An electron is projected horizontally into the uniform electric field directed vertically downward between two parallelplates. The plates are 2.00 cm apart and are of length 4.00 cm. The initial speed of the electron is vi = 8.00 × 106 m/s. As it enters the region between the plates, the electron is midway between the two plates; as it leaves, the electron just misses the upperplate.a)What is the magnitude ofv the electric field?

Answer:

38.6 mi/h

Explanation:

7.4 mi/h = 7.4mi/h * (1/60)hour/min * (1/60) min/s = 0.00206 mi/s

Let v (mi/s) be your original speed, then the time t it takes to go 1 mi/s is

t = 1/v

Since you increase v by 0.00206 mi/s, your time decreases by 15 s, this means

t - 15 = 1/(v+0.00206)

We can substitute t = 1/v to solve for v

[tex]\frac{1}{v} - 15 = \frac{1}{v + 0.00206}[/tex]

We can multiply both sides of the equation with v(v+0.00206)

v+0.00206 - 15v(v+0.00206) = v

[tex]-15v^2 - 0.0308v + 0.00206 = 0[/tex]

[tex]v= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]v= \frac{0.03083\pm \sqrt{(-0.03083)^2 - 4*(-15)*(0.00205)}}{2*(-15)}[/tex]

[tex]v= \frac{0.03083\pm0.35}{-30}[/tex]

v = -0.01278 or v = 0.01 0724 mi/s

Since v can only be positive we will pick v = 0.010724 mi/s or 0.010724*3600 = 38.6 mi/h

Answer:

[tex]1.36\times 10^{-7} C[/tex]

Explanation:

We are given that

Mass of charged tine spheres=m=1 g=[tex]\frac{1}{1000}=0.001 kg[/tex]

1 kg=1000g

The distance between charged tine spheres=r=2 cm=[tex]\frac{2}{100}=0.02 m[/tex]

1 m=100 cm

Acceleration =[tex]a =414 m/s^2[/tex]

Let q be the charge on each sphere.

[tex]k=9\times 10^9Nm^2/C^2[/tex]

The electric force between two charged particle

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

Using the formula

The force between two charged tiny spheres=[tex]F_e=\frac{kq^2}{(0.02)^2}[/tex]

According to  Newton's second law , the net force

[tex]F=ma[/tex]

[tex]F=F_e[/tex]

[tex]0.001\times 414=\frac{9\times 10^9\times q^2}{(0.02)^2}[/tex]

[tex]q^2=\frac{0.001\times 414\times (0.02)^2}{9\times 10^9}[/tex]

[tex]q=\sqrt{\frac{0.001\times 414\times (0.02)^2}{9\times 10^9}}[/tex]

[tex]q=1.36\times 10^{-7} C[/tex]

Hence, the magnitude of charge on each tiny sphere=[tex]1.36\times 10^{-7} C[/tex]

Answer:

[tex]\rho=995.50\ kg.m^{-3}[/tex]

[tex]\bar w=9765.887\ N.m^{-3}[/tex]

[tex]s=0.9955[/tex]

Explanation:

Given:

So, mass of the empty can:

[tex]m_c=\frac{w_c}{g}[/tex]

[tex]m_c=\frac{0.153}{9.81}[/tex]

[tex]m_c=0.015596\ kg[/tex]

Hence the mass of liquid(soda):

[tex]m_l=m_f-m_c[/tex]

[tex]m_l=0.369-0.015596[/tex]

[tex]m_l=0.3534\ kg[/tex]

Therefore the density of liquid soda:

[tex]\rho=\frac{m_l}{v_l}[/tex] (as density is given as mass per unit volume of the substance)

[tex]\rho=\frac{0.3534}{3.55\times 10^{-4}}[/tex]

[tex]\rho=995.50\ kg.m^{-3}[/tex]

Specific weight of the liquid soda:

[tex]\bar w=\frac{m_l.g}{v_l}=\rho.g[/tex]

[tex]\bar w=995.5\times 9.81[/tex]

[tex]\bar w=9765.887\ N.m^{-3}[/tex]

Specific gravity is the density of the substance to the density of water:

[tex]s=\frac{\rho}{\rho_w}[/tex]

where:

[tex]\rho_w=[/tex] density of water

[tex]s=\frac{995.5}{1000}[/tex]

[tex]s=0.9955[/tex]

Explanation:

The given data is as follows.

    Volume of pop in can, V = [tex]355 \times 10^{-6} m^{3}[/tex]

                          W = mg

                               = [tex]0.369 \times 9.81[/tex]

                               = 3.6198 N

Weight of empty can, [tex]w_{1}[/tex] = 0.153 N

                [tex]w_{2} = W - w_{1}[/tex]

                           = 3.6198 - 0.153

                           = 3.467 N

         [tex]\gamma = \frac{\text{weight of liquid}}{\text{volume of liquid}}[/tex]

                      = [tex]\frac{3.467}{355 \times 10^{-6}}[/tex]

                      = 9766.197 [tex]N/m^{3}[/tex]

                [tex]\rho = \frac{\gamma}{g}[/tex]

                          = [tex]\frac{9766.197}{9.81}[/tex]

                          = 995.535 [tex]kg/m^{3}[/tex]

           S.G = [tex]\frac{\text{density of liquid}}{\text{density of water}}[/tex]

                  = [tex]\frac{\rho}{\rho_{w}}[/tex]

                  = [tex]\frac{995.535}{1000}[/tex]

                  = 0.995

Answer:

David will not  be able to catch the bill .

Explanation:

Reaction time = .2 s .

During this period bill will fall vertically between the fingers.

Distance of fall = 1/2 x g x t²

= .5 x 9.8 x 0.2²

= 19.6 cm or 20 cm .

Generally the bill has size of the order of 25 cm . From central point it requires a fall of 12.5 cm for the bill to escape the catch . Since fall is of 20 cm , that means bill will fall below the level of fingers in .2 s .

So David will not be able to catch the bill.

Answer:

[tex]h=9.30m[/tex]

Explanation:

We have an uniformly accelerated motion, due to the gravitational acceleration. So, we use the kinematic equations, since the ball is throw directly upward, g is negative:

[tex]h=v_0t-\frac{gt^2}{2}[/tex]

First, we need to calculate the time taken by the ball to reach the maximum height, in this point its final speed is zero:

[tex]v_f=v_0-gt\\\\\frac{0-v_0}{-g}=t\\t=\frac{v_0}{g}\\t=\frac{13.5\frac{m}{s}}{9.8\frac{m}{s^2}}\\t=1.38s[/tex]

Now, we can calculate h:

[tex]h=v_0t-\frac{gt^2}{2}\\h=13.5\frac{m}{s}(1.38s)-\frac{9.8\frac{m}{s^2}(1.38s)^2}{2}\\h=9.30m[/tex]

Final answer:

Astronomers utilize spectroscopy to analyze the spectrum of a star, identifying unique absorption lines corresponding to different elements, which reveals the star's composition. Spectral lines' broadening indicates temperature and pressure, and shifts in these lines help measure a star's motion, including radial and rotational velocities.

Explanation:

Understanding Stellar Spectroscopy

Astronomers use spectroscopy as a powerful tool to determine various characteristics of stars, including their composition and temperature. When light from a star passes through a prism or diffraction grating, it spreads out into a spectrum of colors. This spectrum contains dark lines known as absorption lines, which are unique to the elements present in the star's atmosphere, as different chemical elements absorb light at specific wavelengths. Therefore, by analyzing these lines, astronomers can identify the elements that make up a star.

Analyzing the broadening of spectral lines can inform us about a star's temperature and pressure. Warmer temperatures and higher pressures in a star's atmosphere tend to broaden the spectral lines. Additionally, the pressure can give clues about the star's size, as stars with lower atmospheric pressure tend to be larger, or giant stars.

Motions of the Stars are also revealed through spectroscopy. The Doppler effect causes spectral lines to shift towards the red end of the spectrum if the star is moving away from us (redshift) or towards the blue end if it is approaching (blueshift). This allows astronomers to measure the star's radial velocity. Spectral line broadening can also indicate the star's rotational velocity, while proper motion is deduced from the movement of the lines over time across the spectrum.

Answer:

(A) Spring constant will be 126.58 N/m

(B) Amplitude will be equal to 0.177 m

Explanation:

We have given mass of the block m = 200 gram = 0.2 kg

Time period T = 0.250 sec

Total energy is given TE = 2 J

(A) For mass spring system time period is equal to [tex]T=2\pi \sqrt{\frac{m}{K}}[/tex]

So [tex]0.250=2\times 3.14 \sqrt{\frac{0.2}{K}}[/tex]

[tex]0.0398=\sqrt{\frac{0.2}{K}}[/tex]

Now squaring both side

[tex]0.00158=\frac{0.2}{K}[/tex]

K = 126.58 N/m

So the spring constant of the spring will be 126.58 N/m

(B) Total energy is equal to [tex]TE=\frac{1}{2}KA^2[/tex], here K is spring constant and A is amplitude

So [tex]2=\frac{1}{2}\times 126.58\times A^2[/tex]

[tex]A^2=0.0316[/tex]

A = 0.177 m

So the amplitude of the wave will be equal to 0.177 m

Answer:

wave speed= constant

frequency = increase

wavelength = decrease

Explanation:

Solution:

- The three basic parameters of a wave are speed, frequency and wavelength. These three parameters are related to each other by an expression:

                                             v = f * λ

Where,

- v is the speed of the wave in m/s.

- f frequency of the wave in Hz.

- λ wavelength of the wave in m

- We are asked how would each of these parameter change if we move the hand up and down faster. The hand moves from a crest to trough faster than before and back again. We can see that the time between a cycle has decreased; hence, frequency f increases. Consequently, we can see that wave speed v remains constant - the medium of transfer of wave energy - remains same. Then from our relation above if we hold speed constant and increase f then the wavelength λ would have to decrease.

Final answer:

To find the transformed vector representations in a rotated coordinate system, the angle of vector A is adjusted by the rotation angle, and the components are calculated using trigonometric functions. Vector B's components in the rotated system are found using a rotation matrix.

Explanation:

The provided question pertains to transforming the representation of vectors in a rotated coordinate system in the subject of physics. The coordinate system is rotated counterclockwise, and the goal is to find the new representations of vectors A and B in unit-vector notation on the x'y' system. Given the initial magnitude and direction angle of vector A and the Cartesian components of vector B on the xy coordinate system, we can calculate their components on the rotated x'y' coordinate system.

The original vector A has a magnitude of 14.4 m and an angle of 51.6° from the positive x-axis. After rotation by 20°, the new angle becomes 51.6° - 20.0° = 31.6° from the new x'-axis. Using the formulas Ax' = A cos θ' and Ay' = A sin θ', where θ' is the new angle, we can find the rotated components of A.

The vector B is already given in Cartesian coordinates as ( 14.3 m )i + (8.52 m )j. To find the components of B in the rotated system, we use a rotation matrix, giving us new components Bx' and By'.

In conclusion, to find the transformed vectors in the rotated system, we apply the rotation to both the magnitude and angle of A, and use a rotation matrix for the components of B.

Answer:

d=894 m

Explanation:

Given that

initial velocity ,u= 35 m/s

Acceleration ,a= 38 m/s²

time ,t= 6 s

Given that at t= 0 s ,x= 0 m

We know that

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

d=Displacement

Now by putting the values

[tex]d=35\times 6+\dfrac{1}{2}\times 38\times 6^2 [/tex]

d=894 m

Therefore the particle position after 6 sec will be 894 m.

