what is the value of the tropic of cancer​

Answer:

The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.

Explanation:

The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.

The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.      

Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.                                  

I hope it helps you!

Answer:

C is the correct answer

Explanation:

Answer:

3.85 percent

Explanation:

From the question,

Percentage error = (error/actual)×100................ Equation 1

Given: actual center of mass = 15 cm, error = 15.6-15 = 0.6 cm

Substitute these values into equation 1

Percentage error = (0.6/15.6)×100

Percentage error = 3.85 percent

Hence the percentage error of the uniform mass = 3.85 percent

Answer:

dr = 0.71 ft/min

the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.

Explanation:

Given;

Rate of inflation dV = 140pift^3/min

Radius r = 7 ft

Change in radius = dr

Volume of a spherical balloon is;

V = (4/3)πr^3

The change in volume can be derived by differentiating both sides;

dV = (4πr^2)dr

Making dr the subject of formula;

dr = dV/(4πr^2)

Substituting the given values;

dr = 140π/(4π×7^2)

dr = 0.714285714285 ft/min

dr = 0.71 ft/min

the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.

Answer:

198.9 x 10^-16

Explanation:

E = hc/ wavelength

E =(6.63 x 10^-34 x 3 x 10^8)/(0.01 x 10^-9)

E = 198.9 x 10^-16

Answer:

a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.

Answer:

(a) 62.3 N

(b) 1.89 N

(c) 0.430 kg m²

Explanation:

(a) Find the acceleration of block B.

Δy = v₀ t + ½ at²

1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²

a = 0.90 m/s²

Draw a free body diagram of block B.  There are two forces:

Weight force mg pulling down,

and tension force Tb pulling up.

Sum of forces in the -y direction:

∑F = ma

mg − Tb = ma

Tb = m (g − a)

Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)

Tb = 62.3 N

(b) Draw a free body diagram of block A.  There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and tension force Ta pulling right.

Sum of forces in the +x direction:

∑F = ma

Ta = ma

Ta = (2.10 kg) (0.90 m/s²)

Ta = 1.89 N

(c) Draw a free body diagram of the pulley.  There are two forces:

Tension force Tb pulling down,

and tension force Ta pulling left.

Sum of torques in the clockwise direction:

∑τ = Iα

Tb r − Ta r = Iα

(Tb − Ta) r = I (a/r)

I = (Tb − Ta) r² / a

I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)

I = 0.430 kg m²

The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is [tex]\rm 0.430 \; kg\;m^2[/tex].

Given :

a) First, determine the acceleration of the B block.

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

[tex]\rm 1.8 = \dfrac{1}{2}\times a\times (2)^2[/tex]

[tex]\rm a = 0.9\; m/sec^2[/tex]

Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.

[tex]\rm \sum F=ma[/tex]

[tex]\rm mg-T_b=ma[/tex]

[tex]\rm T_b = m(g-a)[/tex]

[tex]\rm T_b = 7\times (9.8-0.9)[/tex]

[tex]\rm T_b = 62.3\;N[/tex]

b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.

[tex]\rm \sum F=ma[/tex]

[tex]\rm T_a=ma[/tex]

[tex]\rm T_a = 2.1\times 0.9[/tex]

[tex]\rm T_a = 1.89\;N[/tex]

c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.

[tex]\rm \sum \tau = I\alpha[/tex]

[tex]\rm T_br-T_ar = I\alpha[/tex]

[tex]\rm I = \dfrac{(T_b-T_a)r^2}{a}[/tex]

Now, substitute the values of the known terms in the above expression.

[tex]\rm I = \dfrac{(62.3-1.89)(0.080)^2}{0.90}[/tex]

[tex]\rm I = 0.430 \; kg\;m^2[/tex]

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Answer:

5. The amount of work is the same for both.

Explanation:

Work done is a measure of change in kinetic energy of each egg

For both egg , the initial speed and mass are same , so they have equal initial Kinetic energy

For both egg , the final speed is 0 and mass are same , so they have equal final Kinetic energy which is 0.

So work done is same for both eggs since they have same change in kinetic energy.

Answer:

The force is  [tex]F = 8*10^{-12} \ N[/tex]

Explanation:

From the question we are told that

     The rate at which ATP molecules are used is [tex]R = 80 ATP/ s[/tex]

       The energy provided by a single ATP is  [tex]E_{ATP} = 0.8 * 10^{-19} J[/tex]

       The velocity of the kinesin is  [tex]v = 800 nm/s = 800*10^{-9} m/s[/tex]

The power provided by the ATP in one second is  mathematically represented as

       [tex]P = E_{ATP} * R[/tex]

substituting values

       [tex]P = 80 * 0.8*10^{-19 }[/tex]

       [tex]P = 6.4 *10^{-18}J/s[/tex]

Now  this power is mathematically represented as

       [tex]P = F * v[/tex]

Where  F  is  the force the kinesin is exerting

  Thus  

          [tex]F = \frac{P}{v}[/tex]

substituting values

            [tex]F = \frac{6.4*0^{-18}}{800 *10^{-9}}[/tex]

           [tex]F = 8*10^{-12} \ N[/tex]

The force exerted by the kinesin  is 8  × 10-12 N.

Let us recall that power is defined as the rate of doing work. Hence, power = Energy/Time.

Since;

Energy  =  0.8 × 10-19 J/molecule

Number ATP molecules transported per second = 80 ATP molecules/s

Power =  0.8 × 10-19 J/molecule × 80 ATP molecules/s

Power = 6.4  × 10-18 J

Again, we know that;

Power = Force × Velocity

Velocity of the ATP molecules =  800 nm/s or 8 × 10-7 m/s

Force = Power/velocity

Force =  6.4  × 10-18 J/ 8 × 10-7 m/s

Force = 8  × 10-12 N

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Answer:

C You should deflect the ball back toward your friend.

Explanation:

This is because it would result in a completely inelastic collision, and the final velocity of me would be found using,

with m= mass, V=velocity, i=initial, f=final:

mV(me,i) +mV(ball,i) = [m(me)+m(b)]V(f)

So V(f) would be just the momentum of the ball divided by just MV mass of the ball and it will be higher resulting in inelastic collision

Answer:

A. You should catch the ball.

Explanation:

Catching the ball maximizes your speed by converting most of the momentum of the flying ball into the momentum of you and the ball. Since the ice is smooth, the friction between your feet and the ice is almost negligible, meaning less energy is needed to set your body in motion. Catching the ball means that you and the ball undergoes an inelastic collision, and part of the kinetic energy of the ball is transferred to you, setting you in motion. Deflecting the ball will only give you a relatively small speed compared to catching the ball.

Answer:

The object has an excess of [tex]10^{13}[/tex] electrons.

Explanation:

When an object has a negative charge he has an excess of electrons in its body. We can calculate the number of excessive electrons by dividing the charge of the body by the charge of one electron. This is done below:

[tex]n = \frac{\text{object charge}}{\text{electron charge}}\\n = \frac{-1.6*10^{-6}}{-1.6*10^{-19}} = 1*10^{-6 + 19} = 10^{13}[/tex]

The object has an excess of [tex]10^{13}[/tex] electrons.

Answer: (a) α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]

(b) For r≤R: B(r) = μ_0.[tex](\frac{I.r^{2}}{2.\pi.R^{3}})[/tex]

For r≥R: B(r) = μ_0.[tex](\frac{I}{2.\pi.r})[/tex]

Explanation:

(a) The current I enclosed in a straight wire with current density not constant is calculated by:

[tex]I_{c} = \int {J} \, dA[/tex]

where:

dA is the cross section.

In this case, a circular cross section of radius R, so it translates as:

[tex]I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr[/tex]

[tex]I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr[/tex]

[tex]I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}[/tex]

[tex]\alpha = \frac{3I}{2.\pi.R^{3}}[/tex]

For these circunstances, α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]

(b) Ampere's Law to calculate magnetic field B is given by:

[tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]

(i) First, first find [tex]I_{c}[/tex] for r ≤ R:

[tex]I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr[/tex]

[tex]I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr[/tex]

[tex]I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr[/tex]

[tex]I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}[/tex]

[tex]I_{c} = \frac{I.r^{3}}{R^{3}}[/tex]

Calculating B(r), using Ampere's Law:

[tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]

[tex]B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} )[/tex].μ_0

B(r) = [tex](\frac{Ir^{3}}{R^{3}2.\pi.r})[/tex].μ_0

B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0

For r ≤ R, magnetic field is B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0

(ii) For r ≥ R:

[tex]I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr[/tex]

So, as calculated before:

[tex]I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}[/tex]

[tex]I_{c} =[/tex] I

Using Ampere:

B.2.π.r = μ_0.I

B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0

For r ≥ R, magnetic field is; B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0.

A number, which represents a property, amount, or relationship that does not change under certain situations is constant and further calculations as follows:

The Radius of the cross-section of the wire R

Current passing through the wire I

Current Density [tex]J = \alpha r[/tex]

Constant [tex]\alpha[/tex]

Distance of the point from the center [tex]r[/tex]

For part a)

Consider a circular strip between two concentric circles of radii r and r+dr.

Current passing through the strip [tex]dI =\overrightarrow J \times \overrightarrow{dA}[/tex]

[tex]\to\alpha r (2\pi r dr) cos 0^{\circ}[/tex]

Integration

[tex]\to I =2\pi \alpha \int^R_0 r^2\ dr =2\pi \alpha [r^3]^R_0=2\pi \alpha \frac{r^3}{3}\\\\\to \alpha = \frac{3I}{2\pi R^3}\\\\[/tex]

For part b)

The magnetic field at a point distance [tex]'r'^{(r \ \pounds \ R)}[/tex] from the center is B.

We have the value of the line integral of the magnetic field over a circle of radius ‘r’ given as

[tex]\oint \overrightarrow B \times \overrightarrow{dl} = \mu_0 I\\\\[/tex]

where ‘I’ is the threading current through the circle of radius ‘r’

[tex]\oint B \ dl \cos 0^{\circ} = \mu_0 [2\pi \alpha \frac{r^3}{3}]\\\\ B \int dl = \mu_0 [2\pi \frac{3I}{2\pi R^3} \frac{r^3}{3}]\\\\ B \cdot 2\pi r = \mu_0 I [\frac{r}{R}]^3\\\\ B = \frac{\mu_0}{2\pi} I [\frac{Ir^2}{R^3}]\\\\[/tex]

(ii) Similarly, we can calculate the magnetic field at the point at A distance ‘r’ where

[tex]\to r^3 R\\\\\to \int \overrightarrow{B} \overrightarrow{dl} = \mu_0\ I[/tex] [The threading current is the same]

[tex]\to \beta - 2\pi r = \mu_0 I[/tex] As (I)

[tex]\to \beta =\frac{\mu_o \ I}{2\pi \ r}[/tex]

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Answer:

T₂ = 95.56°C

Explanation:

The final resistance of a material after being heated is given by the relation:

R' = R(1 + αΔT)

where,

R' = Final Resistance = 207.4 Ω

R = Initial Resistance = 154.9 Ω

α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹

ΔT = Change in Temperature = ?

Therefore,

207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]

207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT

1.34 - 1 = (0.0045°C⁻¹)ΔT

ΔT = 0.34/0.0045°C⁻¹

ΔT = 75.56°C

but,

ΔT = Final Temperature - Initial Temperature

ΔT = T₂ - T₁ = T₂ - 20°C

T₂ - 20°C = 75.56°C

T₂ = 75.56°C + 20°C

T₂ = 95.56°C

Answer:

[tex]a_{r} = 1006.382g \,\frac{m}{s^{2}}[/tex]

Explanation:

Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

[tex]a_{r} = \omega^{2}\cdot R[/tex]

Where:

[tex]\omega[/tex] - Angular speed, measured in radians per second.

[tex]R[/tex] - Radius of rotation, measured in meters.

The angular speed is first determined:

[tex]\omega = \frac{\pi}{30}\cdot \dot n[/tex]

Where [tex]\dot n[/tex] is the angular speed, measured in revolutions per minute.

If [tex]\dot n = 3000\,rpm[/tex], the angular speed measured in radians per second is:

[tex]\omega = \frac{\pi}{30}\cdot (3000\,rpm)[/tex]

[tex]\omega \approx 314.159\,\frac{rad}{s}[/tex]

Now, if [tex]\omega = 314.159\,\frac{rad}{s}[/tex] and [tex]R = 0.1\,m[/tex], the resultant acceleration is then:

[tex]a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)[/tex]

[tex]a_{r} = 9869.588\,\frac{m}{s^{2}}[/tex]

If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

[tex]a_{r} = 1006.382g \,\frac{m}{s^{2}}[/tex]

Answer:  If you are traveling at a speed of 60mph, you will go 220 feet.

Explanation: 60mph is a mile a minute. 5280 feet in a mile, 60 seconds in a minute. Divide to find that is 88 feet per second. Multiply by the number of seconds.

Answer:

D

Explanation:

The power equation is P= V^2/R

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As a result of the given scenario, power delivered from the battery to combination decreases. The correct option is D.

A resistor is a two-terminal passive electrical component that uses electrical resistance as a circuit element.

Resistors are used in electronic circuits to reduce current flow, adjust signal levels, divide voltages, and bias active elements.

A resistor is a component of an electronic circuit that limits or regulates the flow of electrical current. Resistors can also be used to supply a fixed voltage to an active device such as a transistor.

The current through resistors is the same when they are connected in series. The battery voltage is divided among resistors.

Adding more resistors to a series circuit increases total resistance and thus lowers current. However, in a parallel circuit, adding more resistors in parallel creates more options while decreasing total resistance.

Thus, the correct option is D.

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Answer:

V = 0.45 Volts

Explanation:

First we need to find the total current passing through the wire. That can be given by:

Total Current = I = (Current Density)(Surface Area of Wire)

I = (Current Density)(2πrL)

where,

r = radius = 1.5/2 mm = 0.75 mm = 0.75 x 10⁻³ m

L = Length of Wire = 6.5 m

Therefore,

I = (4.07 x 10⁻³ A/m²)[2π(0.75 x 10⁻³ m)(6.5 m)]

I = 1.25 x 10⁻⁴ A

Now, we need to find resistance of wire:

R = ρL/A

where,

ρ = resistivity of iron = 9.71 x 10⁻⁸ Ωm

A = Cross-sectional Area = πr² = π(0.75 x 10⁻³ m)² = 1.77 x 10⁻⁶ m²

Therefore,

R = (9.71 x 10⁻⁸ Ωm)(6.5 m)/(1.77 x 10⁻⁶ m²)

R = 0.36 Ω

From Ohm's Law:

Voltage = V = IR

V = (1.25 x 10⁻⁴ A)(0.36 Ω)

V = 0.45 Volts

Answer:

- E = 11.55J

- Q = 0.17C

- E' = (1/4)E

Explanation:

- To calculate the amount of energy stored in the capacitor, you use the following formula:

[tex]E=\frac{1}{2}CV^2[/tex]

C: capacitance = 3800.0*10^-6F

V: potential difference = 78.0V

[tex]E=\frac{1}{2}(3800.0*10^{-6}C)(78.0V)^2=11.55J[/tex]

The energy stored in the capacitor is 11.55J

- If the electrical energy stored in the capacitor is 6.84J, the charge on the capacitor is:

[tex]E=\frac{1}{2}QV\\\\Q=\frac{2E}{V}\\\\Q=\frac{2(6.84J)}{78.0V}=0.17C[/tex]

The charge on the capacitor is 0.17C

- If you take the capacitor as a parallel plate capacitor, you have that the energy stored on the capacitor is:

[tex]E=\frac{1}{2}CV^2=\frac{1}{2}(\frac{\epsilon_oA}{d})V^2=\frac{1}{2}\frac{\epsilon_oAV^2}{d}\\\\[/tex]

A: area of the plates

d: distance between plates

If the distance between plates is increased by a factor of 4, you have:

[tex]E'=\frac{1}{2}\frac{\epsilon_oAV^2}{(4d)}=\frac{1}{4}\frac{\epsilon_oAV^2}{2d}=\frac{1}{4}E[/tex]

Then, the stored energy in the capacitor is decreased by a a factor of (1/4)

Answer:

Explanation:

Beats are produced when two sound waves from two  different sources having slightly different frequencies  interfere . Due to both constructive and destructive interference , high and low intensity sound is heard at some regular interval . This is called rate of beat formation . Amplitudes of sound may be same or different . High intensity sound heard per unit time is called beat . It is equal to difference of frequencies of the sound waves interfering

each other .

Answer:

F' = F/16

So, the force of gravity has become 16 times less than initial value.

Explanation:

The force of gravity between two planets, is given by the following formula:

F = Gm₁m₂/r²   ----------- equation 1

where,

F = Force of gravity between two planets

G = Gravitational Constant

m₁ = Mass of one planet

m₂ = Mass of other plant

r = Distance between two planets

Now, if the distance between the planets (r) is 10 times smaller, then Force of gravity will become:

F' = Gm₁m₂/(4r)²

F' = (1/16) (Gm₁m₂/r²)

using equation 1:

F' = F/16

So, the force of gravity has become 16 times less than initial value.

Answer:

L = 0.60 m

The length in metres should be 0.60 m

Explanation:

A pipe open at both ends can have a standing wave pattern with resonant frequency;

f = nv/2L ........1

Where;

v = velocity of sound

L = length of pipe

n = 1 for the fundamental frequency f1

Given;

Fundamental frequency f1 = 294 Hz

Velocity v = 350 m/s

n = 1

From equation 1;

Making L the subject of formula;

L = nv/2f1

Substituting the given values;

L = 1×350/(2×294)

L = 0.595238095238 m

L = 0.60 m

The length in metres should be 0.60 m

Answer:

q = 2.997*10^-4C

Explanation:

In order to find the required charge that the penny have to have, to acquire an upward acceleration, you take into account that the electric force on the penny must be higher than the weight of the penny.

You use the second Newton law to sum both electrical and gravitational forces:

[tex]F_e-W=ma\\\\qE-mg=ma[/tex]             (1)

Fe: electric force

W: weight of the penny

q: required charge = ?

m: mass of the penny = 3g = 0.003kg

E: magnitude of the electric field = 100N/C

g: gravitational acceleration = 9.8m/s^2

a: acceleration of the penny = 0.19m/s^2

You solve the equation (1) for q, and replace the values of the other parameters:

[tex]q=\frac{ma+mg}{E}=\frac{m(a+g)}{E}\\\\q=\frac{(0.003kg)(0.19m/s^2+9.8m/s^2)}{100N/C}\\\\q=2.997*10^{-4}C[/tex]

It is necessary that the penny has a charge of 2.997*10^-4 C, in order to acquire an upward acceleration of 0.19m/s^2

Answer:

The angular displacement is  [tex]\theta = 28.33 \ rad[/tex]

Explanation:

From the question we are told that

     The speed of the driver is  [tex]v =13 \ m/ s[/tex]

     The acceleration of the driver is  [tex]a = 1.02 \ m/s^2[/tex]

      The time taken is [tex]t = 0.70 \ s[/tex]

      The radius of the tire is  [tex]r = 33 cm = 0.33 \ m[/tex]

The distance covered by the car during this  acceleration can be  calculated using the equation of motion as follows

        [tex]s = v*t +\frac{1}{2} * a * t^2[/tex]

Now substituting values  

       [tex]s = 13 * 0.70 +\frac{1}{2} * 1.02 * (0.700)^2[/tex]

      [tex]s = 9.35 \ m[/tex]

Now the angular displacement of the car with respect to the tire movement can be  represented mathematically as

      [tex]\theta = \frac{s}{r}[/tex]

substituting values

      [tex]\theta = \frac{9.35}{0.33}[/tex]

      [tex]\theta = 28.33 \ rad[/tex]

Answer:

q/6Eo

Explanation:

See attached file pls

Answer:

94.248 g/sec

Explanation:

For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:

[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]

And, the velocity of blood pumping is 30 cm^2

Now apply the following formula to solve the total current

[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]

Q =  94.248 g/sec

Basically we applied the above formula So, that the total current could come

Answer:

T = 15,576 N

Explanation:

The speed of a wave on a string is given by

        v = √ T /ρ rho

also the speed of the wave is given by the relationship

       v = λ f

we substitute

     λ f = √ T /ρ

T = (lam f)² ρ

let's find the wavelength in a string, fixed at the ends, the relation that gives the wavelength is

       L= λ/2 n

       λ= 2L / n

we substitute

      T = (2L / n f)²ρ rho

let's calculate

      T = (2 1.20 / 2 590) 0.022

      T = 15,576 N

Answer:

3g/(8π²) ≈ 0.372 m

Explanation:

Draw a free body diagram.  There is a weight force at the center of the pendulum.

Sum of the torques about the pivot:

∑τ = Iα

mg (½ L sin θ) = (⅓ mL²) α

3g sin θ = 2L α

α = 3g/(2L) sin θ

For small θ, sin θ ≈ θ.

α = 3g/(2L) θ

θ" = 3g/(2L) θ

The solution of this differential equation is:

θ = θ₀ cos(√(3g/(2L)) t)

So the period is:

T = 2π / √(3g/(2L))

If the period is 1 second:

1 = 2π / √(3g/(2L))

√(3g/(2L)) = 2π

3g/(2L) = 4π²

L = 3g/(8π²)

L ≈ 0.372 m

The length of the pendulum rod is 0.37 m.

The time period of a pendulum is defined as the time taken by the pendulum to complete one oscillation.

Here,

The mass of the pendulum, m = 2 kg

Time period of the pendulum, T = 1 s

Since, the pendulum is suspended from the pivot and is oscillating, at the position of the pendulum when it makes an angle θ with the pivot, there is a force of weight acting at the center of the pendulum.

The length of pendulum at that point = L/2

The perpendicular distance at that point, r = (L/2) sinθ

Therefore, the torque acting on the pendulum at that point,

τ = Iα

where I is the moment of inertia of the pendulum and α is the angular acceleration.

mg (L/2 sinθ) = (mL²/3)α

1/2 gsinθ = 1/3 Lα

Therefore,

α = (3g/2L) sinθ

For smaller values of θ, we can take sinθ = θ

So, α = (3g/2L) θ

We know that α = θ''

where θ is the angular displacement.

Therefore,

θ'' = (3g/2L) θ

So, ω = √3g/2L

Therefore, the equation of motion of the pendulum can be written as,

θ = θ₀ cos(ωt)

θ = θ₀ cos [(√3g/2L) t]

So, time period of the pendulum,

T = 2[tex]\pi[/tex]/ω

T = 2[tex]\pi[/tex]/(√3g/2L)

2[tex]\pi[/tex]/(√3g/2L) = 1

(√3g/2L) = 2[tex]\pi[/tex]

Therefore, length of the rod,

L = 3g/8[tex]\pi[/tex]²

L = 0.37 m

Hence,

The length of the pendulum rod is 0.37 m.

To learn more about time period, click:

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Attaching the image file here.

Answer:

6.44 s

Explanation:

Given:

v₀ = 119 m/s

v = 233 m/s

a = 17.7 m/s²

Find: t

v = at + v₀

(233 m/s) = (17.7 m/s²) t + (119 m/s)

t = 6.44 s

Answer:

is this a company name.? java is a computer software right..

Answer:

Towards the west.

Explanation:

The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.

Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.

According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward