A pendulum oscillating on the moon has the same period as a(n) 4.85 m pendulum oscillating on Earth.

If the moons gravity is one-sixth of Earths gravity, find the length of the pendulum on the moon.
Answer in units of m.

Answer:

200 feet

Explanation:

Answer:

Halley's comet will next appear in the night sky in the year 2062. It orbits the sun every 75-76 years, so this is the time between appearances.

Since Two stars M1 and M2 of equal mass make up a binary star system.  their orbital speed is  2.69E4 m/s.

The orbital period of a binary star system is given by the equation:

T = 2 * pi * sqrt(a³ / (G * (M1 + M2)))

where T is the orbital period, a is the semi-major axis of the orbit (half of the distance between the two stars at their closest approach), G is the gravitational constant, and M1 and M2 are the masses of the two stars.

Substituting the given values into this equation, we get:

3.20 days = 2 * pi * sqrt(a³ / (6.67E-11 m³ kg^⁻¹ s^⁻² * (5.45 sm * 1.99E30 kg/sm + 5.45 sm * 1.99E30 kg/sm)))

Solving for the semi-major axis, we find that a = 2.11E11 m.

The speed of each star in the orbit can be found using the equation:

v = sqrt(G * (M1 + M2) / a)

Substituting the values we have calculated, we find that v = 2.69E4 m/s. This is the speed of each star in the orbit.

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The gauge pressure is equal to the absolute pressure minus the atmospheric pressure. We know that the atmospheric pressure is 14.7 psi, then we have:

Therefore, the absolute pressure is 25.7 psi.

Approximately 3,299.9 J of energy was transferred into thermal energy during the fall of the diver.

To determine the amount of energy transferred into thermal energy during the fall of the diver, we can use the principle of conservation of energy.

The initial potential energy of the diver at the maximum height is given by:

Potential Energy (PE) = m * g * h

Where:

m = mass of the diver (48.0 kg)

g = acceleration due to gravity (9.8 m/s^2)

h = maximum height (11.0 m)

Substituting the given values:

PE = 48.0 kg * 9.8 m/s^2 * 11.0 m = 5,219.2 J

The final kinetic energy of the diver just before hitting the pool is given by:

Kinetic Energy (KE) = (1/2) * m * v^2

Where:

m = mass of the diver (48.0 kg)

v = speed of the diver (8.81 m/s)

Substituting the given values:

KE = (1/2) * 48.0 kg * (8.81 m/s)^2 = 1,919.3 J

The difference between the initial potential energy and the final kinetic energy represents the energy transferred into thermal energy:

Energy transferred into thermal energy = PE - KE

                                     = 5,219.2 J - 1,919.3 J

                                     = 3,299.9 J

Therefore, approximately 3,299.9 J of energy was transferred into thermal energy during the fall of the diver.

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The force that your textbook exerts on the Earth is immeasurably small, since the mass of the textbook is much, much smaller than the mass of the Earth.

Force is a push or pull on an object due to its interaction with another object. Force is a vector quantity that is described by its magnitude and direction. It is typically measured in Newtons and represented by the symbol F. Force can cause an object to accelerate, change its direction, or change its shape. Force is an integral part of physics and is governed by the laws of motion.

This force is equal to the downward pull on the textbook but in the opposite direction, meaning that it is trying to pull the Earth up, canceling out the downward force. However, since the force exerted by the textbook is so small, it is not enough to counteract the pull of gravity, and thus the textbook falls.

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Answer:

a) the velocity increases then decreases.

(a) For the density function, the value of B is 2.5 g/cm³   and the value of C is 1.3 g/cm⁴

(b) The total mass of the rod is 1470 g.

The density of the rod is a function of distance and it is given as;

ρ = B + Cx

where;

ρ = 2.5 g/cm³  +  (19.5 g/cm³ ) / ( 15 cm ) x

ρ = 2.5 g/cm³  +  1.3 g/cm⁴  x

The total mass of the rod is calculated by integrating the function;

dm = ( B + Cx)(8 cm² ) dx

m = 8B + 8Cx

m = 8Bx  +  8Cx² / 2

m = ( 8 x 2.5 x 15 )  +  ( 8 x 1.3 x 15² ) / 2

m = 1470 g

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boy do not say that , who knows

Answer:

7 because salt lake and Southis weat

The velocity of the particle after 5.5 seconds is 11.25 m/s.

