a / q

E = (1.67 x 10^-27 x 8.5 x 10^10) / (1.6 x 10^-19)

E = 887.19 V/m

V = E x s

V = 887.19 x 5.62 x 10^-2 = 49.85 V

Answer:

Work done by the gravitational force is 37 mJ.

Explanation:

It is given that,

Mass of the raindrop, [tex]m=3.26\times 10^{-5}\ kg[/tex]

It falls from a height of, h = 115 m

It falls vertically at constant speed under the influence of gravity and air resistance. We need to find the  work done on the raindrop by the gravitational force. It is given by :

[tex]W=mgh[/tex]

[tex]W=3.26\times 10^{-5}\ kg\times 9.8\ m/s^2\times 115\ m[/tex]

W = 0.0367 J

or

W = 0.037 J = 37 mJ

So, the work done on the raindrop by the gravitational force is 37 mJ. Hence, this is the required solution.

Final answer:

The work done on a raindrop of mass 3.26 times [tex]10^{-5}[/tex] kg by the gravitational force.

Explanation:

To calculate the work done on a raindrop by the gravitational force as it falls, we use the formula for work: W = mgh, where W is the work done, m is the mass of the object, g is the acceleration due to gravity (9.8 m/s2), and h is the height the object falls through.

In this case, the mass m of the raindrop is 3.26 times 10-5 kg, and the height h is 115 m. So:

W = (3.26 times 10-5 kg)(9.8 [tex]m/s^2[/tex])(115 m) = 0.0368134 J

Therefore, the work done on the raindrop by the gravitational force as it falls 115 m is approximately 0.0368134 joules.

Answer:

Weight in planet X = 0.413 N

Explanation:

Weight = Mass x Acceleration due to gravity.

W = mg

Mass, m = 100 g = 0.1 kg

We have equation of motion s = ut + 0.5 at²

Displacement, s = 10 m

Initial velocity, u = 0 m/s

Time, t = 2.2 s

Substituting

        s = ut + 0.5 at²

        10 = 0 x 2.2 + 0.5 x a x 2.2²        

        a = 4.13 m/s²

Acceleration due to gravity, a = 4.13 m/s²

W = mg = 0.1 x 4.13 = 0.413 N

Weight in planet X = 0.413 N

In Planet X, a 100-g ball is released from rest from a height of 10.0 m and it takes 2.2 s for it to reach the ground. The weight of the ball on the surface of Planet X is 0.41 N.

In physics, gravitational acceleration is the acceleration of an object in free fall within a vacuum (and thus without experiencing drag).

A 100-g (m) ball is released from rest from a height of 10.0 m (s) and it takes 2.2 s (t) for it to reach the ground. We can calculate the gravitational acceleration using the following kinematic equation.

s = 1/2 × g × t²

g = 2 s / t² = 2 (10.0 m) / (2.2 s)² = 4.1 m/s²

We will use Newton´s second law of motion.

w = m × g = 0.100 kg × 4.1 m/s² = 0.41 N

In Planet X, a 100-g ball is released from rest from a height of 10.0 m and it takes 2.2 s for it to reach the ground. The weight of the ball on the surface of Planet X is 0.41 N.

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Answer:

880 kg / m^3

Explanation:

height of column of oil = 85 cm = 0.85 m

height of column of mercury = 5.5 cm = 0.055 m

Density of mercury = 13600 kg/m^2

Let teh density of oil is d.

A the height of mercury column is balanced by the height of oil column

So, the pressure due to the mercury column = pressure by teh oil column

height of mercury column x density of mercury x g = height of oil column  

                                                                                       x density of oil x g

0.055 x 13600 x g = 0.85 x d x g

748 = 0.85 d

d = 880 kg / m^3

The average current passing through a device is given by:

I = Q/Δt

I is the average current

Q is the amount of charge that has passed through the device

Δt is the amount of elapsed time

Given values:

Q = 4.00C

Δt = 4.00hr = 14400s

Plug in the values and solve for I:

I = 4.00/14400

I = 0.000277777778A

I = 0.278mA

Final answer:

The current produced by the solar cells of a pocket calculator through which 4.00 C of charge passes in 4.00 hours is 0.278 milliamperes.

Explanation:

The current produced by the solar cells of a pocket calculator when 4.00 C of charge passes through it in 4.00 hours can be calculated using the formula for electric current I = Q / t, where I is the current in amperes, Q is the charge in coulombs, and t is the time in seconds.

To find the current in milliamperes (mA), first convert the time to seconds:

4.00 hours × 3600 seconds/hour = 14400 seconds.

