At sea level, at a latitude where , a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m. What is the value of g in m/s2 at a location where the length of such a pendulum is 0.970 m

Answer:

93.3x10^-3s

Explanation:

If

Resistance = 1.8 kΩ

Current = 50 mA

Capacitor = 120 μF

Voltage = 140 V

to calculate the discharge current

Applying the formula of discharge current

io=vo/R

io= 140/ 1.8x 10³

= 0.078A

to calculate the time

Applying the formula of current

io= vo/R e-t/RC

50= 140/1800e-t/RC

0.649= e-t/RC

-t/RC= ln( 0.649)

t = 0.432x 120x10^-6x 1800

t=93.3 x 10^-3seconds

We are being given that:

A pictorial view showing the diagrammatic representation of the information given in the question is being attached in the image below.

From the attached image below, the potential difference across the conducting element of the length (dx) moving with the velocity (v) appears to be perpendicular to the magnetic field (B).

The magnitude of the potential difference induced between the center of the blade in relation to either of its ends can be determined by using the derived formula from Faraday's law of induction which can be expressed as:

[tex]\mathsf{E = B\times l\times v}[/tex]

where;

From the diagram, Let consider the length of the conducting element (dx) at a distance of length (x) from the center O.

Then, the velocity (v) = ωx

The potential difference across it can now be expressed as:

[tex]\mathsf{dE = B*(dx)*(\omega x)}[/tex]

For us to determine the potential difference, we need to carry out the integral form from center point O to L/2.

∴

[tex]\mathsf{E = \int ^{L/2}_{0}* B (\omega x ) *(dx)}[/tex]

[tex]\mathsf{E = B (\omega ) \times \Big[ \dfrac{x^2}{2}\Big]^{L/2}_{0}}[/tex]

[tex]\mathsf{E = B (\omega ) * \Big[ \dfrac{L^2}{8}\Big]}[/tex]

Recall that,

[tex]\mathsf{E = (4\times 10^{-3}) * (17) \Big[ \dfrac{(1.5)^2}{8}\Big]}[/tex]

[tex]\mathsf{E = 19.13 mV}[/tex]

Thus, we can now conclude that the magnitude of the potential difference as a result of the rotation caused by the metal blade from the center to either of its ends is 19.13 mV.

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Answer:

the answer to your question us c honey

Answer:

C

Explanation:

This is so because different materials vary in resistance and conductance of current, heat. Metals are good conductors while none metals like rubber, plastic, glass etc are good insulators or resistors.

Answer:

A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

A) Let;

m be total mass of the man

m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung

m_3 be the mass above the water with the empty lung

m_5 be the mass above the water with full lung

F_b be the buoyant force due to the air in the lung

V_a be the volume of air inside man's lungs

w_p be the weight that the buoyant force opposes as a result of the air.

Now, we are given;

m = 78.5 kg

m_3 = 3.2% × 78.5 = 2.512 kg

m_5 = 4.85% × 78.5 = 3.80725 kg

Now, m_p = m_5 - m_3

m_p = 3.80725 - 2.512

m_p = 1.29525 kg

From archimedes principle, we have the formula for buoyant force as;

F_b = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

F_b = w_p = 1.29525 × 9.81

F_b = 12.7064 N

As earlier said,

F_b = (ρ_water × V_air × g)

Thus;

V_air = F_b/(ρ_water × × g)

V_air = 12.7064/(1000 × 9.81)

V_air = 1.295 × 10^(-3) m³

We want to convert to litres;

1 m³ = 1000 L

Thus;

V_air = 1.295 × 10^(-3) × 1000

V_air = 1.295 L

B) From research, the average lung capacity of an adult human being is 6 litres of air.

Thus, the calculated lung volume is not reasonable

Answer:

physical feature of a wave is related to the depth of the wave base is The circular orbital motion

B. The wave base is the depth, and the still water level is the horizontal level

Explanation:

1 inch = 25.4 mm

1 foot = 12 inches

1 mile = 5260 feet

1 cm = 0.01 m or 10 mm

Now Sammy's height is 5 feet and 5.3 inches.

(a) We need to find Sammy's height in inches.

Since, 1 foot = 12 inches

5 feet = 5 × 12 inches = 60 inches

Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches

Sammy's height is 65.3 inches.

(b) We need to find Sammy's height in feet.

Since, 1 foot = 12 inches

[tex]1\ \text{inch}=\dfrac{1}{12}\ \text{feet}[/tex]

So,

[tex]5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}[/tex]

5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet

Sammy's height is 5.44 feet.

Answer:

5.465 × 10^10 revolutions

Explanation:

Formula for Magnetic Field = m. v/ q . r

M = mass of electron = mass of positron = 9.11 x 10^-31 kg,

radius of the positron = 5.03 mm

We convert to meters.

1000mm = 1m

5.03mm = xm

Cross multiply

x = 5.03/1000mm

x = 0.00503m

q = Electric charge = -1.6 x 10^-19 C

Magnetic field (B) = 0.85 T

Speed of the positron is unknown

0.85 = 9.11 x 10^-31 kg × v/ -1.6 x 10^-19 C × 0.00503

0.85 × 1.6 x 10^-19 C × 0.00503 = 9.11 x 10^-31 kg × v

v = 0.85 × -1.6 x 10^-19 C × 0.00503/9.11 x 10^-31 kg

v = 6.8408 ×10-22/ 9.11 x 10^-31 kg

v = 750911086.72m/s

Formula for complete revolutions =

Speed × time / Circumference

Time = 2.30s

Circumference of the circular path = 2πr

r =0.00503

Circumference = 2 × π × 0.00503

= 0.0316044221

Revolution = 750911086.72 × 2.30/0.0316044221

= 1727095499.5/0.0316044221

= 546541562294 revolutions

Approximately = 5.465 × 10^10 revolutions

Answer:

a. 320.06 Nm b. 2.814 rad/s² c. 2.811 rad/s².

Explanation:

a. The torque exerted τ = Frsinθ where F = tangential force exerted = 185 N, r = radius of grindstone = 1.73 m and θ = 90° since the force is tangential to the grindstone.

τ = Frsinθ

= 185 N × 1.73 m × sin90°

= 320.05 Nm

So, the torque τ = 320.05 Nm

b. Since torque τ = Iα where I = moment of inertia of grindstone = 1/2MR² where M = mass of grindstone = 76 kg and R = radius of grindstone = 1.73 m

α = angular acceleration of grindstone

τ = Iα

α = τ/I = τ/(MR²/2) = 2τ/MR²

substituting the values of the variables, we have

α = 2τ/MR²

= 2 × 320.05 Nm/[76 kg × (1.73 m)²]

= 640.1 Nm/227.4604 kgm²

= 2.814 rad/s²

So, the angular acceleration α = 2.814 rad/s²

c. The opposing frictional force produces a torque τ' = F'r' where F' = frictional force = 20.0 N and r' = distance of frictional force from axis = 1.50 cm = 0.015 m.

So  τ' = F'r' = 20.0 N × 0.015 m = 0.3 Nm

The net torque on the grindstone is thus τ'' = τ - τ' = 320.05 Nm - 0.3 Nm = 319.75 Nm

Since τ'' = Iα

α' = τ''/I where α' = its new angular acceleration

α' = 2τ/MR²

= 2 × 319.75 Nm/[76 kg × (1.73 m)²]

= 639.5 Nm/227.4604 kgm²

= 2.811 rad/s²

So, the angular acceleration α' = 2.811 rad/s²

Answer:

Explanation:

To convert from °C to °F , the formula is

( F-32 ) / 9 = C / 5

F is reading fahrenheit scale and C is in centigrade scale .

F = 205 , C = ?

(205 - 32) / 9 = C / 5

C = 96°C approx .

Let us calculate the heat required .

Total heat required = heat required to heat up the ice at - 20 °C  to 0°C  + heat required to melt the ice + heat required to heat up the water at  0°C to

96°C.

=  5 x 2.04 x (20-0) +  5 x 336 + 5 x ( 96-0 ) x 4.2  kJ .

= 204 + 1680 + 2016

= 3900 kJ .

[tex]\boxed{\sf 1°C=273K}[/tex]

Sol:-1

[tex]\\ \sf\longmapsto 30°C[/tex]

[tex]\\ \sf\longmapsto 273+30[/tex]

[tex]\\ \sf\longmapsto 303K[/tex]

Sol:-2

[tex]\\ \sf\longmapsto 50°C[/tex]

[tex]\\ \sf\longmapsto 50+273[/tex]

[tex]\\ \sf\longmapsto 323K[/tex]

Answer:

A solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix. The coil can be arranged to produce a uniform magnetic field in a volume of space when an electric current is passed through it.

The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.

Centripetal force is described as the force applied to a body that is travelling in a circular motion and is pointed in the direction towards the center of the circular path.

Here,

Mass of the car, m = 0.5 kg

Velocity of the car, v = 6 m/s

Radial friction between the car and the track, f = 7 N

The necessary centripetal force for the car to execute the circular motion is provided by the maximum radial frictional force between the car and the track.

So, the condition that the car cannot continue to travel in the circular path is that the centripetal force required is greater than the maximum radial friction.

So,

mv²/r > f

0.5 x 6²/r > 7

Therefore, the radius of the circular track,

r < 18/7

r < 2.57 m

Hence,

The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.

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Answer:

A) 60 Hz

B) 0.04 T

Explanation:

Given that.

Number of turns, N = 200

Length of the side, l = 20 cm = 0.2 m

Speed if rotation, w = 3600 rpm

Voltage, V = 120 V

First, we try to convert the speed from rpm to rad/s

3600 * (2π/60)

3600 * 0.10473

3600 rpm = 377 rad/s

Now, we use that as our w, speed of rotation

Frequency of output, f =

w/2π

f = 377 / 6.284

f = 59.99 Hz or approximately, 60 Hz.

B

Strength of the magnetic field in which the coil is turning

E• = NABw

Where, A = l² = 0.2² = 0.04, on substituting the values to the equation, we have

120 = 200 * 0.04 * 377 * B

120

Making B subject of formula,

B = 120/ 3016

B = 0.04 T..

The frequency of the output voltage is 60 Hz and the strength of the magnetic field is 0.04 T

Answer:

No

Explanation:

Moon has no light of its own. It just shines because its surface reflects light from the sun and that's what we see.

:-)

Answer:

The wavelength is 2.16 m.

Explanation:

Given the speed of the sound = 343 m/s

Trombone generate the frequency = 159 Hz

Now we have to find the wavelength of the sound. Here, we can find the wavelength by dividing the speed of the sound with frequency.

The wavelength of the sound = Speed of sound/frequency

Wavelength of the sound = 343 / 159 = 2.16 m

Answer:

The average density of the rod is 1.605 kg/m.

Explanation:

The average density of the rod is given by:

[tex] \rho = \frac{m}{l} [/tex]    

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:

[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex]   (1)

Using u = x+1  →  du = dx  → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]

[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]

Therefore, the average density of the rod is 1.605 kg/m.  

I hope it helps you!    

The average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

Given data:

The length of rod is, L = 3 m.

The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3,  The expression for the average density is given as,

[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)

Using u = x+1  

du = dx

u₁= x₁+1 = 0+1 = 1

and

u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]

Thus, we can conclude that the average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

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Answer:

The crowning of Charlemagne by Pope Leo III was significant in a number of ways. For Charlemagne, it was necessary because it encouraged to give him higher reliability. It gave him the rank of a dictator, giving him the only ruler in Europe west of the Byzantine emperor in Constantinople.

Answer:

θ = 16.2 rad

Explanation:

First we find the angular acceleration by using first equation of motion in angular form:

ωf = ωi + αt

where,

ωf =final angular speed = (110 rev/min)(2π rad/1 rev)(1 min/60 s) = 11.5 rad/s

ωi =initial angular speed = (45 rev/min)(2π rad/1 rev)(1 min/60 s) = 4.7 rad/s

α = angular acceleration = ?

t = time = 2 s

Therefore,

11.5 rad/s = 4.7 rad/s + α(2 s)

α = (6.8 rad/s)/(2 s)

α = 3.4 rad/s²

Now, we use 2nd equation of motion:

θ = ωi t + (1/2)αt²

where,

θ = rotation = ?

Therefore,

θ = (4.7 rad/s)(2 s) + (1/2)(3.4 rad/s²)(2 s)²

θ = 9.4 rad + 6.8 rad

θ = 16.2 rad

Explanation:

It took [tex]t_1 =1.5\:\text{s}[/tex] for the sound to reach the 1st wall and at the same time time, the same sound took [tex]t_2 = 2.5\:\text{s}[/tex] to reach the 2nd wall. Assuming that the sound travels at 343 m/s, then let [tex]x_1[/tex] be the distance of the person to the 1st wall and [tex]x_2[/tex] be the distance to the 2nd wall. So the distance between the walls X is

[tex]X = x_1 + x_2 = v_st_1 + v_st_2 = v_s(t_1 + t_2)[/tex]

[tex]\:\:\:\:\:= (343\:\text{m/s})(4.0\:\text{s}) = 1372\:\text{m}[/tex]

Answer:

 I = 1.23 A

Explanation:

In an RL circuit current passing is described by

           I = E / R (1 - [tex]e^{-Rt/L}[/tex])

Let's reduce the magnitudes to the SI system

        L = 35 mH = 35 10⁻³ H

        t = 5.0 ms = 5.0 10⁻³ s

let's calculate

         I = 18/12 (1 - [tex]e^{-12 .. 5 {10}^{-3}/35 .. {10}^{-3} }[/tex]e (- 5 10-3 12/35 10-3))

         I = 1.5 (1- [tex]e^{-1.715}[/tex])

         I = 1.23 A

Answer:

Explanation:

The lens equation is expressed as;

1/f = 1/u+1/v where;

f is the focal length of the lens

u is the object distance

v is the image distance

Since the near sighted person wants focus the starts at nigt, the stars at night are the images located that infinity. Hence the image distance v = ∞.

The object distance u = 130cm

Substituting the given parameters in the formula to get the focal length f

[tex]\frac{1}{f} = \frac{1}{\infty} + \frac{1}{130} \\\\As \ x \ tends \ to \ \infty, \, \frac{a}{x} \ tends \ to \ 0 \ where\ 'a' \ is \ a\ constant \\\\} \\\\[/tex]

[tex]\frac{1}{f} = 0+ \frac{1}{130}\\\\[/tex]

[tex]\frac{1}{f} =\frac{1}{130}\\cross\ multiply\\\\f = 130*1\\\\f = 130cm[/tex]

Hence the focal length of contact lenses that would allow a nearsighted person with a 130 cm far point to focus on the stars at night is 130cm

Answer:

v = 0.89 c = 2.67 x 10⁸ m/s

Explanation:

The time dilation consequence of the special theory of relativity shall be used here, From time dilation formula we have:

t = t₀/√[1 - v²/c²]

where,

t = time measured by the person on earth = 2.50 s

t₀ = rest time of the intergalactic rock star = 1.30 s

v = relative speed of the rock star = ?

Therefore,

2.5 s = (1.3 s)/√[1 - v²/c²]

√[1 - v²/c²] = 1.3/2.5

√[1 - v²/c²] = 0.52

[1 - v²/c²] = 0.52²

[1 - v²/c²] = 0.2074

v²/c² = 1 - 0.2074

v²/c² = 0.7926

v/c = √0.7926

v = 0.89 c

where,

c = speed of light = 3 x 10⁸ m/s

v = (0.89)(3 x 10⁸ m/s)

v = 0.89 c = 2.67 x 10⁸ m/s

Answer:

The number of turns in the secondary coil is 4145 turns

Explanation:

Given;

the induced emf on the primary coil, [tex]E_p[/tex] = 95 V

the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V

the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns

the number of turns in the secondary coil, [tex]N_s[/tex] = ?

The number of turns in the secondary coil is calculated as;

[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]

[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]

Therefore, the number of turns in the secondary coil is 4145 turns.

Answer:

Pressure, P = 1250 Pa

Explanation:

Given that,

Weight of a brick, F = 50 N

Dimension of the brick is 30.0 cm × 10.0 cm × 4.00 cm

We need to find the maximum pressure it can exert on a horizontal surface due to its weight. Pressure is equal to the force acting per unit area. Pressure exerted is inversely proportional to the area of cross section. So, we need to minimize area. Taking to smaller dimensions.