To find the velocity of the particle after 5.5 seconds, we can use the equation of motion:

v = u + at

Where:

v = final velocity

u = initial velocity

a = acceleration

t = time

Given:

u = 8.5 m/s (initial velocity)

a = F/m = 125 N / 250 N = 0.5 m/\(s^2\) (acceleration)

t = 5.5 sec (time)

substitute the values:

v = 8.5 m/s + (0.5 m/\(s^2\))(5.5 sec)

v = 8.5 m/s + 2.75 m/s

v = 11.25 m/s

Therefore, the velocity of the particle after 5.5 seconds is 11.25 m/s.

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Answer:

N cos θ = m g           where N is the normal force

N sin θ = m v^2 / R

tan θ = v^2 / (R g)       dividing equations

tan θ = 20.5^2 / (85 * 9.80) = .505

θ = 26.8 deg       banking angle required

The stage of sleep which is marked by long slow delta brain waves is stage 4 and is therefore denoted as option B.

This is referred to as a sedentary state of mind and body and it is characterized by altered consciousness, reduced muscle activity and reduced interactions with surroundings.

In the deepest level of sleep, stage IV sleep it has been observed that it consists of low frequency and high-amplitude fluctuations called delta waves and the sequence from drowsiness to deep stage IV sleep usually takes about an hour thereby making it the correct choice.

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Explanation: Magma is molten rock found below the earth's surface. ... On the other hand, if the rocks are under greater pressure, they will require higher temperatures to melt. Melting of rocks typically occurs in the lower lithosphere or upper asthenosphere. The earth gets hot pretty quickly as you dig down from the earth's surface.

Resistance of the resistors, R₁ and R₂ are 80Ω and 200Ω respectively.

V(A) = 12 V

ΔV(B) = 2 V

ΔV(C) = 5 V

R₃ = 200Ω

Since, the total voltage-drop along the upper branch must be 12 V, according to loop rule, the voltage-drop across resistor 3 is 5.0V

So, current through the loop,

I = V(C)/R₃

I = 5/200

I = 25 x 10⁻³A

a) Therefore, the resistance,

R₁ = V(B)/I

R₁ = 2/25 x 10⁻³

R₁ = 80Ω

b) Resistor 2 has the same voltage-drop as resistor 3. So, its resistance, R₂ is 200Ω.

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If the impulse is strong enough and lasts for a sufficient amount of time, the go-kart will eventually come to a stop.

Option A is correct.

impulse is described as the integral of a force, F, over the time interval, t, for which it acts. Since force is a vector quantity, impulse is also a vector quantity.

If the force is insufficient to stop the go-kart entirely, it will slow down but continue to move. The force and duration of the impulse, along with the mass and speed of the go-kart, will all affect how much deceleration occurs.

Given that momentum plus impulse equals zero, the go-kart's change in momentum as a result of the impulse will be equal in amount but will move in the opposite direction of its original momentum.

As a result, the go-kart's final momentum will be zero, suggesting that it has either stopped or is travelling very slowly.

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a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

a) The mass flow rate through the nozzle can be calculated with the following equation:

\( \dot{m_{i}} = \rho_{i} v_{i}A_{i} \)

Where:

\(v_{i}\): is the initial velocity = 20 m/s

\(A_{i}\): is the inlet area of the nozzle = 60 cm²  

\(\rho_{i}\): is the density of entrance = 2.21 kg/m³

\( \dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s \)  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

\( \rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f} \)

\( 0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f} \)

\( A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2} \)

Therefore, the exit area of the nozzle is 23.6 cm².

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a) Mass flow rate through the nozzle: 0.265 kilograms per second, b) Exit area of the nozzle: 23.202 square centimeters.

We determine the Mass Flow Rate through the nozzle and the Exit Area of the nozzle by means of the Principle of Mass Conservation. A nozzle is a device that works at Steady State, so that Mass Balance can be reduced into this form:

\(\dot m_{in} = \dot m_{out}\) (1)

Where:

\(\dot m_{in}\) - Inlet mass flow, in kilograms per second.

\(\dot m_{out}\) - Outlet mass flow, in kilograms per second.

Given that air flows at constant rate, we expand (1) by dimensional analysis:

\(\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}\) (2)

Where:

\(\rho_{in}, \rho_{out}\) - Air density at inlet and outlet, in kilograms per cubic meter.

\(A_{in}, A_{out}\) - Inlet and outlet area, in square meters.

\(v_{in}, v_{out}\) - Inlet and outlet velocity, in meters per second.

a) If we know that \(\rho_{in} = 2.21\,\frac{kg}{m^{3}}\), \(A_{in} = 60\times 10^{-4}\,m^{2}\) and \(v_{in} = 20\,\frac{m}{s}\), then the mass flow rate through the nozzle is:

\(\dot m = \rho_{in}\cdot A_{in}\cdot v_{in}\)

\(\dot m = \left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)\)

\(\dot m = 0.265\,\frac{kg}{s}\)

The mass flow rate through the nozzle is 0.265 kilograms per second.

b) If we know that \(\rho_{in} = 2.21\,\frac{kg}{m^{3}}\), \(A_{in} = 60\times 10^{-4}\,m^{2}\), \(v_{in} = 20\,\frac{m}{s}\), \(\rho_{out} = 0.762\,\frac{kg}{m^{3}}\) and \(v_{out} = 150\,\frac{m}{s}\), then the exit area of the nozzle is:

\(\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}\)

\(A_{out} = \frac{\rho_{in}\cdot A_{in}\cdot v_{in}}{\rho_{out}\cdot v_{out}}\)

\(A_{out} = \frac{\left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)}{\left(0.762\,\frac{kg}{m^{3}} \right)\cdot \left(150\,\frac{m}{s} \right)}\)

\(A_{out} = 2.320\times 10^{-3}\,m^{2}\)

\(A_{out} = 23.202\,cm^{2}\)

The exit area of the nozzle is 23.202 square centimeters.

a) It is the error present in the measuring instrument that causes it to register a value even when there is no input or output being measured.

b) An example of when you would have to allow for zero error is when using a measuring instrument like a vernier caliper or micrometer screw gauge.

a) Zero error refers to the deviation or discrepancy in the measurement instrument, where the indication or reading on the instrument is not zero when the quantity being measured is zero. In other words, it is the error present in the measuring instrument that causes it to register a value even when there is no input or output being measured.

Zero error can occur due to various reasons such as manufacturing defects, wear and tear, misalignment, or improper calibration of the instrument. It can be positive or negative, depending on whether the instrument reads higher or lower than the actual value.

b) An example of when you would have to allow for zero error is when using a measuring instrument like a vernier caliper or micrometer screw gauge. These instruments are commonly used to measure the dimensions of objects with high precision.

In a vernier caliper, for instance, zero error can occur when the jaws do not close perfectly when there is no object being measured. If the caliper shows a reading other than zero when the jaws are closed, it indicates the presence of zero error.

To obtain accurate measurements, the zero error needs to be accounted for and compensated. This can be done by adjusting the position of the zero on the scale or by subtracting the zero error value from the measured readings.

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Answer:

    vₐ = v_c  \(( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )\)

Explanation:

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

          Em₀ = K + U = ½ m v_c² - G Mm / R

final point. At a very distant point

         Em_f = U = - G Mm / R₂

energy is conserved

           Em₀ = Em_f

           ½ m v_c² - G Mm / R = - G Mm / R₂

           v_c² = 2 G M (1 /R -  1 /R₂)

if we consider the speed so that it reaches an infinite position R₂ = ∞

           v_c = \(\sqrt{\frac{2GM}{R} }\)

now indicates that the mass and radius of the planet changes slightly

            M ’= M + ΔM = M ( \(1+ \frac{\Delta M}{M}\) )

            R ’= R + ΔR = R ( \(1 + \frac{\Delta R}{R}\) )

we substitute

           vₐ = \(\sqrt{\frac{2GM}{R} } \ \frac{\sqrt{1+ \frac{\Delta M}{M} } }{ \sqrt{1+ \frac{ \Delta R}{R} } }\)

let's use a serial expansion

           √(1 ±x) = 1 ± ½ x +…

we substitute

         vₐ = v_ c ( \((1 + \frac{1}{2} \frac{\Delta M}{M} ) \ ( 1 - \frac{1}{2} \frac{\Delta R}{R} )\))

we make the product and keep the terms linear

        vₐ = v_c  \(( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )\)

The minimum escape speed, vc , for an object launched from the surface of the planet will be    \(v_a=v_c(1+\dfrac{1}{2}(\dfrac{\Delta M}{M}-\dfrac{\Delta R}{R})\)

The escape velocity is defined as the velocity required to send the object out of the gravitational influence of the earth.

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

    \(E_{mo} = K + U = \dfrac{1}{2} m v_c^2 - \dfrac{G Mm} { R}\)

final point. At a very distant point

      \(E_{mf} = U = \dfrac{- G Mm }{ R_2}\)

energy is conserved

         \(E{mo} = E{mf}\)

          \(\dfrac{1}{2}m v_c^2 - \dfrac{G Mm} {R} = \dfrac{- G Mm }{ R_2}\)

\(v_c^2 = 2 G M (\dfrac{1} {R} - \dfrac{ 1 }{R_2})\)

if we consider the speed so that it reaches an infinite position R₂ = ∞

\(v_c = \sqrt{\dfrac{2GM}{R}\)

now indicates that the mass and radius of the planet changes slightly

\(M ’= M + \Delta M = M(1+\dfrac{\Delta M}{M})\)  

\(R ’= R + \Delta R = R (1+\dfrac{\Delta R}{R} )\)

we substitute

   \(vₐ = \sqrt{\dfrac{2GM}{R} }\dfrac{\sqrt{1+\dfrac{\Delta M}{M}}} {\sqrt{1+\dfrac{\Delta R}{R}}}\)

let's use a serial expansion

√(1 ±x) = 1 ± ½ x +…

we substitute

      \(v_a=v_c(1+\dfrac{1}{2}\dfrac{\Delta M}{M})(1-\dfrac{1}{2}\dfrac\Delta R}{R})\)

Hence the minimum escape speed, vc , for an object launched from the surface of the planet will be    \(v_a=v_c(1+\dfrac{1}{2}(\dfrac{\Delta M}{M}-\dfrac{\Delta R}{R})\)

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Answer:

32 degrees Fahrenheit

Snow forms when the atmospheric temperature is at or below freezing  and there is a minimum amount of moisture in the air. If the ground temperature is at or below freezing, the snow will reach the ground.

Explanation:

The new gravitational force between the two planets, when they are moved to two million miles apart, is 250,000 N

The gravitational force between two objects can be calculated using Newton's Law of Universal Gravitation, which states that the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Given:

Initial distance between the planets = 1 million miles

Initial gravitational force = 1,000,000 N

Final distance between the planets = 2 million miles

To determine the new gravitational force, we need to compare the ratios of the distances and apply the inverse square law.

Let's denote the initial distance as d1, the initial gravitational force as F1, the final distance as d2, and the unknown final gravitational force as F2.

According to the inverse square law, the ratio of the gravitational forces is the square of the ratio of the distances:

(F2/F1) = (d1/d2)²

Substituting the given values:

(F2/1,000,000 N) = (1 million miles / 2 million miles)²

Simplifying:

(F2/1,000,000 N) = (1/2)²

(F2/1,000,000 N) = 1/4

F2 = (1/4) * 1,000,000 N

F2 = 250,000 N

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The amount of energy needed to power a 0.20 kW bulb for one minute would be just sufficient to lift a 2.5 kg object through a vertical distance of approximately 29.03 meters.