Next, use the formula to calculate current:

I = 4.00 C / 14400 s = 0.00027778 A,

which is equivalent to 0.278 mA

Answer:

4.4 cm

Explanation:

B = 1.98 mT = 1.98 x 10^-3 T, v = 1.53 x 10^7 m/s, m = 9.1 x 10^-31 kg

q = 1.6 x 10^-19 C

(a) The force due to the magnetic field is balanced by the centrpetal force

mv^2 / r = q v B

r = m v / q B

r = (9,1 x 10^-31 x 1.53 x 10^7) / (1.98 x 10^-3 x 1.6 x 10^-19)

r = 0.044 m = 4.4 cm

Answer:

a) 0.0483 mol

b) 232 °C

Explanation:

Ideal gas law:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

a) Given:

P = 1.20×10⁵ Pa

V = 1.00 L = 1.00×10⁻³ m³

T = 25.8 °C = 298.95 K

PV = nRT

(1.20×10⁵ Pa) (1.00×10⁻³ m³) = n (8.314 m³ Pa / mol / K) (298.95 K)

n = 0.0483 mol

b) Given:

P = 1.013×10⁵ Pa

V = 2.00 L = 2.00×10⁻³ m³

n = 0.0483 mol

PV = nRT

(1.013×10⁵ Pa) (2.00×10⁻³ m³) = (0.0483 mol) (8.314 m³ Pa / mol / K) T

T = 505.73 K

T = 232 °C

Answer:

The cylinder’s total kinetic energy is 1.918 J.

Explanation:

Given that,

Mass = 4.1 kg

Radius = 0.057 m

Speed = 0.79 m/s

We need to calculate the linear kinetic energy

Using formula of linear kinetic energy

[tex]K.E_{l}=\dfrac{1}{2}mv^2[/tex]

[tex]K.E_{l}=\dfrac{1}{2}\times4.1\times(0.79)^2[/tex]

[tex]K.E_{l}=1.279\ J[/tex]

We need to calculate the rotational kinetic energy

[tex]K.E_{r}=\dfrac{1}{2}\times I\omega^2[/tex]

[tex]K.E_{r}=\dfrac{1}{2}\times\dfrac{1}{2}\times mr^2\times(\dfrac{v}{r})^2[/tex]

[tex]K.E_{r}=\dfrac{1}{4}\times m\times v^2[/tex]

[tex]K.E_{r}=\dfrac{1}{4}\times4.1\times(0.79)^2[/tex]

[tex]K.E_{r}=0.639\ J[/tex]

The total kinetic energy is given by

[tex]K.E=K.E_{l}+K.E_{r}[/tex]

[tex]K.E=1.279+0.639[/tex]

[tex]K.E=1.918\ J[/tex]

Hence, The cylinder’s total kinetic energy is 1.918 J.

Answer:

The Reynolds number is 5600.

Explanation:

Given that,

Density = 1400 kg/m³

Viscosity = 0.5 Pa's

Length = 2 m

Speed = 1 m/s

We need to calculate the Reynolds number

Using formula of Reynolds number

[tex]R_{e}=\dfrac{\rho V\times L}{\mu}[/tex]

Where, [tex]\rho[/tex] = density of fluid

v = speed of syrup

l = length of a person

[tex]\mu[/tex]=Viscosity

Put the all value into the formula

[tex]R_{e}=\dfrac{1400\times1\times2}{0.5}[/tex]

[tex]R_{e}=5600[/tex]

Hence, The Reynolds number is 5600.

Answer:

Explanation:

When the pendulum falls freely the net acceleration due to gravity is zero.

As we know that the time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity, thus the time period becomes infinity.

In freefall, the pendulum's effective acceleration due to gravity becomes zero, causing the pendulum to not swing, and its period becomes theoretically infinite and immeasurable.

Effect of Freefall on a Pendulum's Period

When considering simple pendulum motion in an elevator under normal conditions, we can determine its periodic time (T) using the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. This equation illustrates that the period of the pendulum (T) is affected by two variables: the length of the pendulum (L) and the acceleration due to gravity (g).

When the elevator is in free fall, the effective acceleration g becomes zero because the elevator and the pendulum are both in a state of free fall with the same acceleration due to gravity. Therefore, in this scenario, the pendulum would experience weightlessness and would not oscillate, resulting in an infinite theoretical oscillation period, making the concept of a period inapplicable.

The period is normally independent of mass or amplitude for small angles, but since freefall changes the acceleration experienced by the pendulum to zero, it significantly affects the pendulum's oscillation, negating the normal conditions for calculating a pendulum's period.

Answer:

work =p×v =3.27×10^5×1.5×10^-3 =490.5 joule

efficiency =w/q in

:. qin= w/efficiency =490.5/0.225=2180 joule

qout =q in - work =1689.5 joule

q out is work given as heat

The engine must give up 1689.5 J of heat to the low-temperature reservoir after calculating the total work done by the gas and accounting for the engine's efficiency.

To find the amount of work the engine gives up as heat, we first calculate the total work done by the gas using the formula W = PΔV, where W is work, P is pressure, and ΔV is the change in volume. Given a constant pressure of 3.27 x 105 Pa and a change in volume of 1.50 x 10-3 m3, the work done is:

W = PΔV = 3.27 x 105 Pa x 1.50 x 10-3 m3 = 490.5 J.

