A package falls out of an airplane that is flying in a straight line at a constant altitude and speed. If you ignore air resistance, what would be the path of the package as observed by the pilot? As observed by a person on the ground?

The charge on the dust particle is [tex]-2.07 x 10^{-14} C.[/tex]

The dust particle will obtain a charge due to the electric field produced by the charged carpet. Be that as it may, calculating its correct charge requires a few presumptions and steps:

1. Charge density:

To begin with, we ought to calculate the charge density [tex]\sigma[/tex] of the carpet:

[tex]\sigma[/tex] = total Charge / Range = -10 μC / (2.0 m x 4.0 m) = -2.5  μC/m²

2. Electric Field:

The charge thickness creates an electric field (E) over the carpet. Ready to utilize the equation:

E = [tex]\sigma[/tex] / ϵ0

here,  ϵ0 is the permittivity of free space which is equal to [tex]8.85 x 10^{-12[/tex]F/m

Electric field, E = (-2.5 μC/m² / [tex]8.85 x 10^{-12} F/m[/tex]) = ([tex]-2.83 x 10^{5} N/C[/tex])

3. dust particle Charge:

The dust particle will involve an electrostatic force due to the electric field. Since the molecule is suspended, the net force on it must be zero. This implies the electrostatic force must balance the gravitational force acting on the molecule.

Suspicions:

The dust particle could be a circle with uniform charge dissemination.

Discussing resistance is unimportant.

Calculations:

Tidy molecule mass (m): 6.0 μg = [tex]6.0 x 10^{-9} kg[/tex]

Gravitational force (Fg): Fg = m * g (where g is increasing speed due to gravity,= 9.81 m/s²)

Electrostatic force (Fe): Fe = q * E (where q is the charge of the dust particle)

Likening the powers:

Fg = Fe

m * g = q * E

Tackling for q:

q = Fg / E = (m * g) / E = [tex](6.0 x 10^{-9} kg[/tex] * 9.81 m/s²) / ([tex]-2.83 x 10^{5} N/C[/tex]) = [tex]-2.07 x 10^{-14}[/tex] C

Subsequently, the charge on the dust particle is[tex]-2.07 x 10^{-14}[/tex] C.

The electrostatic force between the two charges is [tex]3.4\cdot 10^{-8}N[/tex]

Explanation:

The electrostatic force between two charges is given by Coulomb's law:

[tex]F=k\frac{q_1 q_2}{r^2}[/tex]

where:

[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant

[tex]q_1, q_2[/tex] are the two charges

r is the separation between the two charges

In this problem, we have the following data:

[tex]q_1 = q_2 = 1.00e[/tex] is the magnitude of the two charges, where

[tex]e=1.6\cdot 10^{-19}C[/tex] is the fundamental charge

[tex]r=8.19\cdot 10^{-11}m[/tex] is the separation between the two charges

Substutiting into the equation, we find the force:

[tex]F=(8.99\cdot 10^9)\frac{(1.00\cdot 1.6\cdot 10^{-19})^2}{(8.19\cdot 10^{-11})^2}=3.4\cdot 10^{-8}N[/tex]

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Answer:

λ = 397 nm

Explanation:

given,

Rydberg wavelength equation for Balmer series

[tex]\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})[/tex]

R is the Rydberg constant, R = 1.097 x 10⁷ m⁻¹

n_i = initial energy level  

n_f = final energy level

where as for Balmer series n_f = 2

            n_i = 7

[tex]\dfrac{1}{\lambda}=(1.097\times 10^7)(\dfrac{1}{2^2}-\dfrac{1}{7^2})[/tex]

[tex]\dfrac{1}{\lambda}=(1.097\times 10^7)(\dfrac{1}{2^2}-\dfrac{1}{7^2})[/tex]

[tex]\dfrac{1}{\lambda}=2.5186\times 10^6[/tex]

[tex]\lambda = 3.97\times 10^{-7}[/tex]

Hence, the wavelength is equal to  λ = 397 nm

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.

Answer:

C.

Measure the wear on his treads before and after riding a certain number of laps.

Answer:

Measure the wear on his treads before and after riding a certain number of laps.

Explanation:

By riding in a circular motion the inside of the tire will be in contact with the road more than the outside of the tire. Thus, to see if the constant circular motion had any effect on his tires David should measure the tread depth on both the inside and the outside of the tires before the experiment and measure the inside and the outside of the tires (at the same location on the tires) after the experiment. Then he can compare the tread loss on the inside of the tire to the tread loss on the outside of the tire.

Answer:

s_y = 9.82 m

Explanation:

Given:

- Initial velocity v_i = 43 m/s

- Angle with the horizontal Q = 25 degree

- Initial distance s_o = 1 m

- The distance of the center field fence x_f = 122 m

Find:

- How high above the ground is the ball when it reaches the center field fence

Solution:

- The time taken for the ball to reach the fence t_f:

                             s_x = S(0) + v_x,o*t

                             122 = 0 + (43*cos(25))*t

                              t = 122 / (43*cos(25)) = 3.1305 s

- Compute the height of the ball when it reaches the fence:

                              s_y = S(0) + v_y,o*t + 0.5*g*t^2

                              s_y = 1 + 43*sin(25)*3.1305 - 0.5*(9.81)*(3.1305)^2

                             s_y = 9.82 m

Answer:

A: 50 kg

Explanation:

Answer:

0.08

Explanation:

The alpha particle suffers a head-on collision with the gold nucleus, so it retraces it path after the collision.

Let us take the masses of the particles in atomic mass units.

The initial momentum and kinetic energy of the gold nucleus is 0(since it is stationary). So, applying conservation of momentum and energy, we get the following two equations:

[tex]m_{1}u_{1}=m_{1}v_{1}+m_{2}v_{2}[/tex]      ..........(1)

[tex]\frac{1}{2}m_{1}u_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{2} +\frac{1}{2} m_{2} v_{2}^{2}[/tex]       ..........(2)

where,

[tex]m_{1}[/tex] = mass of the alpha particle = 4 units

[tex]m_{2}[/tex] = mass of the gold nucleus = 197 units

[tex]u_{1}[/tex] = initial velocity of the alpha particle

[tex]v_{1}[/tex] = final velocity of the alpha particle

[tex]v_{2}[/tex] = final velocity of the gold nucleus

Now, we shall substitute the value of [tex]v_{2}[/tex] from equation (1) in equation (2). After some simplifications, we get,

[tex]u_{1}^{2}=v_{1}^{2}+\frac{m_{1}}{m_{2}} (u_{1}^{2}+v_{1}^{2}-2u_{1}v_{1})[/tex]

Dividing both sides by [tex]u_1^2[/tex] and substituting [tex]x=\frac{v_1}{u_1}[/tex] and [tex]k=\frac{m_1}{m_2}[/tex] , we get,

[tex]1=x^2+k(1+x^2-2x)\\[/tex]

or, [tex]x^2(k+1)-2kx+(k-1)=0[/tex]

Here, [tex]k=\frac{m_1}{m_2}=\frac{4}{197}=0.02[/tex]

Therefore, [tex]x=\frac{2(0.02)\pm\sqrt{(2\times0.02)^2-(4\times1.02\times-0.98)} }{2\times1.02}[/tex]

or, [tex]x = 1, -0.96[/tex]

Our required solution is -0.96 because the final velocity([tex]v_1[/tex]) of the alpha particle will be a little less the initial velocity([tex]u_1[/tex]). The negative sign comes as the alpha particle reverses it's direction after colliding with the gold nucleus.

Fractional change in kinetic energy is given by,

[tex]\delta E=\frac{\frac{1}{2} m_1u_1^2-\frac{1}{2}m_1v_1^2 }{\frac{1}{2}m_1u_1^2 }=1-x^2=0.078\approx0.08[/tex]

The alpha particle can lose a significant amount of its kinetic energy in a head-on elastic collision with a gold nucleus due to the gold nucleus's much larger mass. The original kinetic energy of the alpha particle is converted to potential energy before being transferred mostly to the gold nucleus. Specific loss would depend upon the alpha particle's original kinetic energy.