Final answer:

The position of the particle at t = 6 seconds, with an initial velocity of 35 m/s and a constant acceleration of 38 m/s², is 894 meters from the start.

Explanation:

The question asks us to calculate the position of a particle moving in a straight line at t = 6 seconds, given an initial velocity of 35 m/s and a constant acceleration of 38 m/s². To find the position, we can use the kinematic equation:

x = v0t + ½at²

where x is the position, v0 is the initial velocity, a is the acceleration, and t is the time. Plugging in our values we get:

x = (35 m/s)(6 s) + ½(38 m/s²)(6 s)²

x = 210 m + ½(38 m/s²)(36 s²)

x = 210 m + 684 m

x = 894 m

Therefore, the position of the particle at t = 6 s is 894 meters from the starting point.

Answer:

a

Explanation:

Given:

In a mixture of the gases oxygen and helium, which statement is valid:

(a) the helium molecules will be moving faster than the oxygen molecules, on average

(b) both kinds of molecules will be moving at the same speed

(c) the oxygen molecules will, on average, be moving more rapidly than the helium molecules

(d) the kinetic energy of the helium will exceed that of the oxygen

(e) none of the above

Solution:

- We will use Boltzmann distribution to answer this question. The root mean square speed of molecules of a gas gives the average speed as follows:

                                        V_rms = sqrt ( 3 k T / m )

- Where, k is the Boltzmann constant, T is the temperature and m is the mass of a single molecule of a gas.

- In general, a mixture has a constant equilibrium temperature T_eq.

- So the v_rms is governed by the mass of a single molecule.

- We know that mass of single molecule of Oxygen is higher than that of Helium molecule. Hence, the relation of mass is inversely proportional to square of root mean speed. So the helium molecules will be moving faster than the oxygen molecules.

- Note: The kinetic energy of the mixture remains constant because it is due to the interaction of the molecules within i.e oxygen and helium. Which makes the kinetic energy independent of mass.

                                     E_k = 0.5*m*v_rms^2

                                     E_k = 0.5*m*(3*k*T/ m )

                                    E_k = 0.5*3*k*T

Hence, E_k is only the function of Temperature which we already established to remain constant at equilibrium.

OPTION A.

In a mixture of oxygen and helium, the helium molecules move faster on average due to their lighter weight, though the average kinetic energy of both gases remains the same at a given temperature.

In a mixture of gases, the speeds of the molecules of different gases are primarily dependent on their masses. For gases at a given temperature, all have the same average kinetic energy (KEavg) for their molecules. However, gases made up of lighter molecules, such as helium, have more high-speed particles and an average speed (Urms) that is higher than gases composed of heavier molecules, like oxygen.

Therefore, in a mixture of oxygen and helium, statement (a) the helium molecules will be moving faster than the oxygen molecules, on average, is valid. This is due to helium molecules being lighter than oxygen molecules. Moreover, the average kinetic energy of both gases, helium and oxygen, would be the same at a given temperature, meaning statement (d) the kinetic energy of the helium will exceed that of the oxygen, would not be valid.

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Answer:

given,

reaction time.t_r = 0.50 s

deceleration of the car = 6 m/s²

initial speed,v = 20 m/s

distance at which the car stop = ?

distance travel by the car in reaction time

 d= v x t_r

 d = 20 x 0.5 = 10 m

using equation of motion

distance travel to stop the car

v² = u² + 2 a s

0² = 20² - 2 x 6 x s

 12 s = 400

 s = 33.33 m

Total distance travel by the car

D = 10 + 33.33

D = 43.33 m

Hence, the car stops before to avoid collision.

Answer:

The work done is zero.

Solution:

As per the question:

Charge, [tex]q = 1\mu C = 1\times 10^{- 6}\ C[/tex]

Distance moved, d = 5 m

Voltage, V = 6V

Now, we know that an equipotential surface is one where the potential is same everywhere on the surface.

Suppose the the voltage at a distance d = 5 m is V'

Thus

V' = 6 V, (since the surface is equipotential)

Work done in moving a charge is given by:

[tex]W = q\Delta V[/tex]

[tex]W = q(V - V')[/tex]

[tex]W = (1\times 10^{- 6})(V - V')[/tex]

[tex]W = (1\times 10^{- 6})(6 - 6) = 0[/tex]

Thus the work done in moving a charge on an equipotential surface comes out to be zero as the potential difference is zero.

Final answer:

The work required to move a 1 microcoulomb charge by a distance of 5 meters along an equipotential line of 6V is zero because there's no change in potential energy.

Explanation:

The question relates to determining the amount of work needed to move a charge along an equipotential line. When a charge moves along an equipotential, the potential energy of the charge does not change because the voltage (potential difference) across its path remains zero. In other words, the work done on the charge is zero since work is defined as the change in potential energy, which is given by the formula W = qV, where W is work, q is charge in coulombs, and V is potential difference in volts. For movement along an equipotential line, V = 0, hence, Work = 0 Joules.

Final answer:

The length of the phasor in a phasor diagram representing a transverse wave on a string indicates the wave's amplitude, which corresponds to the maximum displacement of the medium's particles from their equilibrium position.

Explanation:

In the phasor representation of a transverse wave on a string, the length of the phasor corresponds to the amplitude of the wave. In a phasor diagram, this amplitude represents the maximum displacement of the wave particles from the equilibrium position as the wave propagates through the medium. The phasor's length will rotate in a circular motion at a rate determined by the wave's frequency, and this motion represents the oscillatory nature of the wave at a certain point in space over time. The amplitude is a crucial parameter as it determines the energy carried by the wave, with a larger amplitude indicating a greater energy transfer.

The phasor length is particularly important when analyzing multiple wave forms together, such as voltage and current in electrical circuits, where the ratio of their lengths can denote relative magnitudes, such as resistance in the circuit. In this context, however, we focus on mechanical waves on a string, and the length of the phasor would only represent the wave amplitude, not voltage or current.

Answer:

V = 156.85 Km/h

Explanation:

Speed of plane = 125 Km/h

angle of plane=  25° N of E

Speed of wind = 36 Km/h

angle of plane = 6° S of W

Horizontal component of the velocity

V_x = 125 cos 25° + 36 cos 6°

V_x = 149 Km/h

Vertical component of the velocity

V_y = 125 sin 25° - 36 sin 6°

V_y = 49 Km/h

Resultant of Velocity

[tex]V = \sqrt{V_x^2 + V_y^2}[/tex]

[tex]V = \sqrt{149^2 + 49^2}[/tex]

  V = 156.85 Km/h

the resulting velocity of the plane is equal to  V = 156.85 Km/h

The surface tension of the liquid in air is 0.8 N/m.

To determine the surface tension of the liquid, we need to use the formula F = yL, where F is the force needed to move the wire, y is the surface tension, and L is the length of the wire. In this case, F = 0.096 N and L = 12 cm. We can rearrange the formula to solve for y: y = F / L. Plugging in the values, we get y = 0.096 N / 0.12 m = 0.8 N/m. So, the surface tension of the liquid in air is 0.8 N/m.

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Answer:

261.307 m

Explanation:

b = Base of triangle = 450 m

p = Perpendicular of the triangle

[tex]\theta[/tex] = Angle of the triangle = [tex]30^{\circ}[/tex]

From trigonometry

[tex]tan\theta=\dfrac{p}{b}[/tex]

[tex]\Rightarrow p=btan\theta[/tex]

[tex]\Rightarrow p=450\times tan30[/tex]

[tex]\Rightarrow p=259.807\ m[/tex]

Height of the building = 1.5+259.807 = 261.307 m

To solve this problem we will apply the concepts related to the electric field based on the laws of Coulomb. Said electric field is equivalent to the product between the Coulomb constant and the rate of change of the charge and the squared distance. Mathematically this is,