To calculate the energy required to lift a 2.5 kg object through a vertical distance, we need to consider the gravitational potential energy formula:

Potential energy (PE) = mass (m) × gravity (g) × height (h)

Where:

m = 2.5 kg (mass of the object)

g = 9.8 m/s² (acceleration due to gravity on Earth)

h = ? (height)

First, let's find the height (h) by rearranging the formula:

h = PE / (m × g)

Now, let's calculate the potential energy (PE) needed to lift the object. We are given that the power of the bulb is 0.20 kW, and we want to find the energy required for one minute. To convert kilowatts (kW) to joules (J), we multiply by the conversion factor of 3,600 (60 seconds × 60 minutes):

Energy (E) = power (P) × time (t)

E = 0.20 kW × 1 min × 3,600 J/kW

Now, we can substitute the values into the equation to find the height:

h = (0.20 kW × 1 min × 3,600 J/kW) / (2.5 kg × 9.8 m/s²)

Calculating the expression on the right side:

h ≈ 0.20 × 1 × 3,600 / (2.5 × 9.8) ≈ 29.03 meters (rounded to two decimal places)

Therefore, the amount of energy needed to power a 0.20 kW bulb for one minute would be just sufficient to lift a 2.5 kg object through a vertical distance of approximately 29.03 meters.

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Answer:

T = 80√3 N ≈ 139 N

W = 160 N

Explanation:

Sum of forces on B in the x direction:

∑F = ma

80 N sin 60° − T sin 30° = 0

T = 80 N sin 60° / sin 30°

T = 80√3 N

T ≈ 139 N

Sum of forces on B in the y direction:

∑F = ma

80 N cos 60° + T cos 30° − W = 0

W = 80 N cos 60° + T cos 30°

W = 40 N + 120 N

W = 160 N

The frequency of oscillation of the swing is 0.25 Hz.

The time period of oscillation of the swing is 4 s.

The length of the vine of the swing is 3.97 m.

The restoring force acting on Mohammed is 692.9 N.

Mass of Mohammed, m = 50 kg

Angle made by the vine with the vertical, θ = 45°

Number of complete swings made by Mohammed, n = 30

Time taken for this swing, t = 2 minutes = 120 seconds

a) The frequency of the swing is defined as the number of complete oscillations in one second.

So, the frequency of oscillation of the swing is,

f = n/t

f = 30/120

f = 0.25 Hz

b) The time period of oscillation of the swing is,

T = 1/f

T = 1/0.25

T = 4 s

c) The expression for the time period is given by,

T = 2π√(l/g)

T² = 4π² x (l/g)

l/g = T²/4π²

Therefore, the length of the vine of the swing is,

l = T²g/4π²

l = 4² x 9.8/4 x (3.14)²

l = 3.97 m

d) The restoring force acting on Mohammed,

F = mg sinθ

F = 50 x 9.8 x sin 45°

F = 490 x 1/√2 = 490/1.414

F = 692.9 N

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The examples of ohmic conductors​ are :

Silver(Ag), copper(Cu), aluminium(Al) etc.

Ohmic conductors means which conductors strictly obey the Ohm's law, are known as Ohmic conductors.  

Resistance= \(\frac{Voltage}{Current}\)

In other words there is a linear relationship between voltage and current for all values. That means all the ohmic conductor materials shows a linear character in the V-I characteristic graph.

So, the examples of ohmic conductors are:

Silver(Ag), copper(Cu), aluminium(Al) etc.

All the conductors we mentions here, they all show a linear line in the V-I characteristic graph.

From the above discussion we can conclude that, Silver(Ag), copper(Cu), aluminium(Al) etc. these are the ohmic conductors.

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Yes, yes it does. It's cool right?

Answer:

The word is "wind".

Explanation:

The sentence would be:

"The sun is also the reason wind exists. The wind is moving air."

Hope this helped !

Answer:

W = 650 [kJ]

Explanation:

The definition of work is denoted by the product of force by the distance traveled by the body, this distance traveled corresponds to the direction of the force.

In this case we have:

d = distance = 2.6[km] = 2600 [m]

F = force = 250 [N]

W = F*d = 250 * 2600 = 650000 [J] or 650 [kJ]