1. Water flows through a hole in the bottom of a large, open tank with a speed of 8 m/s. Determine the depth of water in the tank. Viscous effects are negligible.

Answer:

metre per seconds

Explanation:

because velocity = distance ÷ time

Answer:

E = 1.24MeV

Explanation:

The photon travels at the speed of light, 3.0 × [tex]10^{8}[/tex] m/s, and given that its frequency = 1 picometer = 1.0 × [tex]10^{-12}[/tex] m.

Its energy can be determined by;

E = hf

  = (hc) ÷ λ

where E is the energy, h is the Planck's constant, 6.626 × [tex]10^{-34}[/tex] Js, c is the speed of the light and f is its frequency.

Therefore,

E = (6.626 × [tex]10^{-34}[/tex]× 3.0 × [tex]10^{8}[/tex]) ÷ 1.0 × [tex]10^{-12}[/tex]

  = 1.9878 × [tex]10^{-25}[/tex] ÷ 1.0 × [tex]10^{-12}[/tex]

E = 1.9878 × [tex]10^{-13}[/tex] J

But, 1 eV = 1.6 × [tex]10^{-19}[/tex] J. So that;

E = [tex]\frac{1.9878*10^{-13} }{1.6*10^{-19} }[/tex]

  = 1242375 eV

∴ E = 1.24MeV

The energy of the photon is 1.24MeV.

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

[tex]\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s[/tex]

The induced emf in the shorter coil is calculated as;

[tex]E = NA\frac{\delta B}{\delta t}[/tex]

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

[tex]E = NA\frac{\delta B}{\delta t}[/tex]

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

The induced emf in the coil at the center of the longer solenoid is [tex]1.725\times10^{-4}V[/tex]

The induced emf is produced in a coil when the magnetic flux through the coil is changing. It opposes the change of magnetic flux. Mathematically it is represented as the negative rate of change of magnetic flux at follows:

[tex]E=-\frac{\delta\phi}{\delta t}[/tex]

where E is the induced emf,

[tex]\phi[/tex] is the magnetic flux through the coil.

The changing current varies the magnetic flux through the coil at the center of the long solenoid, which is given by:

[tex]\phi = \frac{\mu_oNIA}{L}[/tex]

so;

[tex]\frac{\delta\phi}{\delta t}=\frac{\mu_oNA}{L} \frac{\delta I}{\delta t}[/tex]

where N is the number of turns of longer solenoid, A is the cross sectional area, I is the current and L is the length of the coil.

[tex]\frac{\delta\phi}{\delta t}=\frac{4\pi \times10^{-7} \times600 \times \pi \times(1.25\times10^{-2})^2}{25\times10^{-2}} \frac{5}{60}\\\\\frac{\delta\phi}{\delta t}=1.23\times10^{-7}Wb/s[/tex]

The emf produced in the coil at the center of the solenoid which has 14 turns will be:

[tex]E=N\frac{\delta \phi}{\delta t}\\\\E=14\times1.23\times10^{-7}V\\\\E=1.725\times10^{-4}V[/tex]

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Answer:

a

  [tex]F =326.7 \ N[/tex]

b

  [tex]M = 6[/tex]

Explanation:

From the question we are told that

          The mass of the rock is  [tex]m_r = 200 \ kg[/tex]

          The  length of the small object from the rock is  [tex]d = 2 \ m[/tex]

          The  length of the small object from the branch [tex]l = 12 \ m[/tex]

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The  force exerted by the weight of the rock is mathematically evaluated as

      [tex]W = m_r * g[/tex]

substituting values

     [tex]W = 200 * 9.8[/tex]

     [tex]W = 1960 \ N[/tex]

 So  at  equilibrium the sum  of the moment about the fulcrum is mathematically represented as

         [tex]\sum M_f = F * cos \theta * l - W cos\theta * d = 0[/tex]

Here  [tex]\theta[/tex] is very small so  [tex]cos\theta * l = l[/tex]

                               and  [tex]cos\theta * d = d[/tex]

Hence

       [tex]F * l - W * d = 0[/tex]

=>    [tex]F = \frac{W * d}{l}[/tex]

substituting values

        [tex]F = \frac{1960 * 2}{12}[/tex]

       [tex]F =326.7 \ N[/tex]

The  mechanical advantage is mathematically evaluated as

          [tex]M = \frac{W}{F}[/tex]

substituting values

        [tex]M = \frac{1960}{326.7}[/tex]

       [tex]M = 6[/tex]

Answer:

20.07ohms

Explanation:

Impedance is defined as the opposition to the flow of current through the elements of the circuit.

Impedance for R-L AC circuit is expressed as Z = √R²+XL²

R is the resistance

XL is the inductive reactance.

Given resistance of the air condition = 20 ohms

Inductive reactance XL = 1.68 ohms

Z = √20²+1.68²

Z = √400+2.8224

Z = √402.8224

Z = 20.07 ohms

Hence the impedance of the air conditioner is 20.07ohms

Answer:

Explanation:

Frequency, in physics, the number of waves that pass a fixed point in unit time; also, the number of cycles or vibrations undergone during one unit of time by a body in periodic motion. ... See also angular velocity; simple harmonic motion.

Answer:

931.00ft-lb

Explanation:

Pls see attached file

The work done in moving the object from x = 1 ft to x = 18 ft is 935  ft-lb.

Work is the product of the displacement's magnitude and the component of force acting in that direction. It is a scalar quantity having only magnitude and Si unit of work is Joule.

Given that force = 6x - 2 pounds.

So, work done in moving the object from x = 1 ft to x = 18 ft is = [tex]\int\limits^{18}_1 {(6x-2)} \, dx[/tex]

= [ 3x² - 2x]¹⁸₁

= 3(18² - 1² ) - 2(18-1) ft-lb

= 935  ft-lb.

Hence, the work done is  935  ft-lb.

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Answer:

I believe it is 91

Explanation:

Answer:

gravitational potential energy at point A.

A) The gravitational potential energy at point A is; 705600 J

B) The velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J is; v = 31.295 m/s

A) Formula for gravitational potential energy is;

PE = mgh

At point A;

mass; m = 900 kg

height; h = 80 m

Thus;

PE = 900 × 9.8 × 80

PE = 705600 J

B) Kinetic energy of the roller coaster at point C is given as;

KE = PE - W

We are given Workdone; W = 264870 J

Thus;

KE = 705600 - 264870

KE = 440730 J

Thus, velocity at point C is gotten from the formula of kinetic energy;

KE = ½mv²

v = √(2KE/m)

v = √(2 × 440730/900)

v = 31.295 m/s

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Explanation:

It is given that, Isaac drop ball from height of 2.0 m, and it bounces to a height of 1.5 m.

We need to find the speed before and after the ball bounce.

Let u is the initial speed of the ball when he dropped from height of 2 m. The conservation of energy holds here. So,

[tex]\dfrac{1}{2}mu^2=mgh\\\\u=\sqrt{2gh} \\\\u=\sqrt{2\times 9.8\times 2} \\\\u=6.26\ m/s[/tex]

Let v is the final speed when it bounces to a height of 1.5 m. So,

[tex]\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 1.5} \\\\v=5.42\ m/s[/tex]

So, the speed before and after the ball bounce is 6.26 m/s and 5.42 m/s respectively.

Answer:

0.0081T

Explanation:

The magnetic field B in the toroid is proportional to the applied current I and the number of turns N per unit length L of the toroid. i.e

B ∝ I [tex]\frac{N}{L}[/tex]

B = μ₀ I [tex]\frac{N}{L}[/tex]                   ----------------(i)

Where;

μ₀ = constant of proportionality called the magnetic constant = 4π x 10⁻⁷N/A²

Since the radius (r = 4.2cm = 0.042m) of the toroid is given, the length L is the circumference of the toroid given by

L = 2π r

L = 2π (0.042)

L = 0.084π

The number of turns N = 1000

The current in the toroid = 1.7A

Substitute these values into equation (i) to get the magnetic field as follows;

B = 4π x 10⁻⁷ x  1.7 x  [tex]\frac{1000}{0.084\pi }[/tex]        [cancel out the πs and solve]

B = 0.0081T

The magnetic field along the central radius is 0.0081T

Answer:

The critical angle is  [tex]i = 41.84 ^o[/tex]

Explanation:

From the  question we are told that

    The index of refraction of the sugar solution is  [tex]n_s = 1.5[/tex]

   The  index of refraction of air is  [tex]n_a = 1[/tex]

Generally from Snell's  law

      [tex]\frac{sin i }{sin r } = \frac{n_a }{n_s }[/tex]

Note that the angle of incidence in this case is equal to the critical angle

Now for total internal reflection the angle of reflection is [tex]r = 90^o[/tex]

So  

      [tex]\frac{sin i }{sin (90) } = \frac{1 }{1.5 }[/tex]

      [tex]i = sin ^{-1} [\frac{ (sin (90)) * 1 }{1.5} ][/tex]

      [tex]i = 41.84 ^o[/tex]

Answer:

The meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds

Explanation:

We are told that the particle moves back and forth along a straight line with velocity v(t).

Now, velocity is the rate of change of distance with time. Thus, the integral of velocity of a particle with respect to time will simply be the distance covered by the particle.

Thus, the meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds

Answer:

Explanation:

The formula for calculating the resistance of a material in terms of its resistivity is expressed as [tex]R = \rho L/A[/tex] where;

R is the resistance of the material

[tex]\rho[/tex] is the resistivity of the material

L is the length of the wire

A is the area = πr² with r being the radius

[tex]R = \rho L/\pi r^{2}[/tex]

If the length of a certain wire is doubled while its radius is kept constant, then the new length of the wire L₁ = 2L

The new resistance of the wire R₁ will be expressed as [tex]R_1 = \frac{\rho L_1}{A_1}[/tex]

since the radius is constant, the area will also be the same i.e A = A₁ and the resistivity also will be constant. The new resistance will become

[tex]R_1 = \frac{\rho(2L)}{A}[/tex]

[tex]R_1 = \frac{2\rho L}{\pi r^2}[/tex]

Taking the ratio of both resistances, we will have;

[tex]\frac{R_1}{R} = \frac{2\rho L/\pi r^2}{\rho L/ \pi r^2} \\\\\frac{R_1}{R} = \frac{2\rho L}{\pi r^2} * \frac{\pi r^2}{ \rho L} \\\\\frac{R_1}{R} = \frac{2}{1}\\\\R_1 = 2R[/tex]

This shoes that the new resistance of the wire will be twice that of the original wire

Answer:

Explanation:

Given the height reached by a balloon after t sec modeled by the equation

h=1/2t²+1/2t

a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t

If h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

b) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec

c) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec

A)  The height of the balloon at the end of 40 sec is 820 feet.

B) The average velocity of the balloon is 30 ft/sec.

C) The velocity of the balloon at the end of 30 sec is

Given :

Part A)

The height of the balloon after 40 secs is :

h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

Part B)

The average velocity of the balloon is  :

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0 sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

When at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon = 30 ft/sec

The average velocity of the balloon  is  30 ft/sec.

Part C)

The velocity of the balloon at the end of 30 sec is :

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec.

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Answer:

The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.

Explanation:

This is because when a capacitor is charged no current or voltage flows through it so it will have a voltage equal to the terminal voltage of the battery

Answer:

5.81 X 10^3 Ns

Explanation:

Given that

F = At² and F at t = 1.25 s is 781.25 N ?

A = F/t² at t = 1.25 s => F = 781.25/(1.25)² = 500 N/s²

d(Impulse) = Fdt

Impulse = ∫Fdt =∫At²dt evaluated in the interval 2.00 s ≤ t ≤ 3.50 s

Impulse = At³/3 = (500/3)(t³) = 166.7t³ between t = 2.00 s and t = 3.50 s

Impulse = 166.7[3.5³ - 2³] = 166.7[42.875 - 8] = 166.7[34.875] = 5813.7 Ns

5.81 X 10^3 N.s

Answer:

B.solar panels generate electricity that keeps the satellites running.

Explanation:

Solar panels are a renewable resource because they take energy from the sun.

Answer:

Explanation:

Total mass m = 18 kg .

Spring are parallel to each other so total spring constant

= 4 x 24 = 96 N/cm = 9600 N/m

Time period of vibration

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

Putting the given  values

[tex]T=2\pi\sqrt{\frac{18}{9600} }[/tex]

= .27 s .

Answer:

Q = -3280J

Explanation:

From the First Law of Thermodynamics, energy cannot be created nor destroyed but it can be converted from one form to another with the interaction of heat. Mathematically, this can be expressed as:

ΔU = Q + W        ----------(i)

Where;

ΔU = total change in internal energy of a system.

Q = heat exchanged between the system and the surrounding

W = work done by or on the system.

If heat is lost into the surrounding, then Q = -ve, else Q = +ve

If work is done on the system, then W = -ve, else W = -ve

=> From the question, Rick is the system and does a work of

W = +1090J  [since Rick does the work, W = +ve]

=>Also, the internal energy decreases by 2190J, therefore,

ΔU = -2190J   [since there is a decrease in internal energy]

Substitute the values of W and ΔU into equation (i) as follows;

-2190 = Q + 1090

=> Q = -2190 - 1090

=> Q = -3280J

Therefore, the value of Q = -3280J

Answer:

Explanation:

Let the acceleration be a of the system

T₁ and T₂ be the tension in the string attached with 3.85 and 2.6 kg of mass

for motion of 3.85 kg , applying newton's law

3.85g - T₁ = 3.85 a

for motion of 2.6 kg

T₂ - 2.6g = 2.6 a

T₂ - T₁ + 1.25 g = 6.45 a

T₁ - T₂ = 1.25 g -  6.45 a

for motion of pulley

(T₁ - T₂ ) x R = I x α where R is radius of pulley , I is its moment of inertia and α is angular acceleration

(T₁ - T₂ ) x R = 1 /2  m R² x a / R

(T₁ - T₂ )  =   m  x a / 2 = .79 x a / 2 = . 395 a

1.25 g -  6.45 a = .395 a

1.25 g = 6.845 a

a = 1.79 m /s²

B )

When heavier mass is given speed of .2 m /s , it comes to rest in 6.4 s

Average deceleration = .2 / 6.4 = .03125 m /s²

Total deceleration created by frictional torque = 1.79 + .03125

= 1.82125 m /s²

If R be the average frictional torque  acting on the pulley

angular deceleration of pulley = a / R

= 1.82125 / .045

= 40.47 rad /s²

Now  R = I x 40.47 , I is moment of inertia of pulley

= 1 /2 x .79 x .045² x 40.47

= .0323 N.m

Torque created = .0323 Nm

The acceleration of the two masses hanging from ends of the pulley is 31 m/s².

The average frictional torque acting on the pulley is 0.55 Nm.

The given parameters;

The acceleration of the two masses is determined by taking net torque acting on the pulley;

[tex]\tau _{net} = I \alpha[/tex]

[tex]T_1R - T_2R = I \alpha\\\\[/tex]

where;

[tex]R(T_1 - T_2) = (\frac{MR^2}{2} )(\frac{a}{R} )\\\\T_1 - T_2 = (\frac{MR^2}{2} )(\frac{a}{R} ) \times \frac{1}{R} \\\\T_1 - T_2 = \frac{M}{2} \times a\\\\a = \frac{2}{M} (T_1 - T_2)[/tex]

Substitute the given parameters, to solve for the acceleration of the masses;

[tex]a = \frac{2}{M} (m_1g - m_2 g)\\\\a = \frac{2g}{M} (m_1 - m_2)\\\\a = \frac{2 \times 9.8}{0.79} (3.85 - 2.6)\\\\a = 31 \ m/s^2[/tex]

The average frictional torque acting on the pulley when the heavier mass speeds down by 0.2 m/s and stop by 6.4 s.

[tex]a = \frac{v}{t} = \frac{0.2}{6.4} = 0.031 \ m/s^2 \\\\ a_t = 31 m/s^2+ 0.031 m/s^2 = 31.031 m/s^2 \\\\\tau = I \alpha\\\\\tau = (\frac{MR^2}{2} )(\frac{a_t}{R} )\\\\\tau = (\frac{0.79 \times 0.045^2 }{2} ) (\frac{31.031}{0.045} )\\\\\tau = 0.55 \ Nm[/tex]

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Answer:

The electric field strength between the plates can be increased by decreasing the length of each side of the plates.

Explanation:

The electric field strength is given by;

[tex]E = \frac{V}{d}[/tex]

where;

V is the electric potential of the two opposite charges

d is the distance between the two parallel plates

[tex]E =\frac{V}{d} = \frac{\sigma}{\epsilon _o} \\\\(\sigma = \frac{Q}{A} )\\\\E = \frac{Q}{A\epsilon_o} \\\\E = \frac{Q}{L^2\epsilon_o}[/tex]

Where;

ε₀ is permittivity of free space

L is the length of each side of the plates

From the equation above, the electric field strength can be increased by decreasing the length of each side of the plates.

Therefore, decreasing the length of each side of the plates, could be made to increase the strength of the electric field between the plates

Answer:

a

    [tex]\alpha = 2327.7 \ rev/s^2[/tex]

b

   [tex]\theta = 9124.5 \ rev[/tex]

Explanation:

From the question we are told that

    The maximum  angular   speed is  [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]

     The  time  taken is  [tex]t = 2.8 \ s[/tex]

     The  minimum angular speed is  [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest

Apply the first equation of motion to solve for acceleration we have that

       [tex]w_{max} = w_{mini} + \alpha * t[/tex]

=>     [tex]\alpha = \frac{ w_{max}}{t}[/tex]

substituting values

       [tex]\alpha = \frac{40950.73}{2.8}[/tex]

       [tex]\alpha = 14625 .3 \ rad/s^2[/tex]

converting to [tex]rev/s^2[/tex]

  We have

           [tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]

           [tex]\alpha = 2327.7 \ rev/s^2[/tex]

According to the first equation of motion the angular displacement is  mathematically represented as

       [tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]

substituting values

      [tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]

      [tex]\theta = 57331.2 \ radian[/tex]

converting to revolutions  

        [tex]revolution = 57331.2 * 0.159155[/tex]

        [tex]\theta = 9124.5 \ rev[/tex]

Answer:

The frequency will be reduced by a factor of √2/2

Explanation:

Pls see attached file

The new frequency of oscillation will be half the original frequency of oscillation of spring-block system.

Let the initial mass of block be m.

And new mass is, 4m.

The frequency of oscillating motion is defined as the number of complete oscillation made during the time interval of 1 second. The mathematical expression for the frequency of oscillation of block-spring system is given as,

[tex]f = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]

Here,

k is the spring constant.

If the mass of block increased to 4m, then the new frequency of oscillation of spring will be,

[tex]f' = \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{4m}}\\\\\\f' =\dfrac{1}{2} \times \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m}}\\\\\\f' =\dfrac{1}{2} \times f[/tex]

Thus, we can conclude that the new frequency of oscillation will be half the original frequency of oscillation of spring-block system.

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one of the answers that i found was   5.83 m i did some more research and it showed the same answer again. good luck with it. hope i was able to help you.

Answer:

A.21.3T

B.1.8x 10^6J/m^3

C.0.27x10^2J

D.6.6x10^-3H

Explanation:

Pls see attached file

Answer:

Explanation:

radius of aorta = 1.5 cm

cross sectional area = π r²

= 3.14 x 1.5²

= 7.065 cm²

volume of blood flowing out per second out of heart

= a x v , a is cross sectional area , v is velocity of flow

= 7.065 x 11.2

= 79.128 cm³

heart beat per second = 67 / 60

= 1.116666

If V be the volume of heart

1.116666 V = 79.128

V = 70.86 cm³.

Answer:

Explanation:

Using E= hc/wavelength

6.63 x10^-34 x3x10^8/ 236nm

19.86*10^-26/236*10^-9

=0.08*10^-35Joules

Answer:

F= 175.5N

Explanation:

Given:

Torque which can also be called moment is defined as rotational equivalent of linear force. It is the product of the external force and perpendicular distance

torque of 38N⋅m

angle of 120∘

Torque(τ): 38Nm

position r relative to its axis of rotation: 25cm , if we convert to metre for consistency we have 0.25m

Angle: 120°

To find the Force, the torque equation will be required which is expressed below

τ = Frsinθ

We need to solve for F, if we rearrange the equation, we have the expression below

F= τ/rsinθ

Note: the torque is maximum when the angle is 90 degrees

But θ= 180-120=60

F= 38/0.25( sin(60) )

F= 175.5N