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Answer:

[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].

The unit of both sides of this equation are [tex]\rm s[/tex].

Explanation:

The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].

The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].

On the right-hand side of this equation:

[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].

Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].

Explanation:

Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]

The sum of the two vectors is

[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]

[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]

The difference between the two vectors can be written as

[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]

[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]

(A) The pressure will be greater than 10% overpressure limit.

(B) The final pressure will be "582.915 kPa".

Given:

Volume,

Initial pressure,

Initial temperature,

            [tex]= 259 \ K[/tex]

Final temperature,

(B)

Number of moles,

→ [tex]n = (\frac{P_o V}{RT_o} )[/tex]

then,

The final absolute pressure,

→ [tex]P = \frac{nRT}{V}[/tex]

      [tex]= (\frac{P_o V}{RT_o} )(\frac{RT}{V} )[/tex]

      [tex]=(\frac{T}{T_o} )P_o[/tex]

      [tex]= (\frac{305}{259} )\times 495[/tex]

      [tex]= 582.915 \ kPa[/tex]

Thus the above approach is correct.

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Answer:

ln the contemporary time , farming can be considered as comparitively important occupation as it can feed the population , So agriculture is having a greater importance than any other occupation.

Answer:

Alpha particles are more massive than beta particles.

Explanation:

The alpha particles are also called double-positive Heilum Nuclei because they have a charge of "+2" and a mass of 4 a.m.u. The properties of the alpha particles are as follows:

1. It possesses high energy due to high velocity. It is 7.7 MeV for most energetic from Rac (i.e: Bismuth-214)

2. It has a very high ionizing power. A 7.7 MeV particle produces about 0.2 x 10⁶ ions.

3. The range of alpha particles is very small. It is about 7 x 10⁻² m and only 4 x 10⁻⁵ m in aluminum for 7.7 MeV alpha-particle.

4. Alpha particles produce fluorescence on striking certain substances, such as zinc sulphide and bariumplatinocynide.

The beta particles are fast-moving electrons, which have a negligible mass.  

Hence, the correct option is:

Alpha particles are more massive than beta particles.

We have that the spring constant is mathematically given as

[tex]k=2.37*10^{11}N/m[/tex]

Generally, the equation for angular velocity is mathematically given by

[tex]\omega=\sqrt{k}{m}[/tex]

Where

k=spring constant

And

[tex]\omega =\frac{2\pi}{T}[/tex]

Therefore

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

Hence giving spring constant k

[tex]k=m((\frac{2 \pi}{T})^2[/tex]

Generally

Mass of earth [tex]m=5.97*10^{24}[/tex]

Period for on complete resolution of Earth around the Sun

[tex]T=365 days[/tex]

[tex]T=365*24*3600[/tex]

Therefore

[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]

[tex]k=2.37*10^{11}N/m[/tex]

In conclusion

The effective spring constant of this simple harmonic motion is

[tex]k=2.37*10^{11}N/m[/tex]

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Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

20

1

+

12

1

⟹

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

−16

1

+

12

1

⟹

v

1

=

48

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

When the source is moving away from the observer the velocity of the source is added to the speed of light. This increases the value of the denominator, decreasing the value of the observed frequency. Frequency corresponds to pitch or tone; a lower observed frequency will result in a lower observed pitch.

Answer:

d = 30kg/cm³

Explanation:

d = m/v

d = 1080kg/(3cm*4cm*3cm)

d = 30kg/cm³

Answer:

Answer:

Following are the solution to the given question:

Explanation:

For charging plates that are connected in a similar manner:

Calculating the total charge:

[tex]\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C[/tex]

Calculating the common potential:

[tex]\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\[/tex]

Calculating the charge after redistribution:

[tex]When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2[/tex]        

[tex]\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\ \\ \to q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C[/tex]

Answer:

[tex]v_1=87.40m/s[/tex]

Explanation:

From the question we are told that:

Mass of arrow [tex]m=52g[/tex]

Mass of rock [tex]m_r=1.50kg[/tex]

Height [tex]h=0.47m[/tex]

Generally the equation for Velocity is mathematically given by

 [tex]v = \sqrt{(2gh)}[/tex]

 [tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]

 [tex]v= 3.035m/s[/tex]

Generally the equation for conservation of momentum is mathematically given by

 [tex]m_1v_1=m_2v_2[/tex]

 [tex]0.052kg * v = 1.5 * 3.03m/s[/tex]

 [tex]v_1=87.40m/s[/tex]

Answer:

P = -1 D

Explanation:

For this exercise we must use the equation of the constructor

       / f = 1 / p + 1 / q

where f is the focal length, p and q is the distance to the object and the image, respectively

The far view point is at p =∞  and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image  

        [tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]

         f = 1 m

         P = 1/f

          P = -1 D

Answer:

a)  [tex]P=0.80[/tex]

b)  [tex]1.25Hz[/tex]

c)  [tex]A=25cm[/tex]

Explanation:

From the question we are told that:

Travel Time [tex]T=0.40s[/tex]

Distance [tex]d=50cm[/tex]

a)

Period

Time taken to complete one oscillation

Therefore

[tex]P=2*T\\\\P=2*0.40[/tex]

[tex]P=0.80[/tex]

b)

Frequency is

[tex]F=\frac{1}{T}\\\\F=\frac{1}{0.80}[/tex]

[tex]1.25Hz[/tex]

c)

Amplitude:the distance between the mean and extreme position

[tex]A=\frac{50}{2}[/tex]

[tex]A=25cm[/tex]

Answer:

Florida is tropical, with a significant rainy seson

Answer:

The total surface area is "90 km²".

Explanation:

Given:

Power from solar installations,

= 27 GW

Other natural installations,

= 740 GW

Intensity,

[tex]\frac{F}{At}=\frac{P}{A}=1000 \ W/m^2[/tex]

%n,

= 30%

Now,

⇒ %n = [tex]\frac{out.}{Inp.}\times 100[/tex]

then,

⇒ [tex]Inp.=\frac{27}{30}\times 100[/tex]

           [tex]=90 \ GW[/tex]

As we know,

⇒ [tex]I=\frac{P}{A}[/tex]

by substituting the values, we get

[tex]1000=\frac{90\times 10^9}{A}[/tex]

    [tex]A = \frac{90\times 10^9}{10^3}[/tex]

        [tex]=90\times 10^6[/tex]

        [tex]=90 \ km^2[/tex]

Answer:

[tex]E=35921.96N/C[/tex]

Explanation:

From the question we are told that:

Radius [tex]r=0.321mm[/tex]

Charge Density [tex]\mu=0.100[/tex]

Distance [tex]d= 5.00 cm[/tex]

Generally the equation for electric field is mathematically given by

[tex]E=\frac{mu}{2\pi E_0r}[/tex]

[tex]E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}[/tex]

[tex]E=35921.96N/C[/tex]

Answer:

PlROCA

Explanation:

Answer:

Ecologist.

Your answer is Ecologist.

(Ecologist) is a scientist who studies the whole environment as a working unit.

Answer:

Charge of the particle is 1 coulomb.

Explanation:

Force, F:

[tex]{ \bf{F=BeV}}[/tex]

F is magnetic force.

B is the magnetic flux density.

e is the charge of the particle.

V is the velocity

[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]

The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex].

The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:

Q(t) = Aeσ[tex]T^{4}[/tex]

where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.

To determine the rate of energy radiated by the man in the given question;

[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ

But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.

So that;

[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]

     = 3.8556 x [tex]10^{-8}[/tex]

     = 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex]

Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex].

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Answer:

In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.

The intensity of the electromagnetic wave in terms of the electric field is 0.00298 W/m² and the intensity of the electromagnetic wave in terms of the magnetic field is 1.193x10⁻⁶  W/m².

The intensity of the electromagnetic wave is related to the electric field as well as to the magnetic field.    

a) Intensity of the electromagnetic wave for the electromagnetic field.

The intensity of the electromagnetic wave (I) in terms of the electromagnetic field is given by:

[tex] I = \frac{E^{2}*c*\epsilon_{0}}{2} [/tex]   (1)

Where:

c: is the speed of light = 3.00*10⁸ m/s  

E: is the magnitude of the electric field = 1.5 V/m

ε₀: is the permittivity of free space = 8.85*10⁻¹² C²/Nm²

Hence, the intensity of the electromagnetic wave (eq 1) is:

[tex] I = \frac{(1.5 V/m)^{2}*3.00 \cdot 10^{8} m/s*8.85 \cdot 10^{-12} C^{2}/(N*m^{2})}{2} = 0.00298 W/m^{2} [/tex]                                                                                          

b) Intensity of the electromagnetic wave for the magnetic field

We can calculate the intensity of the electromagnetic wave (I) in terms of the magnetic field with the following equation:

[tex] I = \frac{cB^{2}}{2\mu_{0}} [/tex]   (2)

Where:

B: is the magnitude of the magnetic field = 10⁻¹⁰ T

μ₀: is the vacuum permeability = 4π*10⁻⁷ m*T/A

Therefore, the intensity of the electromagnetic wave (eq 2) is:

[tex] I = \frac{3.00 \cdot 10^{8} m/s*(1\cdot 10^{-10} m*T/A)^{2}}{2*4\pi \cdot 10^{-7} T/A} = 1.193 \cdot 10^{-6} W/m^{2} [/tex]

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I hope it helps you!

Explanation:

some thing inside a resistor

Answer:

The magnitude of the magnetic field is 1.83 x [tex]10^{-5}[/tex] T.

Explanation:

The flow of an electric current in a straight wire induces magnetic field around the wire. When current is flowing through two wires in the same direction, a force of attraction exists between the wires. But if the current flows in opposite directions, the force of repulsion is felt by the wires.

In the given question, the direction of flow of current through the wires is opposite, thus both wires applies the same field on each other. The result to repulsion between them.

The magnetic field (B) between the given wires can be determined by:

B = [tex]\frac{U_{o}I }{2\pi r}[/tex]

where: I is the current, r is the distance between the wires and [tex]U_{0}[/tex] is the magnetic field constant.

But, I = 11 A, r = 0.12 m and [tex]U_{0}[/tex] = 4[tex]\pi[/tex] x [tex]10^{-7}[/tex] Tm/A

So that;

B = [tex]\frac{4\pi *10^{-7}*11 }{2\pi *0.12}[/tex]

   = 1.8333 x [tex]10^{-5}[/tex]

B = 1.83 x [tex]10^{-5}[/tex] T

Answer:

0.5849Weber

Explanation:

The formula for calculating the magnetic flus is expressed as:

[tex]\phi = BAcos \theta[/tex]

Given

The magnitude of the magnetic field B = 3.35T

Area of the loop = πr² = 3.14(0.24)² = 0.180864m²

angle of the wire loop θ = 15.1°

Substitute the given values into the formula:

[tex]\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb[/tex]

Hence the magnetic flux Φ through the loop is 0.5849Weber