a 250 mH coil of negligible resistance is connected to an AC circuit in which as effective current of 5 mA is flowing. if the frequency is 850Hz, find the inductive reactance

Answer:

The absolute pressure is 28.15 psi.

Explanation:

Gauge pressure = 28 inch of Mercury

Absolute pressure, Po = 14.4 psi

The absolute pressure is the sum of the gauge pressure and the absolute pressure.  

gauge pressure = 28 inch = 0.7112 m of Mercury

= 0.7112 x 13.6 x 1000 x 9.8 = 94788.736 Pa

= 13.75 psi

The absolute pressure is

P = 14.4 + 13.75 = 28.15 psi

This question is incomplete, the complete question is;

Consider the cylindrical weir of diameter 3m and length 6m. If the fluid on the left has a specific gravity of 1.6 and on the right has a specific gravity of 0.8, Find the magnitude and direction of the resultant force.

Answer:

- the magnitude of the resultant force is 557.32 kN

- the direction of resultant force is  48.29°

Explanation:

Given the data in the question and the diagram below,

First we work on the force on the left hand side.

Left Horizontal

[tex]F_{LH[/tex] = βgAr

here, h = 3/2 = 1.5 m, β = 1.6, g = 9.81 m/s², A = 3 m × 6 m = 18 m²

we substitute

[tex]F_{LH[/tex] = βgAh = ( 1.6 × 1000 ) × 9.81 × 18 × 1.5 = 423792 N

Left Vertical

[tex]F_{LV[/tex] = ( βgπh² / 2 ) × W

we substitute

[tex]F_{LV[/tex] = [ ( ( 1.6 × 1000 ) × 9.81  × π(1.5)² ) / 2 ] × 6 = 332845.458 N

Now we go to the right hand side

Right Horizontal

[tex]F_{RH[/tex] = βgAh

here, h' = 1.5/2 = 0.75 m, β = 0.8, g = 9.81 m/s², A = 1.5 m × 6 m = 9 m²

we substitute

[tex]F_{RH[/tex] = ( 0.8 × 1000 ) × 9.81 × 9 × 0.75 ) = 52974 N

Right Vertical

[tex]F_{RV[/tex] = ( βgπh² / 4 ) × W

we substitute

[tex]F_{RV[/tex] = [ ( ( 0.8 × 1000 ) × 9.81  × π(1.5)² ) / 4 ] × 6 =  83211.36 N

Hence

Fx = [tex]F_{LH[/tex] - [tex]F_{RH[/tex] = 52974 N - 423792 N =  370818 N

Fy = [tex]F_{LV[/tex] + [tex]F_{RV[/tex] = 332845.458 N + 83211.36 N = 416056.818 N

R = √( Fx² + Fy² ) = √[ (370818 N)² + (416056.818 N)² ] = 557323.3 N

R = 557.32 kN

Therefore, the magnitude of the resultant force is 557.32 kN

Direction of resultant force;

tanθ = Fy / Fx

we substitute

tanθ = 416056.818 N / 370818 N

tanθ = 1.121997

θ = tan⁻¹( 1.121997 )

θ = 48.29°

Therefore, the direction of resultant force is  48.29°

Answer: The given statement is False.

Explanation:

Boyle's law states that pressure is inversely proportional to the volume of the gas at constant temperature and the number of moles.

Mathematically,

[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)

OR

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1\text{ and }V_1[/tex] are the initial pressure and volume of the gas

[tex]P_2\text{ and }V_2[/tex] are the final pressure and volume of the gas

Hence, the given statement is False.

Answer:

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Explanation:

Given :

Frequencies of the sinusoids,

[tex]$f_{m_1}= 65 \ Hz$[/tex] ,  and

[tex]$f_{m_2}= 95 \ Hz$[/tex]

Sampling rate [tex]f_s = \ 245 \ Hz[/tex]

The positive frequencies at the output of the sampling system are :

[tex]$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s $[/tex]

When n = 0,

[tex]$f_{o_1}= f_{m_1} = 65 \ Hz,\ \ f_{o_2}= f_{m_2} = 95 \ Hz $[/tex]

when n  = 1,

[tex]$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm f_s $[/tex]

[tex]$f_{o_1}= \pm 65 \pm 245,\ \ f_{o_2}=\pm 95 \pm 245$[/tex]

[tex]$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \ f_{o_2}= 150 \ Hz,340 \ Hz$[/tex]

When n = 2,

[tex]$f_{o_1}= \pm 65 \pm 2(245),\ \ f_{o_2}=\pm 95 \pm 2(245)$[/tex]

[tex]$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \ f_{o_2}= 395 \ Hz,585 \ Hz$[/tex]

Therefore, the first six positive frequencies present in the replicated spectrum are :

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Answer:

0 J

Explanation:

Since work done W = PΔV where P = pressure and ΔV = change in volume.

