When we look at an object that is 1,000 light-years away we see it _________.
a. as it is right now, but it appears 1,000 times dimmer
b. as it was 1,000 light-years ago
c. as it was 1,000 years ago
d. looking just the same as our ancestors would have seen it 1,000 years ago

Answer:

Explanation:

a) The magnitude of the net force on the student = 0 N since the student is standing still on level ground and upward reaction force = downward force.

b) the magnitude of the contact force on the student by the backpack = 80 N since the student was backing the backpack

c) the magnitude of the contact force o the student by the ground = 550 N + 80 N = 630 N reactional force on the student

Final answer:

To find the displacement, subtract the distance traveled in one direction from the distance traveled in the opposite direction.

Explanation:

To solve this problem, we need to consider the concept of displacement. Displacement is a vector quantity that represents the change in position of an object. To find the displacement, we can subtract the distance traveled in one direction from the distance traveled in the opposite direction. In this case, the girl rides 5.4 km due east and then turns around and rides back to her starting point. So her displacement is 5.4 km - 5.4 km = 0 km.

Answer:

T = 3.23 s

Explanation:

In the simple harmonic movement of a spring with a mass the angular velocity is given by

               w = √ K / m

With the initial data let's look for the ratio k / m

The angular velocity is related to the frequency and period

           w = 2π f = 2π / T

            2π / T = √ k / m

            k₀ / m₀ = (2π / T)²

            k₀ / m₀ = (2π / 3.0)²

            k₀ / m₀ = 4.3865

The period on the new planet is

          2π / T = √ k / m

           T = 2π √ m / k

In this case the amounts are

           m = 6 m₀

           k = 10 k₀

We replace

            T = 2π√6m₀ / 10k₀

            T = 2π √6/10 √m₀ / k₀

            T = 2π √ 0.6  √1 / 4.3865

            T = 3.23 s

Answer:

a. 20652 m

b. 1239120 m

c. 0.688 m

Explanation:

First we have to find the speed of sound in air of temperature 20°C.

The speed of sound in a medium is dependent on the temperature of that medium. At any given temperature, the speed of sound is given as:

[tex]v(T) = v(T_{0}) + 0.61T[/tex]

where

v(T) = Speed of sound at any temperature T;

[tex]v(T_{0})[/tex] = Speed of sound at 0°C;

T = temperature of the medium.

The speed of sound in air of 0°C is 332 m/s, hence, in air of temperature 20°C, speed will be:

[tex]v(20) = 332 + (0.61 * 20)\\\\v(20) = 332 + 12.2\\\\v(20) = 344.2 m/s[/tex]

a. Distance traveled in 60 seconds:

distance = speed * time

distance = 344.2 * 60

distance = 20652 m

b. Distance traveled in 1 hour:

1 hour in seconds = 1 * 60 * 60 = 3600 seconds

distance = 344.2 * 3600

distance = 1239120 m

c. Distance traveled in 0.002 second:

distance = 344.2 * 0.002

distance = 0.688 m

Answer:

6613.87 lbs

Explanation:

1.2 m³ = 1200000 cm³

Mass = Density * Volume

M = (2.5 g/cm³) * 1200000 cm³ = 3000000 g

1 lb = 453.592 g

3000000 g * (1 lbs / 453.592 g) = 6613.87 lbs

The mass of the rock is 6613.87 lbs.

Mass is the quantity of matter in a physical body and the weigh is used to measure ascertain the heaviness of by or as if by a balance.

Given that volume of the rock is 1.2 meter cube and density is 2.5 gram/cm cube. The height of the rock is 4 feet. The mass can be calculated as given below.

Mass m = Density w [tex]\times[/tex] Volume v

Substituting the values in the above equation.

[tex]m = 2.5 \;\rm g/cm^3 \times 1.2\;\rm m^3[/tex]

[tex]m = 2.5 \times 1.2 \times 1000000[/tex]

[tex]m = 3000000\;\rm g[/tex]

The mass of the rock is 3000000 g. We know that

[tex]1\;\rm lb = 453.592\;\rm g[/tex]

[tex]1\;\rm g = \dfrac {1}{453.592} \;\rm lb[/tex]

Now convert the mass (grams) into mass (lbs).