A = 40 cm × 10 cm = 400 cm² = 0.04 m²

So,

Pressure,

[tex]P=\dfrac{50\ N}{0.04\ m^2}\\\\P=1250\ Pa[/tex]

So, the maximum pressure of 1250 Pa it can exert on a horizontal surface.

The maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure. It is denoted by P.

The given data in the problem is;

W is the weight of a brick = 50 N

The dimension of the brick = 30.0 cm × 10.0 cm × 4.00 cm

A is the area,

The area is found as;

A=40 cm × 10 cm = 400 cm² = 0.04 m²

The pressure is the ratio of the force and area

[tex]\rm P = \frac{F}{A} \\\\ \rm P = \frac{50}{0.04} \\\\ \rm P =1250 \ Pascal[/tex]

Hence the maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

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Answer:

Option (A) : 1m

Explanation:

According to Hooke's law:

F (spring elastic force) =

k( spring const.) * x(displacement)

Case-1

2 N = k * 0.4m

k = 5

Case- 2

5 N = 5 * x

x ( displacement) = 1 m

The displacement of the spring if a 5-N force was applied is equal to 1m. Therefore, option (1) is correct.

The strain and stress are proportional to each other, and this is called Hooke’s Law. Hooke’s law states that the strain is proportional to the stress applied within the elastic limit of the material.

When the materials are stretched, the atoms or molecules deform and when the stress is removed, they will return to their original state.

The mathematical equation for Hooke's law is as follows:

F = –kx

where F is the force, x is displacement, and k is the spring constant in N/m.

Given, F = 2N and x = 0.4m

F = -kx

2 N = - k (0.4m)

k = 5 N/m where the negative sign is omitted.

Now, the spring constant of the spring, k = 5 N/m and F = 5N

F = -kx

5 N = - (5 N/m)(x)

x = - 1m

Therefore, the displacement of the spring is 1 m.

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Answer:

366 N

Explanation:

  τ = Iα

FR = ½mR²α

  F = ½mR(Δω/t)

  F = ½(207)(1.50)(0.75)(2π) /2.00

  F = 365.79919...

Answer:

3. the intermolecular bonding of water

Explanation:

Anomalous behavior of water is an advantage in aquatic habitat during winter. Because of some unique properties of water, it behaves irregularly. Thus, a pond or river does not freeze completely during winter.

Water has its highest density when temperature is 4[tex]^{0}C[/tex] , and lowest volume at 4[tex]^{0}C[/tex]. Thus, the denser layers of water sink accordingly until the upper layer is the least dense during winter. This layer then freeze leaving the layers below it unfrozen.

Answer:

D. The tendency for water to freeze from the bottom up.

Explanation:

Answer:

The water takes longer, because it is the better insulator here.

Explanation:

Conductors and insulators work similarly in "reverse".

If something is a good heat conductor, then it's good at both absorbing heat energy and giving it away. Insulators are good at resisting temperature changes, but also take longer to cool down once they are heated up.

So because copper is the better conductor here, it will cool faster than the water at the same temperature.

Answer:

the answer is nearly 5.655 [tex]ms^{-2}[/tex]

Explanation:

Given,

[tex]v_{x}=2.7 ms^{-1}[/tex]

[tex]v_{y}=13.3 ms^{-1}[/tex]

[tex]t=2.4 s[/tex]

[tex]a_{x}=\frac{2.7}{2.4}=1.125 ms^{-2}[/tex] (as  [tex]a=\frac{v-u}{t}[/tex])

[tex]a_{y}=\frac{13.3}{2.4}=5.542 ms^{-2}[/tex]

[tex]a=\sqrt{a_{x}^{2}+a_{y}^{2} }[/tex]

[tex]=\sqrt{1.125^{2}+5.542^{2} }[/tex]

[tex]=5.655 ms^{-2}[/tex]

hope you have understood this...

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Answer:

Explanation:

Given data:

power rating of the heater P= 1010 W

voltage of the heater V= 115 volts

current taken by the heater I= ?

We can apply the power formula to solve for the current in the heater

i.e P= IV

Making I the current subject of formula we have

I= P/V

Substituting our given data into the expression for I we have

I=1010/115= 8.78 A