Since the volume is constant, ΔV = 0

So, Work done, W = PΔV = P × 0 = 0 J

So, the work done is 0 J.

Answer:

,hgfghjhytreftgyhujijjuhygtfrderftgyh

Explanation:

Answer:

See explanation

Explanation:

The degradation of the drug is a first order process;

Hence;

ln[A] = ln[A]o - kt

Where;

ln[A] = final concentration of the drug

ln[A]o= initial concentration of the drug = 5 gm/100

k= degradation constant = 0.05 day-1

t= time taken

When [A] =[ A]o - 0.5[A]o = 0.5[A]o

ln2.5 = ln5 - 0.05t

ln2.5- ln5 = - 0.05t

t= ln2.5- ln5/-0.05

t= 0.9162 - 1.6094/-0.05

t= 14 days

b) when [A] = [A]o - 0.9[A]o = 0.1[A]o

ln0.5 = ln5 -0.05t

t= ln0.5 - ln5/0.05

t= -0.693 - 1.6094/-0.05

t= 46 days

Answer:

[tex]T=728.9K[/tex]

Explanation:

Power [tex]P=100W[/tex]

Diameter [tex]d=5[/tex]

Radius [tex]r=2.5cm=>2.5*10^{-2}m[/tex]

Emissivity [tex]e=0.8[/tex]

Generally the equation for Area of Spherical bulb is mathematically given by

[tex]A=4\pi r^2[/tex]

[tex]A=4\pi (2.5*10^{-2}m)^2[/tex]

[tex]A=7.85*10^{-3}m^2[/tex]

Generally the equation for Emissive Power bulb is mathematically given by

[tex]E=e\mu AT^4[/tex]

Where

[tex]\mu=Boltzmann constants\\\\\mu=5.67*10^{-8}[/tex]

Therefore

[tex]T^4=\frac{E}{e\mu A}[/tex]

[tex]T^4=\frac{100}{0.8*5.67*10^{-8}*7.85*10^{-3}m^2}[/tex]

[tex]T=^4\sqrt{2.80*10^{11}}[/tex]

[tex]T=728.9K[/tex]

Answer: [tex]107.8\ J[/tex]

Explanation:

Given

Initial position of object is (4.4 i+5 j)

Final position of object is (11.6 i -2 j)

Force acting (4i-9j)

Work done is given by

[tex]\Rightarrow W=F\cdot dx\\\Rightarrow W=(4i-9j)\cdot (11.6i-4.4i-2j-5j)\\\Rightarrow W=(4i-9j)\cdot (7.2i-7j)\\\Rightarrow W=28.8+63\\\Rightarrow W=91.8\ J[/tex]

Initial kinetic energy

[tex]K_i=\dfrac{1}{2}\times 2\times 4^2\\\\K_i=16\ J[/tex]

Change in kinetic energy is equal to work done by object

[tex]\Rightarrow K_f=K_i+W\\\Rightarrow K_f=16+91.8\\\Rightarrow K_f=107.8\ J[/tex]

Answer:

the tension force of the string on the stone is 30 N

Option d) 30 N is the correct answer.

Explanation:

Given the data in the question;

mass m = 0.2 kg

radius r = 0.6 m

θ = 150 revolutions = 300π rad

time t = 60 seconds

we know that; Angular speed ω = θ / t

we substitute

ω = 300π / 60

ω = 5π rad

Linear speed of stone u = ω × r

we substitute

u = 5π × 0.6

u = 3π m/s

The tension force of the string on the stone is equal to centripetal force, which aid it move in circle;

so

T = mv² / r

we substitute

T = [ 0.2 × (3π)² ] / 0.6

T = 17.7652879 / 0.6

T = 29.6 ≈ 30 N

Therefore, the tension force of the string on the stone is 30 N

Option d) 30 N is the correct answer.

Answer:

According to this model, the atom is a sphere of positive charge, and negatively charged electrons are embedded in it to balance the total positive charge.

Explanation:

Hope this helps you

Answer:

The plum pudding model of the atom states that  had negatively-charged electrons embedded within a positively-charged "soup."

Explanation:

Thomson's plum pudding model of the atom had negatively-charged electrons embedded within a positively-charged "soup." Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively-charged nucleus. Based on these results, Rutherford proposed the nuclear model of the atom.

Answer:

the gain in gravitational potential energy is 0.4369 J

Explanation:

Given the data in the question;

from the definition of density, we know that;

ρ = m / V

where ρ is density, m is the mass and V is the volume.

Now, lets make mass the subject of the formula

m = ρV  ------ let this be equation 1

Now, we know that potential energy PE = mgh ------- let this be equation 2

where m is mass, g is acceleration due gravity, and h is the height.

substitute equation 1 into 2

PE = ρVgh

given that; V =  110 mL = 110 × 10⁻⁶ m³, h = 38.6 cm = 38.6 × 10⁻² m, g = 9.8 m/s², ρ = 1.05 × 10³ kg/m³

we substitute

PE = (1.05 × 10³ kg/m³) × (110 × 10⁻⁶ m³) × 9.8 m/s² × 38.6 × 10⁻² m

PE =  0.4369 J

Therefore,  the gain in gravitational potential energy is 0.4369 J

This is the answer hope it helps

Answer:

θ = 34.77°

Explanation:

From diffraction equation:

[tex]m\lambda = dSin\theta[/tex]

where,

m = order of diffraction

λ = wavelength of light used

d = slit separation

θ = angle

Therefore, for initial case:

m = 2

λ = 600 nm = 6 x 10⁻⁷ m

d = slit seperation = ?

θ = angle 20°

Therefore,

[tex](2)(6\ x\ 10^{-7}\ m)=d(Sin\ 20^o)\\\\d = \frac{12\ x 10^{-7}\ m}{0.342}\\\\d = 3.5\ x\ 10^{-6}\ m[/tex]

Now, for the second case:

m = 5

λ = 600 nm = 6 x 10⁻⁷ m

d = slit seperation = (1.5)(3.5 x 10⁻⁶ m) = 5.26 x 10⁻⁶ m

θ = angle = ?

Therefore,

[tex](5)(6\ x\ 10^{-6}\ m) = (5.26\ x\ 10^{-6}\ m)Sin\theta\\\\Sin\theta = \frac{(5)(6\ x\ 10^{-7}\ m)}{(5.26\ x\ 10^{-6}\ m)}\\\\\theta = Sin^{-1}(0.5703)[/tex]

θ = 34.77°

Answer:

15.44 m/s

Explanation:

Below is the given values:

The bell in rings with a sound = 474 Hz

If the frequency of pigeon hears = 453 Hz

Speed = ?

Use below formula:

Frequency = [(Vs - Vo) / Vs ] x fo

Vs = Speed of sound

Vo = Speed of observer

fo = Sound frequency

Frequency = [(Vs - Vo) / Vs ] x fo

453 = [(343 - Vo) / 343 ] x 474

453 / 474 =  [(343 - Vo) / 343 ]  

0.955 = (343 - Vo) / 343

0.955 x 343 = 343 - Vo

327.56 = 343 - Vo

Vo = 343 - 327.56

Vo = 15.44 m/s

Answer:

so in a given orbital there can be 3 electrons.

Explanation:

The Pauli exclusion principle states that all the quantum numbers of an electron cannot be equal, if the spatial part of the wave function is the same, the spin part of the wave function determines how many electrons fit in each orbital.

In the case of having two values, two electrons change. In the case of three allowed values, one electron fits for each value, so in a given orbital there can be 3 electrons.

(C)

Explanation:

From Ohm's law,

V = IR

Solving for I,

I = V/R

= (230 V)/(25 ohms)

= 9.2 A

Answer:what the choices

Explanation:

Answer:

the magnetic force on the proton is zero.

Explanation:

Given;

speed of the proton, v = 10 m/s

magnitude of the magnetic field, B = 3 T

The magnitude of the magnetic force on the particle is calculated as;

F = qvBsinθ

where;

θ is the angle between the velocity of the particle and the magnetic field

Since the particle is moving parallel to the magnetic field, θ = 0

F = qvBsin(0)

F = 0

Therefore, the magnetic force on the proton is zero.

Answer:

This is a trick question, in that the numbers do not matter.

Study the dependence of the magnetic force on the direction of the magnetic field and the direction of motion of the particle. When is it a maximum? When is it a minimum?

Hint: In vector notation, this is often expressed as q v x B, where q is the electric charge of the particle, v is its velocity (a vector), and B is the magnetic field vector.

Answer:

The net torque is 0.98 Nm.

Explanation:

The torque is given by

Torque = force x perpendicular distance

The clock wise torque is taken as negative while the counter clock wise torque is taken as positive.

Take the torques about the fulcrum.

Torque =  1 x 9.8 x 0.4 - 0.5 x 9.8 x 0.6

Torque = 3.92 - 2.94 = 0.98 Nm

Answer:

Explanation:

ádasdasdasd

potang ina mooooooo bubu hayop kaaaaapestrng yawa

Explanation:

peste kakkkkaaaaaaaa bubu pesteng yawaaaayaka kaayu kaaaaaaa

Answer:

228

Explanation:

Answer:

There are no options, so....

Explanation:

Chlorophyll a absorbs its energy from the Violet-Blue and Reddish orange-Red wavelengths, and little from the intermediate (Green-Yellow-Orange) wavelengths.

Answer: sea-level rise, extreme weather events, and drought, and water scarcity.

Explanation: