Given that the internal energy of water at 28 bar pressure is 988 kJ kg–1 and that the specific volume of water at this pressure is 0.121 × 10–2 m3 kg–1, calculate the specific enthalpy of water at 55 bar pressure.

Answer:

(a) 92 %

(b) 6.76 %

Explanation:

length, l = 4 m, height, h = 5 m, width, w = 3 m, density of water = 1000 kg/m^3

density of ice = 920 kg/m^3, density of mercury = 13600 kg/m^3

(a) Let v be the volume of ice below water surface.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of water x g = Total volume of ice block x density

                                                                      of ice x g

v x 1000 x g = V x 920 x g

v / V = 0.92

% of volume immersed in water = v/V x 100 = 0.92 x 100 = 92 %

(b) Let v be the volume of ice below the mercury.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of mercury x g = Total volume of ice block x  

                                                                      density of ice x g

v x 13600 x g = V x 920 x g

v / V = 0.0676

% of volume immersed in water = v/V x 100 = 0.0676 x 100 = 6.76 %

Answer:

False

Explanation:

Actually, the converse is true. The mass number would be lower than the sum of the mass of the individual nucleons combined. According to Einstein’s equation of E=MC², this will be due to a phenomenon called mass defect. This ‘anomaly’ is due to the loss of some energy (now the nuclear binding energy) when the nucleons were brought in together to form the nucleus.

Answer:

- 2.7 rad/s^2

Explanation:

f0 =  + 5.85 rotations per second (counter clockwise)

t = 21.3 s

f =  - 3.31 rotations per second (clockwise)

w0 = 2 x 3.14 x 5.85 = + 36.738 rad/s

w = - 2 x 3.14 x 3.31 = - 20.79 rad/s

Let α be teh angular acceleration.

α = (w - w0) / t

α = (-20.79 - 36.738) / 21.3

α = - 2.7 rad/s^2

The flywheel's average angular acceleration is -0.43 rad/s².

The average angular acceleration of the flywheel is determined by applying the following kinematic equation as shown below;

α = (ωf - ωi)/t

where;

α = (-3.31 - 5.85)/21.3

α = -0.43 rad/s²

Thus, the flywheel's average angular acceleration is -0.43 rad/s².

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Answer:

[tex]11304 \frac{in^{3}}{s}[/tex]

Explanation:

r = radius of right circular cone = 150 in

h = height of right circular cone = 144 in

[tex]\frac{dr}{dt}[/tex] = rate at which radius increase = 1.5 in/s

[tex]\frac{dh}{dt}[/tex] = rate at which height decrease = - 2.4 in/s

Volume of the right circular cone is given as

[tex]V = \frac{\pi r^{2}h}{3}[/tex]

Taking derivative both side relative to "t"

[tex]\frac{dV}{dt} = \frac{\pi }{3}\left ( r^{2}\frac{dh}{dt} \right + 2rh\frac{dr}{dt})[/tex]

[tex]\frac{dV}{dt} = \frac{3.14 }{3}\left ( (150)^{2}(- 2.4) + 2(150)(144)(1.5))[/tex]

[tex]\frac{dV}{dt} = 11304 \frac{in^{3}}{s}[/tex]

Answer:

a)4.36*10^-12W/m^2

b)L=6.39dB

c)ξ=5.73×10⁻¹¹m

Explanation:

Take speed of sound in air as 344m/s and density of air 1.2kg/m3 .

a)

[tex]I=\frac{P^2}{2pv}[/tex]

where

P=pressure

p=density of air

v=velocity

[tex]I=\frac{6*10^-5^2}{2*1.2*344} = 4.36*10^-12[/tex] W/m^2

b)

Sound intensity level in dB is defined as:  

L = 10∙log₁₀(I/I₀)  

with  

I₀ = 1.0×10⁻¹² W/m²  

Hence;  

L = 10∙log₁₀( 4.36x10^-12 W/m² / 1.0×10⁻¹² W/m²) = 6.39dB

c)

Displacement is given by :  

ξ = p/(Z∙ω) = p/(Z∙2∙π∙f)  

where  

Z = 416.9 N∙s/m³ = 416.9 Pa∙s/m

f frequency and ω angular frequency of the sound wave.  

So the amplitude of this sound wave is:  

ξ = 6×10⁻⁵Pa / (416.9 Pa∙s/m ∙ 2∙π∙ 400s⁻¹) = 5.73×10⁻¹¹m

To calculate the intensity and sound intensity level of a sound wave with a frequency of 400 Hz and a pressure amplitude of 6.0 * 10^(-10) Pa, use the given formulas. The intensity is 1.5 * 10^(-16) W/m², the sound intensity level is -60 dB, and the displacement amplitude is 5.3 * 10^(-9) m.

To calculate the intensity and sound intensity level of a sound wave with a frequency of 400 Hz and a pressure amplitude of 6.0 * 10^(-10) Pa, we can use the formulas:

(a) Intensity (I) = (Pressure Amplitude)^2 / (2 * Z)

(b) Sound Intensity Level (SIL) = 10 * log10(I / Io)

(c) Displacement Amplitude = Pressure Amplitude / (2 * π * f * v)

Using the given values, we can calculate:

(a) Intensity = (6.0 * 10^(-10))^2 / (2 * (1.21 * 10^(-3))) = 1.5 * 10^(-16) W/m²

(b) Sound Intensity Level = 10 * log10((1.5 * 10^(-16)) / (10^(-12))) = -60 dB

(c) Displacement Amplitude = (6.0 * 10^(-10)) / (2 * π * 400 * 343) = 5.3 * 10^(-9) m

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Answer:

16.96 m/s

Explanation:

It is the case of projectile motion in which the projectile fires from the ground and again it hit the ground.

The angle of projection is 32 degree and the velocity of projection is 20 m/s

he velocity at the maximum height is equal to the horizontal component of velocity which always remains constant as there is no acceleration along x axis.

Velocity at the highest point = u Cos 32 = 20 x Cos 32 = 16.96 m/s

Answer:  The gauge pressure of the air in the tires is 179.5 kPa.

Solution :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = Atmospheric pressure + gauge pressure = 101 kPa + 165 kPa = 266 kPa  

[tex]P_2[/tex] = final pressure of gas = ?

[tex]V_1[/tex] = initial volume of gas = [tex]1.63\times 10^{-2}m^3[/tex]

[tex]V_2[/tex] = final volume of gas = [tex]1.70\times 10^{-2}m^3[/tex]

[tex]T_1[/tex] = initial temperature of gas = [tex]0^oC=273+0=273K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]27.3^oC=273+27.3=300.3K[/tex]

Now put all the given values in the above equation, we get the final pressure of gas.

[tex]\frac{266\times 1.63\times 10^{-2}}{273K}=\frac{P_2\times 1.70\times 10^{-2}}{300.3K}[/tex]

[tex]P_2=280.5kPa[/tex]

Gauge pressure = Absolute pressure - atmospheric pressure  = (280.5 - 101) kPa= 179.5 kPa

Therefore, the gauge pressure of the air in the tires is 179.5 kPa.

Answer:

The current in the resistor is 56.44 mA.

Explanation:

Given that,

Capacitor Q= 2.04 nF

Initial charge [tex]q= 4.55\ \mu C[/tex]

Resistor [tex] R= 1.28 k\omega[/tex]

Time [tex]t = 9.00\times10^{-6}[/tex]

We need to calculate the current

Using formula of current

[tex]I=\dfrac{Q}{RC}e^{\dfrac{-t}{RC}}[/tex]

Where, C = capacitor

R = resistor

t = time

Q = charge

Put the value into the formula

[tex]I=\dfrac{4.55\times10^{-6}}{1.28\times10^{3}\times2.04\times10^{-9}}e^{\dfrac{-9.00\times10^{-6}}{1.28\times10^{3}\times2.05\times10^{-9}}}[/tex]

[tex]I=0.05644\ A[/tex]

[tex]I=56.44\ mA[/tex]

Hence, The current in the resistor is 56.44 mA.

Answer:

The original speeds of the two cars were :

[tex]6.36\frac{m}{s},12.72\frac{m}{s}[/tex]

Explanation:

Let's start reading the question and making our equations in order to find the speeds.

The first equation is :

[tex]m_{1}=2m_{2}[/tex] (I)

The kinectic energy can be calculated using the following equation :

[tex]K=(\frac{1}{2}).m.v^{2}[/tex] (II)

Where ''K'' is the kinetic energy

Where ''m'' is the mass and where ''v'' is the speed.

By reading the exercise we find that :  

[tex]K_{1}=\frac{K_{2}}{2}[/tex] (III)

If we use (II) in (III) :

[tex](\frac{1}{2}).m_{1}.v_{1}^{2}=(\frac{1}{2}).m_{2}.v_{2}^{2}.(\frac{1}{2})[/tex]

[tex]m_{1}.v_{1}^{2}=\frac{m_{2}.v_{2}^{2}}{2}[/tex] (IV)

If we replace (I) in (IV) ⇒

[tex]2.m_{2}.v_{1}^{2}=\frac{m_{2}.v_{2}^{2}}{2}[/tex]

[tex]4.v_{1}^{2}=v_{2}^{2}[/tex]

[tex]2.v_{1}=v_{2}[/tex] (V)

'' When both cars increase their speed by [tex]9.0\frac{m}{s}[/tex], they then have the same kinetic energy ''

The last equation is :

[tex](\frac{1}{2}).m_{1}.(v_{1}+9)^{2}=(\frac{1}{2}).m_{2}.(v_{2}+9)^{2}[/tex] (VI)

If we use (I) in (VI) ⇒

[tex]2.m_{2}.(v_{1}+9)^{2}=m_{2}.(v_{2}+9)^{2}[/tex]

[tex]2.(v_{1}+9)^{2}=(v_{2}+9)^{2}[/tex]

If we use (V) in this last expression ⇒

[tex]2.(v_{1}+9)^{2}=(2.v_{1}+9)^{2}[/tex]

[tex]2.(v_{1}^{2}+18v_{1}+81)=4v_{1}^{2}+36v_{1}+81[/tex]

[tex]2v_{1}^{2}+36v_{1}+162=4v_{1}^{2}+36v_{1}+81[/tex]

[tex]2v_{1}^{2}=81[/tex]

[tex]v_{1}^{2}=40.5[/tex]

[tex]v_{1}=\sqrt{40.5}=6.36[/tex]

We find that the original speed [tex]v_{1}[/tex] is [tex]6.36\frac{m}{s}[/tex]

If we replace this value in the equation (V) ⇒

[tex]2.(6.36\frac{m}{s})=v_{2}[/tex]

[tex]v_{2}=12.72\frac{m}{s}[/tex]

Answer:

7.6 m/s

Explanation:

m = 2.3 kg, h = 4.6 m, x = 25 cm = 0.25 m

Use the conservation of energy

Potential energy of the block = Elastic potential energy of the spring

m g h = 1/2 k x^2

Where, k be the spring constant

2.3 x 9.8 x 4.6 = 0.5 x k x (0.25)^2

k = 3317.88 N/m

Now, let v be the velocity of the block, when the compression is 15 cm.

Again use the conservation of energy

Potential energy of the block = Kinetic energy of block + Elastic potential  

                                                                                         energy of the spring

m g h = 1/2 m v^2 + 1/2 k x^2

2.3 x 9.8 x 4.6 = 0.5 x 2.3 x v^2 + 0.5 x 3317.88 x (0.15)^2

103.684 = 1.15 v^2 + 37.33

v = 7.6 m/s

Answer:

One characteristic of mass wasting processes is that they move materials relatively short distances compared to streams. - d.

A. 409 Hz

The fundamental frequency of a string is given by:

[tex]f_1=\frac{1}{2L}\sqrt{\frac{T}{m/L}}[/tex]

where

L is the length of the wire

T is the tension in the wire

m is the mass of the wire

For the piano wire in this problem,

L = 0.400 m

T = 1070 N

m = 4.00 g = 0.004 kg

So the fundamental frequency is

[tex]f_1=\frac{1}{2(0.400)}\sqrt{\frac{1070}{(0.004)/(0.400)}}=409 Hz[/tex]

B. 24

For this part, we need to analyze the different harmonics of the piano wire. The nth-harmonic of a string is given by

[tex]f_n = nf_1[/tex]

where [tex]f_1[/tex] is the fundamental frequency.

Here in this case

[tex]f_1 = 409 Hz[/tex]

A person is capable to hear frequencies up to

[tex]f = 1.00 \cdot 10^4 Hz[/tex]

So the highest harmonics that can be heard by a human can be found as follows:

[tex]f=nf_1\\n= \frac{f}{f_1}=\frac{1.00\cdot 10^4}{409}=24.5 \sim 24[/tex]

The fundamental frequency of the piano wire is approximately 409 Hz, based on the given mass, length, and tension. The highest harmonic that could be heard by a person capable of hearing frequencies up to 10,000 Hz would be the 24th harmonic.

To find the fundamental frequency (f1) of the piano wire, we can use the formula for the fundamental frequency of a stretched string, which is f1 = (1/2L) * sqrt(T/μ), where L is the length of the string, T is the tension, and μ is the mass per unit length (linear mass density).

First, we need to calculate the linear mass density of the piano wire:

Now, we plug in the values into the formula:

f1 = (1 / (2 * 0.400 m)) * sqrt(1070 N / 0.010 kg/m) = (1 / 0.8 m) * sqrt(107000 N/m) = (1 / 0.8) * 327.16 Hz ≈ 409 Hz.

For part B, we need to determine the highest harmonic that can be heard if the upper limit of human hearing is 10,000 Hz. Since the fundamental frequency is 409 Hz, the harmonics will be integer multiples of this value. The highest harmonic number (n) will be the largest integer such that:

n * f1 ≤ 10,000 Hz

Doing the division gives us n ≤ 10,000 / 409, and we round down to the nearest whole number because we cannot have a fraction of a harmonic. So, the highest harmonic heard would be:

n = floor(10,000/409) = 24 (since 24 * 409 ≈ 9816 Hz which is below the 10,000 Hz threshold).

Explanation:

For this question the magnetic force provides the force required for the circular motion

Equating

Magnetic force = centripetal force

Bqvsinx = mv^2/r

as magnetic field is perpendicular x=90

Bqvsin90 = mv^2/r

Bq = mv/r

B= mv/rq

Then replace r= 0.6 , v= 2.1ms and mp from formula sheet and you can obtain B which is magnetic field intensity

The amount of energy to reach the boiling point is [tex]50*80*4.184 J=16,736J[/tex]. To pass the boiling point, [tex]40.79*\frac{50}{18.02}kJ=113,180J[/tex] are necessary (18.02 is the molar mass of water). This means that [tex]150kJ-113.180kJ-16.736kJ=20,084J[/tex] are left. This allows the steam to heat another [tex]\frac{20,084}{50*1.99}=201.8^{\circ}C[/tex]. Therefore, it ends as steam at temperature [tex]100^{\circ}C+201.8^{\circ}C=301.8^{\circ}C[/tex]

Answer:

Option (b)

Explanation:

Let the ball takes time t to reach the maximum height. The final velocity at maximum height is zero.

Use I equation of motion, we get

V = u + gt

0 = 3 - 9.8 x t

t = 0.306 s

As the air friction is negligible, so the ball takes same time to reach the hand of boy. Thus the time taken by the ball in air is

T = 2 t = 2 x 0.306 = 0.61 s

Answer:

(a): s(t)= hi + Vo * t - g* t²/2

(b): Will take the ball to reach the ground t= 11 seconds.

Explanation:

hi= 880 ft

Vo= 96 ft/s

g= 32 ft/s²

equating to 0 the equation of s(t) and clearing t, we find the time it takes for the ball to fall to the ground.

Answer:

a) [tex]s(t) = 96t - 16t^{2} + 880[/tex]

b) It will take 11 seconds for the ball to reach the ground.

Explanation:

We have an initial height of 880 feet.

And

[tex]v(t) = 96 - 32t[/tex]

a) Find​ s(t), the function giving the height of the ball at time t

The position, or heigth, is the integrative of the velocity. So

[tex]s(t) = \int {(96 - 32t)} \, dt[/tex]

[tex]s(t) = 96t - 16t^{2} + K[/tex]

In which the constant of integration K is the initial height, so [tex]K = 880[/tex]

So

[tex]s(t) = 96t - 16t^{2} + 880[/tex]

b) How long will the ball take to reach the​ ground

This is t when [tex]s(t) = 0[/tex]

So

[tex]s(t) = -16t^{2} + 96t + 880[/tex]

This is t = -5 or t = 11.

However, t is the instant of time, so it has to be a positive value.

So it will take 11 seconds for the ball to reach the ground.

Answer:

Explanation:

As we know that magnetic flux is given by

[tex]\phi = B.A[/tex]

[tex]\phi = B.\pi r^2[/tex]

now from Faraday's law

[tex]EMF = \frac{d\phi}{dt}[/tex]

[tex]EMF = \frac{d(B. \pi r^2)}{dt}[/tex]

[tex]EMF = 2\pi r B \frac{dr}{dt}[/tex]

now we have

[tex]r = 40/2 = 20 cm[/tex]

B = 12 T

[tex]\frac{dr}{dt} = 5 \times 10^{-3} m/s[/tex]

Part a)

now at t = 1 s

r = 20 + 0.5 = 20.5 cm

[tex]EMF = (2\pi (0.205))(12)(5 \times 10^{-3})[/tex]

[tex]EMF = 0.077 Volts[/tex]

Part b)

now at t = 10 s

r = 20 + 0.5(10) = 25 cm

[tex]EMF = (2\pi (0.25))(12)(5 \times 10^{-3})[/tex]

[tex]EMF = 0.094 Volts[/tex]

Answer:

0.45 m

Explanation:

m = 1.23 kg, k = 37.4 N/m, vmax = 2.48 m/s

velocity is maximum when it passes through the mean position.

[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]

[tex]T = 2\pi \sqrt{\frac{1.23}{37.4}}[/tex]

T = 1.139 sec

w = 2 π / T

w = 2 x 3.14 / 1.139

w = 5.51 rad / s

Vmax = w A

Where, A be the amplitude

2.48 = 5.51 A

A = 2.48 / 5.51 = 0.45 m

Final answer:

The amplitude of oscillation for the given spring oscillator is 0.58 m.

Explanation:

The amplitude of oscillation can be calculated using the equation:

A = vmax/ω

where A is the amplitude, vmax is the maximum speed, and ω is the angular frequency.

The angular frequency can be calculated using the equation:

ω = sqrt(k/m)

where k is the spring constant and m is the mass of the oscillator.

Substituting the given values into the equations:

ω = sqrt(37.4 N/m / 1.23 kg) = 4.29 rad/s

A = 2.48 m/s / 4.29 rad/s = 0.58 m

Therefore, the amplitude of oscillation is 0.58 m.

Answer:

The distance between adjacent antinodes of the standing wave is 1.25 m.

Explanation:

Given that,

Frequency f= 120 MHz

We need to calculate the distance between adjacent antinodes of the standing wave

Using formula of distance

[tex]\Delta x=\dfrac{\lambda}{2}[/tex].....(I)

We know that,

[tex]\lambda=\dfrac{c}{f}[/tex]

Put the value of [tex]\lambda[/tex] in to the equation (I)

[tex]\Delta x=\dfrac{c}{2f}[/tex]

Where, c = speed of light

f = frequency

Put the all value into the formula

[tex]\Delta x=\dfrac{3\times10^{8}}{2\times120\times10^{6}}[/tex]

[tex]\Delta x=1.25\ m[/tex]

Hence, The distance between adjacent antinodes of the standing wave is 1.25 m.

The airplane's speed relative to the air mass is 38.51 m/s. To find the airplane's speed relative to the air mass, you can use the formula sqrt(VAx² + VAy²) by calculating the magnitude of the resultant vector of VAx and VAy.

In order to determine the airplane's speed relative to the air mass, a comprehensive analysis involves breaking down velocities into their horizontal and vertical components.

The wind velocity components are established as VWx = magnitude of wind velocity * cos(angle of wind direction) in the 'x' direction and VWy = magnitude of wind velocity * sin(angle of wind direction) in the 'y' direction.

Simultaneously, the airplane's velocity relative to the air mass components are defined as VAx = magnitude of airplane velocity * cos(angle of travel direction relative to air) in the 'x' direction and VAy = magnitude of airplane velocity * sin(angle of travel direction relative to air) in the 'y' direction.

The speed of the airplane relative to the air mass is ascertained by calculating the magnitude of the resultant vector of VAx and VAy using the formula: Speed of airplane relative to air mass = sqrt(VAx² + VAy²).

After substituting the given values in sqrt(VAx² + VAy²): the airplane's speed relative to the air mass is 38.51 m/s.

Therefore, the airplane's speed relative to the air mass is 38.51 m/s.

Answer:

4.9 m/s²

Explanation:

F = reading of the elevator while running = 594 N

W = weight of person and elevator together when at stop = 396 N

m = mass of person and elevator together

a = acceleration of elevator

Weight of the person is given as

W = mg

inserting the values

396 = m (9.8)

m = 40.4 kg

Force equation for the motion of the elevator is given as

F - W = ma

Inserting the values

594 - 396 = (40.4) a

a = 4.9 m/s²

Answer:

Tangential speed = 157 m/s

Explanation:

Tangential speed = Angular speed x Radius

Angular speed = 7.85 x 10⁴ rad/s

Radius = 2 mm = 2 x 10⁻³ m

Tangential speed = Angular speed x Radius = 7.85 x 10⁴ x 2 x 10⁻³

Tangential speed = 15.7 x 10 = 157 m/s

Tangential speed = 157 m/s

The tangential speed of a point on the outer edge of the disk attached to a high-speed drill, given an angular velocity of 7.85 x 10^4 rad/s and a radius of 2.00 mm, is 157 m/s.

To determine the tangential speed of a point on the outer edge of the disk which is attached to a high-speed drill, we apply the formula that relates tangential speed (v) to the angular velocity ($$omega$) and the radius (r) of the disk: v = r  $$omega$. Given the angular velocity of the disk is 7.85  $$x$ 10^4 rad/s and the radius is 2.00 mm (which we need to convert to meters for standard units), the calculation is straightforward.

v = r  $$omega$
v = 0.002 m  $$x$ 7.85  $$x$ 10^4 rad/s
v = 157 m/s.

Thus, the tangential speed of a point on the outer edge of this disk is 157 m/s.

The energy stored by the electric field between two plates, known as a parallel-plate capacitor, is calculated using the respective formulas for voltage, capacitance and energy in a capacitor, factoring in the charge of the plates, surface area, and separation distance.

The energy stored by the electric field between two parallel plates (known as a parallel-plate capacitor) can be calculated using the formula for the energy stored in a capacitor, which is U = 0.5 * C * V^2. Here, V is the voltage across the plates and C is the capacitance of the capacitor.

To calculate V, we use the relation V = E * d, where E is the electric field, which can be found using the formula E = Q/A (Charge per Area), and d is the distance separating the plates. In this case, the charge Q is 14 nC, the area A is (8.5 cm)^2 and the distance d is 2.5 mm.

Finally, to find C, we use the formula C = permittivity * (A/d), where the permittivity of free space is approximately 8.85 × 10^-12 F/m. With these variables, C, V and U can be calculated accordingly.

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The energy stored in the electric field of a parallel-plate capacitor can be calculated using the dimensions of the plates, their separation, and charges. For this particular scenario, the energy stored would be approximately 0.44 μJ.

In this scenario, we have a parallel-plate capacitor where each plate is a square with side 8.5 cm and has a charge of 14 nC. The plates are separated by a 2.5mm gap. Using this information, we can calculate the energy stored in the capacitor.

Firstly, the surface area A of the plates can be calculated using the formula A = s², where s is the side of the square. Converting 8.5 cm to meters, we get s = 0.085 m. So, A = (0.085 m)² = 0.007225 m².

Next, we calculate the electric field E between the plates using the formula E = Q/ε₀A, where Q is the charge on one plate and ε₀ is the permittivity of free space (8.85 x 10^-12 F/m). Converting 14 nC to coulombs gives Q = 14 x 10^-9 C. Substituting these values into the formula, we find E ≈ 224.717 kV/m.

The energy U stored in an electric field can be calculated using the formula U = 0.5ε₀EA². Substituting the earlier calculated values, we find U ≈ 0.44 μJ.

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Answer:

The height of the building is 180 meters.

Explanation:

It is given that,

Time taken for a ball to strike the ground, t = 6 sec

It is released from the rest. We need to find the height of the building from which it is released. It can be calculated using the equation as follows :

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

Here, a = g and u = 0

So, [tex]s=0+\dfrac{1}{2}gt^2[/tex]

[tex]s=\dfrac{1}{2}\times 10\ m/s^2\times (6\ s)^2[/tex]

s = 180 meters

So, the height of the building is 180 meters. Hence, this is the required solution.

Answer:

[tex]R=16.8[/tex]Ω

Explanation:

l=1000m

a=1mm2

p=1.68x10^-8

Electrical Resistivity Equation

[tex]R=p(\frac{L}{A} )\\[/tex]

where

p= proportional constant ρ (the Greek letter “rho”) is known as Resistivity.

L = is the length in metres (m)

A= is the area in square metres (m2),

R=1.68×10^-8×(1000/1×10^-6)

[tex]R=16.8[/tex]Ω

(a) The focal length of the lens is 6.86 cm, and the lens is converging.

(b) The height of the image is 6.375 mm, and it is inverted.

To solve this problem, we need to use the lens formula and magnification formula, which are given below:

The lens formula is:

[tex]\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}[/tex]

where:

The magnification formula is:

[tex]m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}[/tex]

where:

Now we can solve the problem step-by-step.

Part (a): Finding the focal length

Given:

Object distance, [tex]d_o = 16.0[/tex] cm

Image distance, [tex]d_i = 12.0[/tex] cm (since the image is on the same side as the object, [tex]d_i[/tex] is positive)

Using the lens formula:

[tex]\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}[/tex]

Substituting the given values:

[tex]\frac{1}{f} = \frac{1}{16.0} + \frac{1}{12.0}[/tex]

Calculating the right-hand side:

[tex]\frac{1}{f} = \frac{3 + 4}{48} = \frac{7}{48}[/tex]

Therefore, the focal length of the lens, [tex]f[/tex], is:

[tex]f = \frac{48}{7} = 6.86[/tex] cm

Since the focal length is positive, the lens is converging.

Part (b): Finding the height and orientation of the image

Given:

Object height, [tex]h_o = 8.50[/tex] mm

Object distance, [tex]d_o = 16.0[/tex] cm

Image distance, [tex]d_i = 12.0[/tex] cm

Using the magnification formula:

[tex]m = -\frac{d_i}{d_o}[/tex]

Substituting the given values:

[tex]m = -\frac{12.0}{16.0} = -0.75[/tex]

Using the magnification formula for height:

[tex]m = \frac{h_i}{h_o}[/tex]

Therefore:

[tex]-0.75 = \frac{h_i}{8.50}[/tex]

Giving:

[tex]h_i = -0.75 \times 8.50 = -6.375[/tex] mm

Since the height of the image, [tex]h_i[/tex], is negative, the image is inverted.

The appropriate solution is "-87.6 cm/s". A complete solution is given below.

According to the question, the given values are:

Amplitude,

a = 11.3 cm

Time,

t = 0.702 sec

→ The velocity (v) will be:

= [tex]\pm w\sqrt{a^2-x^2}[/tex]

or,

= [tex]-\frac{2 \pi}{t} \sqrt{a^2-(\frac{a}{2} )^2}[/tex]

= [tex]-\frac{2 \pi}{t} \frac{\sqrt{3} }{2}a[/tex]

By substituting the given values, we get

= [tex]-\frac{2 \pi}{0.702} \frac{\sqrt{3}\times 11.3 }{2}[/tex]

=  [tex]-8.946\times 9.7861[/tex]

= [tex]-87.6 \ cm/s[/tex]

Learn more:

https://brainly.com/question/23267480

Final answer:

To find the velocity when the position is half the amplitude in the positive x direction and moving toward x=0, we need to find the time at that position and plug it into the velocity formula.

In simple harmonic motion, the velocity is given by the formula v = ωA sin(ωt), where A is the amplitude and ω is the angular frequency. To find the velocity when the position is half the amplitude in the positive x direction and moving toward x=0, we need to find the time at that position and plug it into the velocity formula.

First, we find the time at which the particle is at the given position. Since the period is 0.702 sec, the time at half the amplitude in the positive x direction is 0.351 sec (half the period).

Next, we plug this time into the velocity formula: v = ωA sin(ωt). The angular frequency ω is given by ω = 2π / T, where T is the period. Plugging in the values, we can calculate the velocity.

Answer:

a) [tex]\mu_s =0.40[/tex]

b) [tex]\mu_k =0.20[/tex]

Explanation:

Given:

Mass of the cabinet, m = 51 kg

a) Applied force = 200 N

Now the force required to move the from the state of rest (F) = coefficient of static friction ([tex]\mu_s[/tex])× Normal reaction(N)

N = mg

where, g = acceleration due to gravity

⇒N = 51 kg × 9.8 m/s²

⇒N = 499.8 N

thus, 200N  = [tex]\mu_s[/tex] × 499.8 N

⇒[tex]\mu_s[/tex] = [tex]\frac{200}{499.8}=0.40[/tex]

b) Applied force = 100 N

since the cabinet is moving, thus the coefficient of kinetic([tex]\mu_k[/tex]) friction will come into action

Now the force required to move the from the state of rest (F) = coefficient of kinetic friction ([tex]\mu_k[/tex])× Normal reaction(N)

N = mg

where, g = acceleration due to gravity

⇒N = 51 kg × 9.8 m/s²

⇒N = 499.8 N

thus, 100N  = [tex]\mu_k[/tex] × 499.8 N

⇒[tex]\mu_k[/tex] = [tex]\frac{100}{499.8}=0.20[/tex]

Answer:

Gauge Pressure required = 606.258 kPa

Explanation:

Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.

Thus at a depth of 60m we have pressure of water equals

[tex]P(z)=P_{0}+\rho _wgh[/tex]

Now the gauge pressure is given by

[tex]P(z)-P_{0}=\rho _wgh[/tex]

Applying values we get

[tex]P(z)-P_{0}=\rho _wgh\\\\P_{gauge}=1.03\times 1000\times 9.81\times 60Pa\\\\P_{gauge}=606258Pascals\\\\P_{gauge}=606.258kPa[/tex]

Final answer:

To keep water from entering an open diving chamber at 60 meters depth with sea water of specific gravity 1.03, the air pressure inside the chamber must be equal to the water pressure at that depth, which is 603.84 kPa.

Explanation:

To calculate the air pressure (gage pressure) inside the diving chamber that prevents water from entering, we will use the pressure-depth relationship in fluids. The gage pressure is the pressure relative to atmospheric pressure. Given that the chamber is at a depth of 60 meters underwater and the specific gravity (SG) of the sea water is 1.03, we know that the sea water is denser than pure water (whose SG is 1). To prevent water from entering the chamber, the air pressure inside the chamber must be equal to the water pressure at that depth. The pressure due to a column of liquid is given by the formula P = ρgh, where P is the pressure, ρ is the density of the liquid, g is the acceleration due to gravity (9.8 m/s²), and h is the height (or depth) of the liquid column.

Since the water's specific gravity is 1.03, its density (ρ) can be calculated as ρ = SG * ρwater, where ρwater is the density of pure water (1000 kg/m³). So, ρ = 1.03 * 1000 kg/m³ = 1030 kg/m³. The depth (h) is 60 meters.

Now, we can calculate the pressure:

P = ρgh = (1030 kg/m³)(9.8 m/s²)(60 m) = 603,840 Pa or N/m².

To convert this to kilopascals (kPa), we divide by 1,000:

P = 603.84 kPa

This is the gage pressure inside the chamber required to prevent water from entering.

Answer:

R = 0.27 ohm

i = 11.25 A

Explanation:

Resistance of a given cylindrical rod is

[tex]R = \rho \frac{L}{A}[/tex]

here we know that

[tex]\rho = \frac{1}{3 \times 10^4} = 3.33 \times 10^{-5} ohm-m[/tex]

Now we have

[tex]R = (3.33 \times 10^{-5})\times \frac{4 \times 10^{-3}}{0.5 \times 10^{-6}}[/tex]

[tex]R = 0.27 ohm[/tex]

now we know by ohm's law

[tex]\Delta V = i R[/tex]

[tex]14 - 11 = i ( 0.27)[/tex]

[tex]i = 11.25 A[/tex]