The question pertains to the concept of elastic collisions, specifically between an alpha particle and a gold nucleus. In an elastic collision, both momentum and kinetic energy are conserved. However, while total energy is conserved, individual kinetic energies of colliding particles may change. Since the gold nucleus, which was initially at rest, is significantly more massive (197 units) than the alpha particle (4.0 units), the alpha particle can lose a significant amount of its kinetic energy in a head-on collision.

To calculate the fractional loss of kinetic energy for the alpha particle in this instance, we would use the principle of conservation of kinetic energy and momentum. The kinetic energy of an alpha particle before the collision is transformed into both kinetic and potential energy during the collision as it approaches the gold nucleus until its original energy is converted to potential energy.

Upon collision, a good proportion of this energy is transferred to the gold atom, given its much larger mass. However, necessary calculations would require specific knowledge of the kinetic energy of the alpha particle before the collision, which may vary depending upon the specific nuclear decay process involved.

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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 and Explanation

• Heterotopic protons are those that when substituted by the same substituent, are structurally different. They are not similar, diastereotopic or enantiotopic.

• Diastreotopic protons refers to two protons in a molecule which, if replaced by the same substituent, would generate compounds that are diastereomers. Diastereotopic groups are often, but not always, identical groups attached to the same atom in a molecule containing at least one chiral center.

For example, the two hydrogen atoms of the C3 carbon in (S)-2-bromobutane are diastereotopic (shown in the attached image). Replacement of one hydrogen atom with a bromine atom will produce (2S,3R)-2,3-dibromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the diastereomer (2S,3S)-2,3-dibromobutane.

• Homotopic protons in a compound are equivalent protons. Two protons A and B are homotopic if the molecule remains the same (including stereochemically) when the protons are interchanged with some other atom (substituent) while the remaining parts of the molecule stay fixed. Homotopic atoms are always identical, in any environment.

For example, ethane, the two H atoms on C1 and C2 carbons on the same side (as shown in the attached image) are homotopic as they exhibit the phenomenon described above.

• Enantiotopic protons are two protons in a molecule which, if one or the other were replaced (by the same substituent), would generate a chiral compound. The two possible compounds resulting from that replacement would be enantiomers.

For example, in the attached image to this answer, the two hydrogen atoms attached to the second carbon in butane are enantiotopic. Replacement of one hydrogen atom with a bromine atom will produce (R)-2-bromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the enantiomer (S)-2-bromobutane.

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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:

B. +5.75 m/s

Explanation:

When there are two bodies, a and b, whose velocities measured by a third observer (in this case, the ground) are [tex]V_a[/tex] and [tex]V_b[/tex] respectively, the relative velocity of B with respect to A is given by:

[tex]V_{ba}=V_b-V_a[/tex]

Thus, the velocity of the girl relative to the lawnmower is:

[tex]V_{ba}=6.5\frac{m}{s}-0.75\frac{m}{s}\\V_{ba}=5.75\frac{m}{s}[/tex]

I took the test and got B) +5.75 m/s correct

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: The continuation and the last part of the question is  (3) Determine the expression for E0 by substitution of r0 into the above equation for EN. What is the equation that represents the expression for E0?

Explanation:

The detailed steps and appropriate derivation and by differentiation is shown in the attachment.

The knowledge of differential calculus is applied.

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:

Work Done by the earth's gravitational force on the satellite as it travels from apogee to perigee is

W = F*D*Cos90° = 0

Explanation:

Although there is a change in the kinetic energy of the satellite at the apogee and perigee, the work done by the earth's gravitational force on the satellite is Zero.

W = F.D, F is the gravitational force, D is the displacement. Both F and D are vectors and perpendicular to each other. That is, the angle between F and D is 90°.

Answer:

If the radius of the disk is much greater than the point where the electric field is calculated, then the field of the disk matches the field of the infinite sheet.

Explanation:

First, we have to calculate the electric field of the disk.

We should choose an infinitesimal area, 'da', on the disk and calculate the E-field of this small portion, 'dE'. Then we will integrate dE over the entire disk using cylindrical coordinates.

According to the cylindrical coordinates: da = rdrdθ

The small portion is chosen at a distance r from the axis. Let's find the dE at a point on the axis and a distance z from the center of the disk.

[tex]dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{r^2 + z^2}[/tex]

Here dQ can be found by the following relation: The charge density of the disk is equal to the total charge divided by the total area of the disk. The small portion of the disk will have the same charge density, therefore:

[tex]\frac{Q}{\pi R^2} = \frac{dQ}{da}\\dQ = \frac{Qda}{\pi R^2}[/tex]

Furthermore, we need to separate the vertical and horizontal components of dE, because it is a vector and cannot be integrated without separating the components. By symmetry, the horizontal components of dE will cancel out each other, leaving only the vertical components in the z-direction.

[tex]dE_z = dE\sin(\alpha) = dE \frac{z}{\sqrt{z^2+r^2}}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qzda}{\pi R^2(z^2+r^2)^{3/2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}[/tex]

We have to use a double integral over the radius and the angle to find the total electric field due to a uniformly charged disk:

[tex]E = \int \int dE = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 {\int\limits^R_0 {\frac{1}{(z^2+r^2)^{3/2}}} \, rdr} \, d\theta\\E = \frac{1}{2\epsilon_0}\frac{Q}{\pi R^2}[1 - \frac{1}{\sqrt{(R^2/z^2) + 1}}][/tex]

If the radius of the disk is much greater than the point z, R >> z, than the term in the denominator becomes very large, and the fraction becomes zero. In that case electric field becomes

[tex]E = \frac{1}{2\epsilon_0}\frac{Q}{\pi R^2}[/tex]

This is equal to the electric field of an infinite sheet.

As a result, the condition for the field of a disk to be equal to that of a infinite sheet is R >> z.

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

Answer:

The same

Explanation:

Charges of the same sign repel, while those of different sign attract. So, the magnitude of both electrostatic forces is the same but in the opposite direction. On the other hand, when the force on the charge is exerted by an electric field: If the charge is positive, it experiences a force in the direction of the field; If the load is negative, it experiences a force in the opposite direction to the field. Therefore, the magnitude of both forces is the same but in the opposite direction.

Final answer:

The magnitude of the force is the same on both a negative and positive charge due to Coulomb's Law, but the forces act in opposite directions with attractions between opposite charges and repulsions between like charges.

Explanation:

The magnitude of the force experienced by a negative charge is the same as that experienced by a positive charge when they are acting upon each other. This is because the electric force between two charged particles is dictated by Coulomb's Law, which states that the magnitude of the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This law can be summarized by the equation:

F = k * |q₁ * q₂| / r²

where F is the magnitude of the force, k is Coulomb's constant, q₁ and q₂ are the amounts of the charges and r is the distance between the charges. Importantly, the law indicates only that the magnitudes of the forces are equal; however, the directions will be opposite due to the nature of attraction and repulsion between the charges. Thus, negative and positive charges attract each other, while like charges repel each other.

The force on the negatively charged object, using the formula F = qE, where q is the charge and E is the electric field, will be equal in magnitude but opposite in direction to the force on the positively charged object assuming the charges have the same magnitude. For example, if the electric field is directed eastward, a negative charge will experience a force to the west, while a positive charge will experience a force to the east.

Because of their different masses, a proton and an electron will experience different accelerations due to their different inertia, even though the forces acting on them are of the same magnitude.

Answer:

x = 2 x 10¹⁶ Km

Explanation:

distance between two star,d = 10⁹ Km

separation between them, θ = 10⁻⁵ radians

focal length of the eyepiece = 1.5 cm = 0.015 m

focal length of the objective = 3 m

observer distance from star, x = ?

we know,

[tex]tan \theta = \dfrac{d}{x}[/tex]

for small angle

[tex]\theta = \dfrac{d}{x}[/tex].......(1)

angular magnification of telescope

[tex]M = \dfrac{f_{objective}}{f_{eyepiece}}=\dfrac{3}{0.015} = 200[/tex]

Angular magnification of the telescope is also calculated by

[tex]M = \dfrac{observed\ angle}{original\ angle}[/tex]

[tex]M = \dfrac{\theta_0}{\theta}[/tex]

now,

[tex]\dfrac{\theta_0}{\theta}=200[/tex]

[tex]\dfrac{\theta_0}{200}=\theta[/tex]

from equation (1)

[tex]\dfrac{\theta_0}{200}=\dfrac{d}{x}[/tex]

[tex]x=\dfrac{200d}{\theta_0}[/tex]

[tex]x=\dfrac{200\times 10^9}{10^{-5}}[/tex]

x = 2 x 10¹⁶ Km

Distance between the observer and the star is x = 2 x 10¹⁶ Km

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:

hₘₒₒₙ = 6.05 h

Rₘₒₒₙ = 6.05 R

Explanation:

Let θ be the angle of jump.

Let h and R be maximum height and horizontal range attained on earth respectively.

Let hₘₒₒₙ and Rₘₒₒₙ be the maximum height and horizontal range on the moon respectively

The range for a projectile is given as

R = v₀(x)T = v₀ cos(θ) T

T = (2v₀ sinθ)/g

Range, R = (v₀ cos θ)(2v₀ sinθ)/g = v₀²(2sinθcosθ)/g = v₀² (sin2θ)/g

The maximum range occurs at θ = 45°

Maximum range R = v₀²/g = v₀²/9.8 = 0.102v₀²

On the moon, g = 1.62 m/s²

Maximum range, Rₘₒₒₙ = v₀²/gₘₒₒₙ = v₀²/1.62 = 0.617v₀²

Rₘₒₒₙ = 6.05 R

Maximum Height of a projectile is given as = (v₀² Sin²θ)/2g

θ = 45°; sin 45° = (√2)/2; sin²45° = 2/4 = 1/2

h = v₀²(1/2)/2g = v₀²/4g

On earth, g = 9.8 m/s²

h = v₀²/(4×9.8) = v₀²/39.2 = 0.0255v₀²

On the moon, gₘₒₒₙ = 1.62 m/s²

hₘₒₒₙ = v₀²/(4×1.62) = v₀²/6.48 = 0.154v₀²

hₘₒₒₙ = 6.05 h

Answer:

She must be launched with a speed of 74.2 m/s.

Explanation:

Hi there!

The equations of the horizontal component of the position vector and the vertical component of the velocity vector are the following:

x = v0 · t · cos θ

vy = v0 · sin θ + g · t

x = horizontal distance traveled at time t.

v0 = initial velocity.

t = time.

θ = launching angle.

vy = vertical component of the velocity vector at time t.

g = acceleration due to gravity (-9.8 m/s²).

To just cross the 520-m gap, the maximum height of the flight must be reached halfway of the gap at 260 m horizontally (see attached figure).

When she is at the maximum height, her vertical velocity is zero. So, when x = 260 m, vy = 0. Using both equations we can solve the system for v0:

x = v0 · t · cos θ

Solving for v0:

v0 = x/ (t · cos θ)

Replacing v0 in the second equation:

vy = v0 · sin θ + g · t

0 = x/(t·cos(56°)) · sin(56°) + g · t

0 = 260 m · tan (56°) / t - 9.8 m/s² · t

9.8 m/s² · t = 260 m · tan (56°) / t

t² = 260 m · tan (56°) / 9.8 m/s²

t = 6.27 s

Now, let's calculate v0:

v0 = x/ (t · cos θ)

v0 = 260 m / (6.27 s · cos(56°))

v0 = 74.2 m/s

She must be launched with a speed of 74.2 m/s.