[tex]m = \dfrac {1}{453.592}\times 3000000\;\rm lbs[/tex]

[tex]m = 6613.87 \;\rm lbs[/tex]

Hence we can conclude that the mass of the rock is 6613.87 lbs.

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The displacement of thoughts, feelings, fears, wishes, and conflicts from past relationships onto new relationships is called transference.

Answer: Option (b) is the correct answer.

Explanation:

Heat is defined as the energy obtained by the molecules of an object that travels from a hotter body to a colder body.

For example, a metal spoon placed in a cup full of hot tea. Then transfer of heat will take place from the tea to the spoon.

And, this transfer of heat will continue till the temperature of both the objects become equal.

Thus, we can conclude that heat is the energy that moves from a hotter object to a colder object.

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

Resistance of copper wire=R=[tex]13\Omega[/tex]

Resistance is directly proportional to length.

If a wire has greater resistance then,the wire will be greater in length.

Therefore,resistance of long piece of wire=7r

Total resistance of copper  wire=Sum of resistance of two piece of wires

[tex]r+7r=13[/tex]

[tex]8r=13[/tex]

[tex]r=\frac{13}{8}[/tex]ohm

Resistance of long piece of wire=[tex]7\times\frac{13}{8}=\frac{91}{8}\Omega[/tex]

Resistance of short piece of wire =[tex]\frac{13}{8}\Omega[/tex]

Resistivity of wire and cross section area of wire remains same .

Let L be the total  length  of wire and L' be the length of short  piece of wire.

We know that

[tex]R=\frac{\rho L}{A}[/tex]=[tex]\frac{\rho}{A}L=KL[/tex]

[tex]\frac{R}{L}=K[/tex]

Where K=[tex]\frac{\rho}{A}[/tex]=Constant

Using the formula

[tex]\frac{13}{L}=\frac{\frac{13}{8}}{L'}[/tex]

[tex]\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}[/tex]

[tex]L'=\frac{L}{8}[/tex]

Length of short piece of wire=L'=[tex]\frac{L}{8}[/tex]

Length of long piece of  wire=[tex]L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L[/tex]

% of length of short piece of   wire=[tex]\frac{\frac{L}{8}}{L}\times 100=12.5%[/tex]%

The resistance of the short piece=[tex]\frac{13}{8}\Omega[/tex]

The resistance of the long piece=[tex]\frac{91}{8}\Omega[/tex]

The length of short piece wire is [tex]12.5[/tex] % of total wire length.

The resistance of the short piece wire is [tex]\frac{13}{8}ohms[/tex]

The resistance of the long piece wire is, [tex]7*\frac{13}{8}=\frac{91}{8}ohms[/tex]

Let us consider that resistance of short piece wire is r.

So that, resistance of long piece wire is [tex]7r[/tex].

Total resistance of wire is [tex]13[/tex] ohms.

              [tex]7r+r=13\\\\8r=13\\\\r=\frac{13}{8}Ohm[/tex]

Let us consider that total length of wire is 13L.

So that, length of short piece wire [tex]=\frac{13}{8} L[/tex]

length of short piece wire is,

                 [tex]=\frac{\frac{13}{8}L }{13L}*100 =\frac{1}{8} *100=12.5[/tex]

The length of short piece wire is 12.5 % of total wire length.

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Answer: False

Explanation:

It is dependent on the intensity of the radiation i.e the photo electron should increase with the light amplitude.

Answer

a) -2842 J

b) 2842 J

Explanation:

The escalator is moving the man of mass 58 kg till the height of 5 meters

So here;

m=58 kg

distance = height = h= 5 m

gravitational acceleration = g= 9.8 m/sec2

a) We know work is scalar product of force and displacement

i-e Work = F.S= FS cosθ, where θ is the angle between force vector and displacement vector

When escalator is moving up - the gravitational force is acting downward i- opposite to displcacement i-e angle between force and displacement is 180 degrees.

we have data

m= 58 kg

F=mg = 58×9.8=568.4 N

S=5 m

Work = F.S=FS cos 180°=568.4×5×(-1)= -2842 J

So the work done by gravitational force  is -2842 J and -ve sign here indicates that  distance is traveled in opposite direction of force.

b) When escalator is moving up the,   force is exerted by the escalator equal to gravitational force and the displacement is in the same direction so the angle between force and displacement is 0 degrees

Work = F.S=FS cos (0)=568.4×5×1=2842 J

So work done by the gravitational force then will be 2842 J

So in both cases work is same but in opposite directions.

Answer:

C. At the instant the ball reaches its highest point.

Explanation:

When a body is thrown up, it tends to come down due to the influence of gravitational force acting on the body. The body will be momentarily at rest at its maximum point before falling. At this maximum point, the velocity of the body is zero and since force acting on a body is product of the mass and its acceleration, the force acting on the body at that point will be "zero"

Remember, F = ma = m(v/t)

Since v = 0 at maximum height

F = m(0/t)

F = 0N

This shows that the force acting on the body is zero at the maximum height.

Answer:

4.15 m/s

Explanation:

As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:

ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = [tex]\frac{1}{2} *m*vf^{2}[/tex]  (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

[tex]vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s[/tex]

So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.

Final answer:

Betty is moving at a speed of approximately 2.86 m/s when the swing passes through its lowest position.

Explanation:

To calculate the speed at the lowest position of Betty on the swing, we can use the principle of conservation of mechanical energy. At the highest point, all of Betty's potential energy is converted into kinetic energy. So we can write:

Initial potential energy = Final kinetic energy

mgh = 0.5mv^2

where m is the mass of Betty, g is the acceleration due to gravity, h is the initial height, and v is the velocity at the lowest point.

Plugging in the values, we get:

(400 N)(0.21 m) = 0.5(400 N)v^2

After simplifying and solving for v, the velocity at the lowest point is approximately 2.86 m/s.

Answer:

[tex]v_y=-1.2m/s,a_y=-0.76m/s^2[/tex]

Explanation:

We are given that

[tex]y=(0.05x^2)[/tex]

Where x (in m)

x component of velocity, [tex]v_x=-4m/s[/tex]

x- component of acceleration,[tex]a_x=-1.2 m/s^2[/tex]

x=3 m

We have to find the y-component of velocity and acceleration at this instant.

Differentiate w.r.t. time

[tex]\frac{dy}{dt}=0.05\times 2x\frac{dx}{dt}[/tex]

[tex]\frac{dy}{dt}=0.1x\frac{dx}{dt}=0.1xv_x[/tex]

By using the formula

[tex]\frac{dx^n}{dx}=nx^{n-1}[/tex]

[tex]v_x=\frac{dx}{dt}[/tex]

Substitute the values

[tex]v_y=\frac{dy}{dt}=0.1(3)\times (-4)=-1.2m/s[/tex]

Again differentiate w.r.t. t

[tex]a_y=\frac{dv_y}{dt}=0.1v_x+0.1x\frac{d(v_x)}{dt}[/tex]

Substitute the values

[tex]a_y=0.1(-4)+0.1(3)(-1.2)=-0.4-0.36=-0.76m/s^2[/tex]

Where [tex]a_x=\frac{dv_x}{dt}[/tex]

Hence, the y- component of velocity and acceleration

[tex]v_y=-1.2m/s,a_y=-0.76m/s^2[/tex]

To find the y components of velocity and acceleration, differentiate the equation y = 0.05x^2 with respect to time. Plug in the given values to find vy and ay at x = 3m.

To find the corresponding y components of velocity and acceleration at x = 3m, we need to differentiate the equation y = 0.05x^2 with respect to time. Differentiating y with respect to time gives vy = dy/dt = d/dt(0.05x^2) = 0.1x(dx/dt). Plugging in the given values dx/dt = -4m/s and x = 3m, we can find vy. Similarly, to find the y-component of acceleration, we differentiate vy with respect to time, which gives ay = dvy/dt = d/dt(0.1x(dx/dt)) = 0.1(dx/dt)(dx/dt) + 0.1x(d^2x/dt^2). Plugging in the given values dx/dt = -4m/s and d^2x/dt^2 = -1.2m/s^2, we can find ay.

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Answer:

b) twice the energy of each photon of the red light.

Explanation:

[tex]\lambda[/tex] = Wavelength

h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

Energy of a photon is given by

[tex]E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}[/tex]

Let [tex]\lambda_1[/tex] = 700 nm

[tex]\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}[/tex]

For red light

[tex]E_1=\dfrac{hc}{\lambda_1}[/tex]

For UV light

[tex]E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}[/tex]

Dividing the equations

[tex]\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1[/tex]

Hence, the answer is  b) twice the energy of each photon of the red light.

Answer:

b) twice the energy of each photon of the red light.

Explanation:

Given:

wavelength of red light, [tex]\lambda_r=7\times 10^{-7}\ m[/tex]

wavelength of ultraviolet light, [tex]\lambda_{uv}=3.5\times 10^{-7}\ m[/tex]

We know that the energy of a photon is given as:

[tex]E=h.\nu[/tex] ..............(1)

where:

h = plank's constant [tex]=6.626\times 10^{-34}\ J.s[/tex]

[tex]\nu=[/tex]  frequency of the wave

we have the relation:

[tex]\nu=\frac{c}{\lambda}[/tex] ...................(2)

From (1) and (2) we have:

[tex]E=\frac{h.c}{\lambda}[/tex]

Energy of photons for ultraviolet light:

[tex]E_{uv}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{3.5\times 10^{-7}}[/tex]

[tex]E_{uv}=5.6794\times 10^{-19}\ J[/tex]

Since the energy of photons is inversely proportional to the wavelength of the light hence the photon-energy of ultraviolet light is double of the photon-energy of the red light as the wavelength of the red light is twice to that of ultraviolet light.

Answer:

option B.

Explanation:

The correct answer is option B.

when the ball drops, the velocity of the ball before the collision is v

After the collision, the velocity of the ball is the same but in the opposite direction.

Impulse delivered to the ball and the floor, in this case, is not zero.

The magnitude of the momentum remains the same but the direction of the ball changes.

The true statement about the collision between the ball and the floor is that the ball's momentum is conserved.

The term momentum is physics is the product of mass and velocity. Momentum is a vector quantity hence when discussing momentum, we have to take the magnitude and direction into account.

The true statement about the collision between the ball and the floor is that the ball's momentum is conserved.

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Answer:

The magnitude of the displacement is 127.43m and the direction is 38°North of East

Explanation:

Final answer:

To find the magnitude and direction of the fourth displacement in a cave survey, vector components can be used. By breaking down the given displacements into their north and east components, calculating the net displacements, and using the Pythagorean theorem and trigonometry, the magnitude and direction of the fourth displacement can be determined.

Explanation:

To find the magnitude and direction of the fourth displacement, we can use vector components. First, let's break down the given displacements into their north and east components:

Displacement 1: 190 m west (0 north, -190 east)

Displacement 2: 220 m at 45.0° east of south (-220 sin(45°) north, 220 cos(45°) east)

Displacement 3: 270 m at 30.0° east of north (270 sin(30°) north, 270 cos(30°) east)

Next, we can add up the north and east components separately to find the net north and east displacements. Lastly, we can use these net displacements to calculate the magnitude and direction of the fourth displacement using the Pythagorean theorem and trigonometry.

Answer:

Average speed: 0.1 km/min

Explanation:

The average speed of a body is the ratio between the total distance covered and the time taken:

[tex]v=\frac{d}{t}[/tex]

where

d = distance covered

t = time taken

In this problem, we have:

- Distance covered is the sum of the two distances:

d = 6.0 + 1.0 = 7.0 km

- Time taken is the sume of the two times:

t = 54 + 16 = 70 min

Therefore, the average speed is

[tex]v=\frac{7.0}{70}=0.1 km/min[/tex]

Answer:

Coefficient of friction is

Ū = 0.31

Explanation:

T2 = T1* e^(ūơ)

Where T2 = 7500n = tension in the belt, T1 = 150n = reaction force,

ū = coefficient of friction

Ơ = 2pai * N

Where N = number of turns = 2

Ơ = 4pai

7500 = 150e^(ū*4pai)

50 = e^(ū *4pai)

lin 50 = 4pai * ū

Ū = 3.91/4pai

Ū = 0.31

Final answer:

The force exerted by rope R1 on block 1 is equal to the tension in that rope, and if both blocks have the same acceleration and there are no other external forces, this tension is also equal to the force R1 exerts on block 2. However, the force applied through rope R2 on block 2 may be larger because it must account for the total mass and acceleration of both blocks.

Explanation:

When two blocks are connected by a rope and a force is applied to one of the blocks (via a second rope), the tensions in the ropes are subjects of interest in Physics, specifically classical mechanics. In this scenario, the force applied to rope R2 is used to accelerate both blocks 1 and 2. The tension in rope R1, which connects blocks 1 and 2, should be analyzed using a free-body diagram for each block.

Rope R1 exerts a force on block 1, and rope R2 exerts a force on block 2. The force exerted by rope R2 must be sufficient not only to accelerate block 2 but also to account for the force needed to accelerate block 1. Therefore, the force exerted by R2 on block 2 is equal to the sum of the forces necessary to accelerate both blocks. This total force must be greater than or equal to the force that R1 exerts on block 1, as rope R1 only has to provide the force needed to accelerate block 1. However, if we assume the acceleration of the two blocks is the same and the system is not experiencing any external forces other than the applied force, the tension in rope R1 will be equal to the force R1 exerts on block 1 because the massless ropes can only transmit the force equally along their lengths.

Therefore, it can be concluded that the magnitude of the force exerted by rope R1 on block 1 is equal to the magnitude of the tension on block 2 by rope R1, but the magnitude of the applied force through R2 on block 2 could be larger as it has to account for the combined mass of block 1 and block 2.

Answer:

(a) Since net charge remains same,after immersion Q is same

(b) I. 14.56pF ii. 3.05V

(c) ΔU = 5.204nJ

Explanation:

a)

C = kεA/d

k=1 for air

ε is 8.85x10-12F/m

A = .0025m2

d = .125m

C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF

Q = CV = .177pF * 244V = 43.188pC

Since net charge remains same,after immersion Q is same

b)

C = kεA/d, for distilled water k is approx. 80

Cwater = Cair x k

=0.177pF x 80 = 14.16pF

Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V

c) Change in energy: ΔU = Uwater - Uair

Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ

Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ

ΔU = 5.204nJ

Answer:

If there were no greenhouse effect, liquid water would not exist on the surface of the Earth.

The Earth has reached thermal equilibrium, emitting the same amount of energy into space as it absorbs from the Sun.

The more carbon dioxide there is in an atmosphere, the stronger the greenhouse effect will be.

Explanation:

Without greenhouse gases like carbon dioxide, the entire Earth would be well below the freezing temperature of water. These gases slow the radiation of infrared light into space, warming the planet until it reaches thermal equilibrium.

The greenhouse effect is a natural phenomenon that plays a major role in the Earth's climate. It involves the trapping of heat by greenhouse gases in the atmosphere, which raises the Earth's surface temperature. Without the greenhouse effect, the Earth would be much colder.

The greenhouse effect is a natural phenomenon that plays a major role in the Earth's climate. It is similar to the heating of a greenhouse or a car left in the sun with the windows rolled up. The greenhouse gases in the atmosphere, such as carbon dioxide and methane, trap heat from the sun, preventing it from escaping back into space. Without the greenhouse effect, the Earth's average surface temperature would be well below freezing and Earth would be locked in a global ice age.

Greenhouse gases like carbon dioxide and water vapor absorb and emit radiation, which is an important factor in the greenhouse effect. Approximately half of the radiation from the sun passes through these gases in the atmosphere and strikes the Earth. This radiation is converted into thermal radiation on the Earth's surface, which is then partially reflected back to the Earth's surface by greenhouse gases. The more greenhouse gases there are in the atmosphere, the more thermal energy is reflected back to the Earth's surface, causing it to heat up.

The greenhouse effect is responsible for raising the Earth's surface temperature by about 23 °C. Changes in global climate due to the greenhouse effect can lead to more intense storms, changes in precipitation patterns, reduction in biodiversity, and rising sea levels.

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Answer:106.6N

Explanation:

Centripetal force is a force that cause a body to move along a circular path in a field. The body possesses a mass and velocity(since it's moving) and also radius alone the circular path on which it moves.

Given mass (m) = 2kg, velocity (v) = 8m/s and radius (r) = 1.2m

Centripetal acceleration(a)= v²/r

Centripetal force = ma = mv²/r

Substituting this values in the formula above we have;

Centripetal force = 2×8²/1.2m

Centripetal acceleration = 128/1.2

= 106.6N

Answer:

19 800 lbm of carbon dioxide more.

Explanation:

Taurus

Amount of gasoline used in 5 years = 650 ga/year * 5 years = 3250 ga

amount of carbon dioxide released = 19.8 lbm/ga * 3250 ga = 64 350 lbm

Explorer

Amount of gasoline used in 5 years = 850 ga/year * 5 years = 4250 ga

amount of carbon dioxide released = 19.8 lbm/ga * 4250 ga = 84 150 lbm

Extra amount of CO2 released = 84 150 lbm - 64 350 lbm = 19 800 lbm

The extra amount of carbon dioxide emission by a man during his 5 years of trading in Taurus and Explorer is 19,800 lbm.

Taurus  

Amount of gasoline used in 5 years =

[tex]\bold {650\ gallon/year \times 5 = 3250\ gallon}[/tex]

Amount of carbon dioxide released =

[tex]\bold {19.8\ lbm/gallon \times 3250 = 64 350 lbm}[/tex]

Explorer  

Amount of gasoline used in 5 years =

[tex]\bold {850\ gallons/year \times 5 = 4250\ gallons }[/tex]

Amount of carbon dioxide released =[tex]\bold { 19.8\ lbm/gallons \times 4250 gallons = 84 150\ lbm}[/tex]

Extra amount of  [tex]\bold {CO_2}[/tex] released =

[tex]\bold {= 84 150 - 64 350 = 19 800\ lbm}[/tex]

Therefore, the extra amount of carbon dioxide emission by a man during his 5 years of trading in Taurus and Explorer is 19,800 lbm.

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Answer

given,

mass of the rock, m = 30 Kg

Length of the string, r = 0.25 m

angular speed of the rock, ω = 12 rad/s

Tension in the string at the top of the vertical circle = ?

At the top of the string the tension is

[tex]T_{top}=\dfrac{mv^2}{r} - m g[/tex]

we know     v = r ω

[tex]T_{top}= mr\omega^2 - m g[/tex]

[tex]T_{top}= 30\times 0.25\times 12^2-30\times 9.8[/tex]

[tex] T_{top} = 786 N[/tex]

if the given mass is 0.3 Kg then tension in the rope

[tex]T_{top}= 0.30\times 0.25\times 12^2-0.30\times 9.8[/tex]

[tex]T_{top} = 7.86 N[/tex]

If the mass is 0.3 then the correct answer is option A.

Answer: A CD rotate at a varying angular speed

A CD rotates at varying angular speed despite having a constant linear speed. This variation in rotation speed is necessary for the CD to operate efficiently.

Unlike a phonograph record, a CD rotates at varying angular speed even though it scans information at a constant linear speed due to the layout of data on the disc. The rotation speed decreases as the laser moves from the inside to the outside of the CD, maintaining a constant linear speed and adjusting the angular speed accordingly. This change in angular speed is essential for the CD to function effectively.

Answer:From whom should a departing VFR aircraft request radar traffic information during ground operations? ... Answer: Sequencing to the primary Class C airport, traffic advisories, conflict resolution, and safety alerts.

Explanation:

Final answer:

During ground operations, a departing VFR aircraft should request radar traffic information from ground control. Once airborne, the responsibility switches to tower control for traffic updates.

Explanation:

A departing VFR (Visual Flight Rules) aircraft should request radar traffic information from the ground control (also known as ground movement control) before taxiing. Ground control is responsible for the safety on the taxiways and runways, except the active runway. Once airborne or when holding short of the runway, the aircraft switches to tower frequency, at which point the aircraft may receive radar traffic updates relevant to their departure.

Technology utilizing a network of radio frequency waveguides is part of managing air traffic, which is crucial for maintaining a safe separation between aircraft. Furthermore, radar technology can provide directional information, and as noted in research such as Hasselmann, 1991, traffic information can be directly calculated from radar data even during the ground operations of VFR aircraft.

Answer:

As magnification increases, the area of the field of view increases, the depth of the field of view decreases, the working distance decreases, and the amount of light required increases.

As magnification increases, the area of the field of view decreases, the depth of the field of view decreases, the working distance decreases, and the amount of light required increases.

Area of the field of vision: As magnification is increased, the field of view narrows, and you can see less of the object or scene that you are watching. The area you can see reduces as a result of the field of view is smaller.