The efficiency of the engine is the ratio of the useful work output to the total work input, given by  ext_eta = useful work / total work. The equation that relates efficiency, work done (W), and heat given up (Q) is  ext_eta = W / (Q + W). We rearrange the equation to solve for Q:

Q = W /  ext_eta - W

Substituting the known values:

Q = 490.5 J / 0.225 - 490.5 J = 2180 J - 490.5 J = 1689.5 J.

Therefore, the engine must give up 1689.5 J of heat to the low-temperature reservoir.

Answer:

The distance between the person and the building is 68.48 meters.

Explanation:

It is given that,

Angle of elevation, θ = 30.6 degrees

Height of building, MP = 42 m

Height of person, AB = 1.5 m

We need to find the distance between person and building. It is given by BP.

Since, MN + NP = 42

So, MN = 40.5 m

Using trigonometric equation as :

[tex]tan\theta=\dfrac{MN}{AN}[/tex]

[tex]tan(30.6)=\dfrac{40.5}{AN}[/tex]

AN = 68.48 meters.

So, the distance between the person and the building is 68.48 meters. Hence, this is the required solution.

To determine the distance from a building, we use trigonometry and the formula adjacent = opposite / tangent(angle), taking into account the height of the building minus your height. The distance is calculated to be approximately 68.88 meters.

To find out how far you are from the building, we need to calculate the horizontal distance from the building's base to the point where you are standing. To do this, we can use trigonometry, specifically the tangent function which relates the angle of elevation to the opposite side and the adjacent side of a right-angle triangle. We need to consider the height of the building minus your height to find the correct opposite side.

Since the building is 42.0 meters tall, and you are 1.50 meters tall, the effective height we are looking at is 42.0 m - 1.50 m = 40.5 m. The angle of elevation you are looking at is 30.6 degrees. By using the formula tangent (angle) = opposite / adjacent, we can rearrange this to find the adjacent side (the distance from you to the building): adjacent = opposite / tangent (angle).

Therefore, the distance from you to the building is approximately adjacent = 40.5 m / tan(30.6°). Plugging in the values, we get:

Distance = 40.5 m / tan(30.6°) ≈ 40.5 m / 0.588 ≈ 68.88 m.

So, you are approximately 68.88 meters away from the building.

Answer:

115 m/s, 414 km/hr

Explanation:

There are two forces acting on a skydiver: gravity and air resistance (drag).  At terminal velocity, the two forces are equal and opposite.

∑F = ma

D − mg = 0

D = mg

Drag force is defined as:

D = ½ ρ v² C A

where ρ is the fluid density,

v is the velocity,

C is the drag coefficient,

and A is the cross sectional surface area.

Substituting and solving for v:

½ ρ v² C A = mg

v² = 2mg / (ρCA)

v = √(2mg / (ρCA))

We're given values for m and A, and we know the value of g.  We need to look up ρ and C.

Density of air depends on pressure and temperature (which vary with elevation), but we can estimate ρ ≈ 1.21 kg/m³.

For a skydiver falling headfirst, C ≈ 0.7.

Substituting all values:

v = √(2 × 80.0 kg × 9.8 m/s² / (1.21 kg/m³ × 0.7 × 0.140 m²))

v = 115 m/s

v = 115 m/s × (1 km / 1000 m) × (3600 s / hr)

v = 414 km/hr

The terminal velocity of the skydiver in m/s and km/h is;  115m/s  and  414 km/h

Using Given data :

mass of skydiver ( M ) = 80 kg

Cross sectional surface area ( A ) = 0.14 m^2

p ( fluid density ) ≈ 1.21 kg/m³.

C ( drag coefficient ) = 0.7

Determine the terminal velocity of the skydiver

At terminal velocity drag force and gravity is equal and opposite therefore canceling out each other

∑ F = ma

Drag force - Mg = 0

therefore;  D = Mg ----- ( 1 )

where D ( drag force ) = 1/2 pv² C A ---- ( 2 )

p = fluid density , C = drag coefficient , A = cross sectional area

v = velocity

Back to equations 1 and 2  ( equating them )

1/2 pv² CA = Mg ---- ( 3 )

v² = 2mg / ( p C A )

∴ V = √ ( 2mg / (p C A ))

V = √ ( 2 * 80 * 9.8 ) / ( 1.21 * 0.7 * 0.140 ))

    = 115 m/s  

also  V = 414 km/h

Hence we can conclude that the terminal velocity of the skydiver is in m/s and km/h are 115m/s and  414 km/h

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Answer:

0.94 m/s^2 downwards

Explanation:

m = 70 kg, m g = 70 x 9.8 = 686 N

R = 620 N

Let the acceleration be a, as the apparent weight decreases so the elevator is moving downwards with an acceleration a.

mg - R = ma

686 - 620 = 70 x a

a = 0.94 m/s^2

Answer:

366 K

Explanation:

T₀ = Constant Temperature of deep lake = 293.0 K

T = Temperature of hot reservoir  connected to carnot engine = ?

η = Efficiency of Carnot engine during the operation

Efficiency of Carnot engine is given as

[tex]\eta = 1-\frac{T_{o}}{T}[/tex]

Inserting the values

[tex]0.200 = 1-\frac{293.0}{T}[/tex]

T = 366 K