Answer:

it must be launched at a speed of 74.2 m/s

Explanation:

I really hope this helps

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:

W = -1.97KJ, Q = 1.97KJ, Delta U = 0

Delta U = -1.03KJ, Q = -1.03KJ, Delta H = -1.72KJ

Explanation:

The deatiled step by step calculation using the ideal gas equation (Pv =nRT), The first law of thermodynamics ( dQ =dW + dU) as applied is as shown in the attached file.

In the first step, q = -157.29 R mol and w = -2.00 x 10^4 V Pa. In the second step, q = -141.45 R mol and w = 0. The total heat transfer (q_total) is -298.74 R mol and the total work done (w_total) is -2.00 x 10^4 V Pa.

The first step in the process is an isothermal expansion. In an isothermal process, the temperature remains constant, which means the change in internal energy (∆U) is zero. Since ∆U = q + w, this means that q = -w. We can calculate q using the equation q = nCv,m∆T, where n is the number of moles, Cv,m is the molar heat capacity at constant volume, and ∆T is the change in temperature. In this case, q = -w = nCv,m∆T = (1.65 mol)(3R/2)(-35.6 + 14.5) = -157.29 R mol.

The work done during an expansion or contraction process can be calculated using the equation w = -P∆V, where P is the external pressure and ∆V is the change in volume. In this case, the volume doubles, so ∆V = 2V, and the pressure is constant at 1.00 x 10^4 Pa. Therefore, w = -P∆V = -(1.00 x 10^4 Pa)(2V) = -2.00 x 10^4 V Pa.

In the second step, the gas is cooled at constant volume, so no work is done (w = 0). The heat transfer (q) can be calculated using the same equation as before, q = nCv,m∆T. In this case, q = (1.65 mol)(3R/2)(-35.6 - 14.5) = -141.45 R mol.

Putting it all together, for the first step, q = -w = -157.29 R mol and for the second step, q = -141.45 R mol. The total heat transfer for the overall process is the sum of the heat transfers for each step, so q_total = q1 + q2 = (-157.29 R mol) + (-141.45 R mol) = -298.74 R mol. As for the total work done (w_total), it is the sum of the work done in the first step and the work done in the second step, so w_total = w1 + w2 = (-2.00 x 10^4 V Pa) + 0 = -2.00 x 10^4 V Pa.

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Answer:

 Q = 47.06 degrees

Explanation:

Given:

- The transmitted intensity I = 0.464 I_o

- Incident Intensity I = I_o

Find:

What angle should the principle axis make with respect to the incident polarization

Solution:

- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

                                     I = I_o * cos^2 (Q)

- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

                                     Q = cos ^-1 (sqrt (I / I_o))

- Plug the values in:

                                     Q = cos^-1 ( sqrt (0.464))

                                     Q = cos^-1 (0.6811754546)

                                     Q = 47.06 degrees

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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The question deals with the calculation of the magnitude and direction of an electric field necessary to keep a charged object motionless. The two forces acting on the object, namely the gravitational force and the electric force, cancel out making it motionless. The electric field direction is upward as it must counteract the gravitational pull.

In this question, we're examining an object that stays motionless in a uniform electric field. This can be resolved using the equilibrium of forces acting on the object. Given that the object stays motionless, the gravitational force and the electric force on the object should balance each other.

The gravitational force (Fg) experienced by the object is the object mass (m) times the acceleration due to gravity (g), which equals 3.82g * 9.81 m/s². The electric force (Fe) is equal to the charge (q) times the electric field (E), which equals -16.5µC * E.

To find the electric field E, we equate these forces - this gives us

E = Fg / |q|,

where |q| means the absolute value of the charge. The direction of the electric field is taken as the direction of the force that a positive test charge would experience.

Thus, the electric field direction is upwards since the force needed to balance gravity must act against it.

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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:

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