Depth of field: The range of distances inside an object or scene that simultaneously seem in focus is referred to as depth of field. The depth of field tends to get shallower as magnification gets bigger. This makes it more difficult to simultaneously examine objects at various depths because only a limited range of distances will ever be in focus.

The working distance is the separation between the thing being observed and the objective lens (or magnifying device). The working distance often reduces as magnification rises. To keep the object in focus, you must move it closer to the lens or microscope.

Higher magnification frequently necessitates additional light in order to preserve image quality and visibility. The features become more noticeable when you zoom in on a smaller region, although they might need more light to be clearly visible. At greater magnifications, insufficient light can produce a hazy or dim image, therefore more light would be required.

Hence, As magnification increases, the area of the field of view decreases, the depth of the field of view decreases, the working distance decreases, and the amount of light required increases.

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Answer:

[tex]s=5.79\ km[/tex]

[tex]\theta=47^{\circ}[/tex] east of south

Explanation:

Given:

Now refer the schematic for visualization of situation:

[tex]y=d'.\sin47^{\circ}-d[/tex]

[tex]y=8.4\times \sin47-5.3[/tex] ...............(1)

[tex]x=d'.\cos47^{\circ}[/tex]

[tex]x=8.4\times \cos47^{\circ}[/tex] .................(2)

Now the direction of the desired position with respect to south:

[tex]\tan\theta=\frac{y}{x}[/tex]

[tex]\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}[/tex]

[tex]\theta=47^{\circ}[/tex] east of south

Now the distance from the current position to the desired position:

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

[tex]s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}[/tex]

[tex]s=5.79\ km[/tex]

Answer:

85 N

Explanation:

Given that crate mass = 20kg

Distance = 6m

Time = 3 seconds

Coefficient of kinetic friction = 0.3

We begin by calculating for acceleration

Which was gotten as 1.33 m/s sq

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To determine the tension developed in the cable, we consider forces acting on the crate, including the tension in the cable and the force of kinetic friction. We use Newton's second law of motion and equations for tension and force of kinetic friction to calculate the tension. By substituting values into the equations, we can solve for the tension.

To determine the tension developed in the cable, we need to consider the forces acting on the crate. Since the crate is moving with constant acceleration, we can use Newton's second law of motion, which states that the sum of the forces acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the forces acting on the crate are the tension in the cable and the force of kinetic friction. The tension in the cable can be calculated using the equation:

Tension = mass of the crate * acceleration + force of kinetic friction

Given that the mass of the crate is 20 kg and the acceleration is the change in velocity divided by the time taken (6 m / 3 s = 2 m/s), we can calculate the tension using the equation:

Tension = 20 kg * 2 m/s^2 + force of kinetic friction

We also need to determine the force of kinetic friction. The force of kinetic friction can be calculated using the equation:

Force of kinetic friction = coefficient of kinetic friction * normal force

The normal force is equal to the weight of the crate, which can be calculated using the equation:

Normal force = mass of the crate * gravitational acceleration

Given that the coefficient of kinetic friction is 0.3 and the gravitational acceleration is 9.8 m/s^2, we can calculate the force of kinetic friction using the equation:

Force of kinetic friction = 0.3 * (20 kg * 9.8 m/s^2)

Now we can substitute this value back into the equation for tension:

Tension = 20 kg * 2 m/s^2 + (0.3 * (20 kg * 9.8 m/s^2))

Solving this equation will give us the tension developed in the cable.

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Answer:

4 conditions

Explanation:

The National Electrical Code (NEC) provides safety guidelines for the installation of electrical wiring and electrical equipment in the United States. The purpose behind NEC is to standardize the safe electrical installation practices.

There is always a confusion among people to distinguish between inside or outside of a building or structure. Therefore, NEC has listed 4 conditions in the Article 230.6 where it has mentioned conductors to be considered outside a building or structure.

Conductors are considered outside a building when they are installed:

1. Under not less than 2 inches of concrete beneath a building or structure.

2. Within a building or structure in a raceway that is encased in no less than 2 inches thick of concrete or brick.

3. Installed in a vault that meets the construction requirements of Article 450, Part III.

4. In conduit under not less than 18 inches of earth beneath a building or structure.