[tex]E = \frac{kq}{r^2}[/tex]

Here,

k = Coulomb's constant

q = Charge

r = Distance

Replacing we have that

[tex]E = \frac{kq}{r^2}[/tex]

[tex]1.08*10^2 = \frac{(9*10^{9})q}{(6.4*10^{6})^2}[/tex]

Solving for q,

[tex]q = 491520 C[/tex]

Therefore the net charge of the Earth under the previous condition is 491520 C

Answer:

F=480.491 N

Explanation:

Given that

mass ,m = 22 kg

Angular speed ω = 40 rev/min

[tex]\omega=\dfrac{2\pi \times 40}{60}\ rad/s[/tex]

ω =4.18 rad/s

The radius  r= 1.25 m

We know that centripetal force is given as

F=m ω² r

Now by putting the values in the above equation we get

[tex]F=22\times 4.18^2\times 1.25\ N[/tex]

F=480.491 N

Therefore the centripetal force on the child will be 480.491 N.

The spring constant (k) can be obtained by employing the principle of conservation of energy. Here, the kinetic energy of the metal sphere at the beginning of the motion equals the potential energy at the maximum stretch of the spring. Substituting the given values into the energy equation, solving for 'k' yields the spring constant.

The subject of this question lies within the domain of Physics, specifically the domain of mechanics and dynamics dealing with springs and oscillations. The spring constant (k) can be derived from the principle of conservation of energy. Here, we are ignoring friction and air resistance, meaning that the sum of kinetic energy and potential energy remains constant throughout the motion of the metal sphere.

At the beginning, all the energy is kinetic, and at the maximum stretch, all the energy is potential. This can be represented by the equation 0.5*m*v1^2 = 0.5*k*x2^2. By substituting the given values of m (mass = 0.73 kg), v1 (initial velocity = 7.2 m/s), and x2 (maximum displacement = 0.23 m), we can solve for k (spring constant). Here, the calculation would be as follows: k = m*v1^2/x2^2 = (0.73 kg*(7.2 m/s)^2)/(0.23 m)^2. After performing the required calculations, you can obtain the numerical value of the spring constant.

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Answer:

on increasing pressure, temperature will also increase.

Explanation:

Considering the ideal gas equation as:

[tex]PV=nRT[/tex]

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Thus, at constant volume and number of moles, Pressure of the gas is directly proportional to the temperature of  the gas.

P ∝ T

Also,  

Also, using Gay-Lussac's law,

[tex]\frac {P_1}{T_1}=\frac {P_2}{T_2}[/tex]

Thus, on increasing pressure, temperature will also increase.

Answer:

415,224,000

Explanation:

a person blinks 10 times per minute ,60 minutes in a hour so 600 per hour,24 hours per day so 14,400 blinks per day and there are 365 days in a year so 5,256,000 blinks per year and an average person lives to 79 years so 415224000 in an average lifetime

Answer:

P_max = 278 KN

Explanation:

Given:

- The ultimate normal stress S = 350 MPa

- Thickness of both pipes t = 8 mm

- Pipe AB: D_o = 250 mm

- Pipe BC: D_o = 150 mm

- Factor of safety FS = 4.5

Find:

Find the maximum load P_max

Solution:

- Compute cross sectional areas A_ab and A_bc:

                                    A_ab = pi*(D_o^2 - (D_o - 2t)^2) / 4

                                    A_ab = pi*(0.25^2 - 0.234^2) / 4

                                    A_ab = 6.08212337 * 10^-3 m^2

                                    A_bc = pi*(D_o^2 - (D_o - 2t)^2) / 4

                                    A_bc = pi*(0.15^2 - 0.134^2) / 4

                                    A_bc = 3.568212337 * 10^-3 m^2

- Compute the Allowable Stress for each pipe:

                                    sigma_all = S / FS

                                    sigma_all = 350 / 4.5

                                    sigma_all = 77.77778 MPa

- Compute the net for each member P_net,ab  and P_net,bc:

                                    P_net,ab =  sigma_all * A_ab

                                    P_net,ab = 77.77778 MPa*6.08212337 * 10^-3

                                    P_net,ab = 473054.0399 N

                                    P_net,bc =  sigma_all * A_bc

                                    P_net,bc = 77.77778 MPa*3.568212337 * 10^-3

                                    P_net,bc = 277577.1721 N

- Compute the force P for each case:

                                    P_net,ab = P + 50,000

                                    P = 473054.0399 - 50,000

                                    P = 423 KN

                                   P_net,bc = P = 278 KN

- P_max allowed is the minimum of the two load P:

                                   P_max = min (423, 278) = 278 KN

Answer:

2.80321285141

Explanation:

[tex]L_g[/tex] = Thickness of glass = 4.5 mm

[tex]k_g[/tex] = Thermal conductivity of glass = 0.8 W/mK

[tex]R_0[/tex] = Combined thermal resistance = [tex]0.15\times m^2K/W[/tex]

[tex]L_a[/tex] = Thickness of air = 6.6 mm

[tex]k_a[/tex] = Thermal conductivity of air = 0.024 W/mK

The required ratio is the inverse of total thermal resistance

[tex]\dfrac{2(L_g/k_g)+R_0+(L_a/k_a)}{(L_g/k_g)+R_0}\\ =\dfrac{2(4.5\times 10^{-3}/0.8)+0.15+(6.6\times 10^{-3}/0.024)}{(4.5\times 10^{-3}/0.8)+0.15}\\ =2.80321285141[/tex]

The ratio is 2.80321285141

Answer:

[tex]\frac{\dot Q}{\dot Q'} =2.6668[/tex]

Explanation:

Given:

Heat loss through single pane window:

Using Fourier's law of conduction,

[tex]\dot Q=A.dT\div (R_{th}+\frac{t_g}{k} )[/tex]

[tex]\dot Q=0.15\times dT\div (0.15+\frac{4.5\times 10^{-3}}{0.8})[/tex]

[tex]\dot Q=0.9638\ dT\ [W][/tex]

Heat loss through double pane window:

[tex]\dot Q'=dT\times A\div(R_{th}+2\times \frac{t_g}{k}+\frac{t_a}{k_a} )[/tex]

where:

[tex]dT=[/tex] change in temperature

[tex]k_a=[/tex] coefficient of thermal conductivity of air [tex]= 0.026\ W.m^{-1}.K^{-1}[/tex]

[tex]\dot Q'=dT\times 0.15\div (0.15+2\times \frac{4.5\times 10^{-3}}{0.8}+\frac{6.6\times 10^{-3}}{0.026})[/tex]

[tex]\dot Q'=0.3614\ dT\ [W][/tex]

Now the ratio:

[tex]\frac{\dot Q}{\dot Q'} =\frac{0.9638(dT)}{0.3614(dT)}[/tex]

[tex]\frac{\dot Q}{\dot Q'} =2.6668[/tex]

Answer:

A: 50 kg

Explanation:

Answer:

Explanation:

Given

Velocity of Huck w.r.t to raft [tex]v_{H,raft}=0.7\ m/s[/tex]

Perpendicular to the motion of raft

Velocity of Raft in the river [tex]v_{raft,river}=1.6\ m/s[/tex]

As Huck is traveling Perpendicular to the raft so he possess two velocities i.e. vertical velocity and horizontal velocity of River when observed from bank

[tex]v_{Huck,river\ bank}=0.7\hat{j}+1.6\hat{i}[/tex]

So magnitude of velocity is given by

[tex]|v|=\sqrt{0.7^2+1.6^2}[/tex]

[tex]|v|=\sqrt{0.49+2.56}[/tex]

[tex]|v|=\sqrt{3.05}[/tex]

[tex]|v|=1.74\ m/s[/tex]

For direction [tex]\tan =\frac{0.7}{1.6}=0.4375[/tex]

[tex]\theta =23.63^{\circ}[/tex] w.r.t river bank

Answer:

27.3 m/s

Explanation:

We are given that

Distance travel by ball=x=60 m

[tex]\theta=26^{\circ}[/tex]

We have to find the initial speed([tex]v_0)[/tex] of the ball.

[tex]x=v_0cos\theta t[/tex]

Using the formula

[tex]60=v_0cos 26 t[/tex]

[tex]t=\frac{60}{v_ocos 26}=\frac{60}{v_0\times 0.899}=\frac{66.7}{v_0}[/tex]

The value of y at point of foot  of the vertical distance

y=0

[tex]y=v_0sin\theta t-\frac{1}{2}gt^2[/tex]

Using [tex]g=9.8m/s^2[/tex]

Using the formula

[tex]0=v_0sin 26\times \frac{66.7}{v_0}-4.9\times (\frac{66.7}{v_0})^2[/tex]

[tex]4.9\times \frac{(66.7)^2}{v^2_0}=0.44\times 66.7[/tex]

[tex]v^2_0=\frac{4.9\times (66.7)^2}{0.44\times 66.7}[/tex]

[tex]v^2_0=742.8[/tex]

[tex]v_0=\sqrt{742.8}=27.3 m/s[/tex]

Hence, the initial speed of the ball=27.3 m/s

Answer:

27.3 m/s

Explanation:

Horizontal range, R = 60 m

angle of projection, θ = 26°

Let the velocity of projection is vo.

Use the formula of range of the projectile

[tex]R = \frac{u^{2}Sin2\theta} {g}[/tex]

[tex]60 = \frac{v_{0}^{2}Sin52}{9.8}[/tex]

vo = 27.3 m/s

Thus, the velocity of projection is 27.3 m/s.

Answer:

Trigonal planar

Explanation:

Trigonal planar - it is referred to the molecular shape of atom in which three bonds exist around any central atom. As there is no lone pair at the center hence all three atoms have taken the form of a triangle. All three atom lies at same plane and known as peripheral atoms

The angle between all the three atoms is 120 degree

Answer:

a) 66.4 relative to the west in the south-west direction

b) 5.455 hours

Explanation:

a)If the wind is blowing east-ward at a speed of 40km/h, then the west component of the geese velocity must be 40km/h in order to counter balance it. Geese should be flying south-west at an angle of

[tex]cos(\alpha) = 40 / 100 = 0.4[/tex]

[tex]\alpha = cos^{-1}(0.4) = 1.16 rad = 180\frac{1.16}{\pi} = 66.4^0[/tex] relative to the West

b) The south-component of the geese velocity is

[tex]100sin(\alpha) = 100sin(66.4^0) = 91.65 km/h[/tex]

The time it would take for the geese to cover 500km at this rate is

t = 500 / 91.65 = 5.455 hours

The frequency can be defined as the inverse of the period, that is, it can be expressed as

[tex]T = \frac{1}{f}[/tex]

Here,

T = Period

f = Frequency

For each value we only need to replace the value and do the calculation:

PART A)

[tex]T = \frac{1}{f}[/tex]

[tex]T = \frac{1}{60Hz}[/tex]

T = 0.0166s

PART B)

[tex]T = \frac{1}{f}[/tex]

[tex]T = \frac{1}{8*10^6}[/tex]

[tex]T = 1.25*10^{-7} s[/tex]

PART C)

[tex]T = \frac{1}{f}[/tex]

[tex]T = \frac{1}{140*10^{3}}[/tex]

[tex]T = 7.14*10^{-6}s[/tex]

PART D)

[tex]T = \frac{1}{f}[/tex]

[tex]T = \frac{1}{2.4*10^{9}}[/tex]

[tex]T = 4.166*10^{-10}s[/tex]