an elastic cord vibrates with a frequency of 3.0 hz when a mass of 0.60 kg is hung from it. what is its frequency if only 0.38 kg hangs from it?

Answer:

a) The vertical component of the helicopters speed relative to the air = 0 m/s

The horizontal component of the helicopters speed relative to the air = -42.5 m/s

The vertical component of the wind speed relative to the ground = -21.65 m/s

The horizontal component of the wind  speed relative to the ground = 12.5 m/s

b) The x-component of the helicopter speed relative to the wind speed =  -30 m/s

The y-component of the helicopter speed relative to the wind speed =  -12.5 m/s

c) The resultant speed of the helicopter relative to the ground is 32.5 m/s in the direction of W 22.6° S

Explanation:

The speed of the helicopter = 42.5 m/s

The direction of flight of the helicopter = West

The wind speed = 25.0 m/s

The direction of the wind = E30°S

a) The velocity components of the helicopters speed relative to the air are;

Vertical component = 0 m/s

Horizontal component = -42.5 m/s

The velocity components of the wind speed relative to the ground are;

Vertical component = -25 × cos(30) = -21.65 m/s

Horizontal component = 25 × sin(30) = 12.5 m/s

b) To find the speed of the helicopter relative to the wind we have;

x-component of the helicopter speed = -42.5 m/s

y-component of the helicopter speed = -0

x-component of the wind speed relative to the ground = 25×sin(30) = 12.5 m/s

y-component of the wind speed relative to the ground = -25×cos(30) = -21.65 m/s

Therefore, the x-component of the helicopter speed relative to the wind speed = -42.5 + 12.5= -30 m/s

the y-component of the helicopter speed relative to the wind speed = 0 +  (-12.5) = -12.5 m/s

c) The resultant speed of the helicopter relative to the ground is given by the relation;

\(v_{h, ground} =\sqrt{(v_{h,horizontal})^2 + (v_{h,vertical})^2}\)

= √((-30)² + (-12.5)²) =32.5 m/s

The direction of motion from the x-axis is given as follows;

\(tan(x) = \dfrac{Vertical \ velocity \ component}{Horizontal \ velocity \ component} = \dfrac{-12.5}{-30} = \dfrac{5}{12}\)

\(x = tan^{-1} \left (\dfrac{5}{12} \right ) = 22.62 ^{\circ}\)

Therefore, the helicopter is flying at 32.5 m/s in the direction of W 22.6° S.

The sound pressure ratio is 10¹⁰

A decibel is a unit of sound pressure level. It can be determined as

LP = log( P / Po )²

where LP is sound pressure level (dB), P is sound pressure (N/m²) and Po is reference sound pressure (N/m²).

From the question above, we know that :

LP1 = 80 dB

LP2 = 100 dB

Po is same

To determine the sound pressure ratio we can use the sound pressure level equation:

LP1 = log( P1 / Po )²

80 = log( P1 / Po )²

log( 10 )⁸⁰ = log( P1 / Po )²

10⁸⁰ = ( P1 / Po )²

P1/Po = 10⁴⁰ ........................... (1)

LP2 = log( P2 / Po )²

100 = log( P2 / Po )²

log( 10 )¹⁰⁰ = log( P2 / Po )²

10¹⁰⁰ = ( P2 / Po )²

P2/Po = 10⁵⁰ ........................... (2)

Hence, we can determine the ratio of sound pressure

P2 / P1 = 10⁵⁰.Po /  10⁴⁰.Po

P2/P1 = 10¹⁰

Thus, the sound pressure ratio is 10¹⁰.

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The substance will be reduced to 1/16 of A present, that is 1 lb, in 95 hours.

Let the initial amount of A present be X lb. The rate of decomposition of A is proportional to the amount of A present. Therefore, the rate of decomposition = k * X where k is the proportionality constant. We know that 1 lb of A will reduce to 4 lb in 38 hours. So, the rate of decomposition = X/38.

Also, the rate of decomposition = k * X. Comparing both the equations, k = 1/38. Therefore, the rate of decomposition = X/38A substance will reduce to 1/16 of A present i.e., X/16. Using the equation for the rate of decomposition, we get, X/16 = (1/38)*X*(t). Simplifying, we get t = 95 hrs. Hence, the substance will be reduced to 1/16 of A present, that is 1 lb, in 95 hours.

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Answer:

D. A contrast between the power of a black hole and the power of a whirlpool is being made.

Answer:

D

Explanation:

Answer:

both statements are truth

Explanation:

a-The Joule effect, also called Joule's law, is the thermal manifestation of electrical resistance. ... In all these cases, it is intended to generate thermal energy with electricity passing through its conductors. This heat they give off is due to the Joule effect.

b-sure of a liquid tank depends only on the density of the liquid and depth from the free surface. It is a scalar quantity and is same in all directions, at a point.

Answer:

t= 278/17 =16.35s

Explanation:

p=w/t

t=w/p

∴t=278/17

Answer:

t=16.4

Explanation:

power = energy/time

p=e/t

17=278/t

t=278/17

t=16.352941

t=16.35

t=16.4

The centre of mass of the region is bounded by y=9-x^2 y=5/2x, and the z-axis is (3.5, 33/8). Formulae used to find the centre of mass are as follows:x bar = (1/M)*∫∫∫x*dV, where M is the total mass of the system y bar = (1/M)*∫∫∫y*dVwhere M is the total mass of the system z bar = (1/M)*∫∫∫z*dV, where M is the total mass of the systemThe region bounded by y=9-x^2 and y=5/2x, and the z-axis is shown in the attached figure.

The two curves intersect at (-3, 15/2) and (3, 15/2). Thus, the total mass of the region is given by M = ∫∫ρ*dA, where ρ = density. We can assume ρ = 1 since no density is given.M = ∫[5/2x, 9-x^2]∫[0, x^2+5/2x]dAy bar = (1/M)*∫∫∫y*dVTherefore,y bar = (1/M)*∫[5/2x, 9-x^2]∫[0, x^2+5/2x]y*dA= (1/M)*∫[5/2x, 9-x^2]∫[0, x^2+5/2x]ydA...[1].

The limits of integration in the above equation are from 5/2x to 9-x^2 for x and from 0 to x^2+5/2x for y.To evaluate the above integral, we need to swap the order of integration. Therefore,y bar = (1/M)*∫[0, 3]∫[5/2, (9-y)^0.5]y*dxdy...[2].

The limits of integration in the above equation are from 0 to 3 for y and from 5/2 to (9-y)^0.5 for x.Substituting the values and evaluating the integral, we get y bar = (1/M)*[(9-5/2)^2/2 - (9-(15/2))^2/2]= (1/M)*(25/2)...[3].

Also, the x coordinate of the center of mass is given by,x bar = (1/M)*∫∫∫x*dVTherefore,x bar = (1/M)*∫[5/2x, 9-x^2]∫[0, x^2+5/2x]x*dA= (1/M)*∫[5/2x, 9-x^2]∫[0, x^2+5/2x]xdA...[4].

The limits of integration in the above equation are from 5/2x to 9-x^2 for x and from 0 to x^2+5/2x for y.To evaluate the above integral, we need to swap the order of integration. Therefore, x bar = (1/M)*∫[0, 3]∫[5/2, (9-y)^0.5]xy*dxdy...[5].

The limits of integration in the above equation are from 0 to 3 for y and from 5/2 to (9-y)^0.5 for x.

Substituting the values and evaluating the integral, we get x bar = (1/M)*[63/8]= (1/M)*(63/8)...[6]Thus, the centre of mass of the region is bounded by y=9-x^2 y=5/2x, and the z-axis is (3.5, 33/8).

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The low-point station has an elevation of 110.32 feet in the given vertical curve with a g1 grade of -2% and a g2 grade of +3%.

To find the elevation of the low-point station in the given vertical curve, we start with the provided data. The g1 grade is -2%, indicating a downward slope, while the g2 grade is +3%, indicating an upward slope. The BVC Station is located at 16+50, with an elevation of 112.00 feet. The length of the curve is given as 400.00 feet. To calculate the elevation at the low-point station, we consider the change in grade from g1 to g2 along the curve. The low-point station represents the transition point where the slope changes from descending to ascending. Using vertical curve calculations, we determine the elevation at the low-point station to be 110.32 feet. This means that the road reaches its lowest point at this station before it starts to ascend again.

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The new position of the Arctic and Antarctic Circles would be located at 30° North and 30° South, respectively.

If the tilt of Earth's equator relative to its orbit were 30° instead of 23.5°, the location of the Arctic and Antarctic Circles would be affected. The Arctic Circle is defined as the latitude above which the sun does not set on the summer solstice and does not rise on the winter solstice. The same applies to the Antarctic Circle but in the Southern Hemisphere.

Currently, the Arctic Circle is located at 66.5° North and the Antarctic Circle is located at 66.5° South. If the tilt of Earth's equator relative to its orbit were 30°, the position of the Arctic and Antarctic Circles would be closer to the equator.

To determine the new position of the Arctic and Antarctic Circles, we can use the following formula:

θ = 90° - ϕ

where θ is the angle between the axis of rotation and the line perpendicular to the plane of the ecliptic, and ϕ is the tilt of the Earth's axis relative to the plane of the ecliptic.

If the tilt were 30°, then θ would be:

θ = 90° - 30° = 60°

Therefore, the new latitude of the Arctic and Antarctic Circles would be:

ϕ = 90° - 60° = 30°

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The new position of the Arctic and Antarctic Circles would be located at 30° North and 30° South, respectively.

If the tilt of Earth's equator relative to its orbit were 30° instead of 23.5°, the location of the Arctic and Antarctic Circles would be affected. The Arctic Circle is defined as the latitude above which the sun does not set on the summer solstice and does not rise on the winter solstice. The same applies to the Antarctic Circle but in the Southern Hemisphere.

Currently, the Arctic Circle is located at 66.5° North and the Antarctic Circle is located at 66.5° South. If the tilt of Earth's equator relative to its orbit were 30°, the position of the Arctic and Antarctic Circles would be closer to the equator.

To determine the new position of the Arctic and Antarctic Circles, we can use the following formula:

θ = 90° - ϕ

where θ is the angle between the axis of rotation and the line perpendicular to the plane of the ecliptic, and ϕ is the tilt of the Earth's axis relative to the plane of the ecliptic.

If the tilt were 30°, then θ would be:

θ = 90° - 30° = 60°

Therefore, the new latitude of the Arctic and Antarctic Circles would be:

ϕ = 90° - 60° = 30°

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Answer:

B

Explanation:

In this process on merry go round there is not any external torque so angular momentum will be conserve. Mass is always conserved.

The net charge, or the total of the positive and negative charges, is constant in an isolated system because electric charge is a conserved attribute. Electrical charge is transported by subatomic particles. In ordinary matter, the protons in the atoms' nucleus contain positive charge whereas the electrons carry negative charge.

Plate's surface area is 0.13 x 0.13.

= × m²

400 m is the distance between the plates (0.40 mm).

3.0 MV/m is the critical field strength in the case of air.

Let's calculate the voltage when using air as the dielectric.

V = 3××400×V = 1200 V

We now discover capacitance.

C = εA/d

C = (8.854×)(16.9×)/(400×)

= 374× F = 374pF

Let's generate energy.

E = (1/2)(374×)(1200) (1200)

E = 269.28 × J

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Answer:

0.20 m.

Explanation:

Given: R = 30 Ω and L = 1 m

Resistance of 1m wire is R = V/I = 30Ω/0.1A

6Ω resistor will have the same voltage drop across it as 1m wire, so we require I to be 6A. Therefore, the required length of wire  is:

L = (6A)/(30Ω) = 0.2m

Answer:

LETTER C IS THE ANSWER

Explanation:

this is my answer

The wavelength of the sound produced by the car's siren is 0.5 m.

Frequency of the siren = 680 hertz

Speed of sound = 340m per second

The separation between two comparable (identical) points on adjacent waves are referred to as a wavelength. The distance between succeeding crests of an electromagnetic wave, more especially its wavelength, is measured in points. Wavelength and frequency have a strong relationship. The wavelength gets smaller as the frequency increases.

Calculating the wavelength -

Wavelength = Speed of Sound / Frequency

Substituting the values -

= 340 / 680

= 0.5

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Answer:

The nuclear division processes of mitosis and meiosis occur during cell division. During mitosis, body cells are divided, while during meiosis, s3x cells are divided. At mitosis, a cell divides once; at meiosis, it divides twice.

Explanation:

An appropriate speed for glacial movement generally is six centimeters a year.

A glacier is a vast, slow-moving mass of snow and ice that collects in mountain valleys and spreads outwards, frequently flowing like a very slow river. It forms when snow accumulation exceeds snowmelt, and the compacted snow transforms into ice, a process known as "firnification." This ice subsequently flows downhill under the weight of additional snow accumulation, occasionally for hundreds of miles.

Glaciers can move up to several meters per day, but they typically move at a much slower pace. Even though the speed of a glacier might vary widely based on factors like slope, basal conditions, temperature, and ice thickness, a reasonable speed for glacial movement is six centimeters a year. Furthermore, the rate of movement can vary depending on the time of year and the time of day.

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No work is done to move the ball to the right.

Initial velocity = -  30 m/s

Final velocity = + 30 m/s

Kinetic energy :

ΔK = 1/2 mv² - 1/2 mu²

Δk = 1/2m ( v² - u² )

ΔK = 1/2 × m ( 30² - 30² )

ΔK = 0

thus no work is done on the ball

However, the work done to bring the ball to stop is :

W = 0.5 × 0.02 × - 900

W = - 9 J

Work done for the acceleration of the ball to come to rest is :

W2 = + 9 J

The work done again  would be :

Work done = - 9 + 9 = 0 Joule

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The energy transferred when the potential difference is 3000mV and the charge is 2 C is 6 J.

It is given in the question that:

The potential difference, V = 3000 mV

The charge, q = 2 C.

We have, 1000 mV = 1 V

Therefore, V = 3000 mV = 3 V

Energy is related to potential difference and charge by the formula,

Energy = Charge \(\times\) Potential Difference

Or, \(E=qV\)

Substituting the values of charge and potential difference as given,

E = 2 \(\times\) 3 = 6 J

Therefore, the energy is transferred when the potential difference is 3000 mV volts and the charge is 7.5 C is 90 Joules.

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a) When an identical resistor is added in series with the first resistor and the cell, the current through the resistors will remain the same as A amps.

When an identical resistor is added in series with the first resistor and the cell, the current through the resistors will remain the same as A amps.

This is because in a series circuit, the same current flows through each component. The total resistance of the circuit increases when a resistor is added in series, but the voltage across the resistors remains the same as the voltage across the cell.

According to Ohm's law, V = IR, where V is voltage, I is current, and R is resistance.

Since the voltage is constant and the resistance has increased, the current through the resistors must remain the same to satisfy Ohm's law. Therefore, the current through the second resistor will also be A amps.

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The correct statement that explains the observation using Newton's laws is objects at rest tend to remain in their current state of motion unless acted upon by an unbalanced force, but objects in motion require a continual application of force to stay in motion. Here option A is correct.

According to Newton's first law of motion, an object will continue moving at a constant velocity in a straight line unless acted upon by an external force. In this case, when the cannonball is shot horizontally from the cannon, it initially possesses a forward velocity due to the force applied by the cannon. However, once the cannonball is in motion, the only forces acting on it are gravity and friction.

Gravity acts vertically downward, causing the cannonball to accelerate downward. Friction acts horizontally in the opposite direction to the motion of the cannonball. As the cannonball moves forward, friction opposes its motion and gradually slows it down.

Since there is no force continuously propelling the cannonball forward, and the forces of friction and gravity act on it, the cannonball eventually comes to a stop and falls to the ground. Hence option A is correct.

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The resistance of the circuit is 24 ohms.

When an electric current flows through a conductor, it encounters resistance, which causes some of the electrical energy to be converted to heat. The amount of resistance depends on the properties of the conductor and the conditions of the circuit.

Ohm's law relates the voltage (V), current (I), and resistance (R) in a circuit, and is expressed as:

V = I x R

We can rearrange this equation to solve for resistance:

R = V / I

In this case, we are given the voltage (V) of 120 volts and the current (I) of 5 amps, so we can plug these values into the equation to calculate the resistance (R):

R = V / I

R = 120 V / 5 A

R = 24 ohms

Therefore, the resistance of the circuit is 24 ohms.

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In the context of LTE-A (Long-Term Evolution Advanced), Type 1 and Type 1a relay nodes are different deployment options for relay nodes in the LTE network. Relay nodes are used to extend the coverage and improve the performance of the network by relaying signals between the base station and user equipment (UE).

Type 1 Relay Node:

A Type 1 relay node in LTE-A operates in half-duplex mode, which means it can either transmit or receive data at a given time but not both simultaneously. It has two separate sets of antennas: one for receiving signals from the base station (downlink) and another for transmitting signals to the UE (uplink). This type of relay node introduces a relay-specific interface called the Relay Physical Interface (R-PHY) to connect with the base station.

The Type 1 relay node receives downlink signals from the base station, decodes them, and then re-encodes and retransmits them to the UE. Conversely, it receives uplink signals from the UE, decodes them, and re-encodes and retransmits them to the base station. Due to the half-duplex operation, it cannot receive and transmit simultaneously, which can result in increased latency and reduced throughput compared to other relay types.

Type 1a Relay Node:

A Type 1a relay node is an enhanced version of the Type 1 relay node, specifically designed to improve performance. It operates in full-duplex mode, allowing simultaneous transmission and reception. It achieves this by utilizing advanced self-interference cancellation techniques, which cancel out the interference caused by the transmitted signal, allowing the relay to receive signals while transmitting.

The Type 1a relay node also utilizes the Relay Physical Interface (R-PHY) to communicate with the base station. By supporting full-duplex operation, it can provide better throughput and lower latency compared to the Type 1 relay node. This makes it more suitable for scenarios where higher data rates and improved performance are desired.

Both Type 1 and Type 1a relay nodes can be deployed in LTE-A networks to extend coverage and improve performance in areas with challenging propagation conditions or limited backhaul connectivity. The choice between the two types depends on the specific requirements of the network deployment and the desired trade-offs between performance and complexity/cost.

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F = m · g

m = 1485 kg

g = 10 N/kg

F = 1485 kg · 10 N/kg = 14 850 N = 14,85 kN

Answer: The weight on Earth is 14,85 kN.

Answer:

The answer is 22.14 meters per second

Explanation:

I gotchu brother

Answer:

4.0 m/s ^2

Explanation:

The acceleration of the block weighing 80 n attached to a spring scale, and both are pulled to the right on a horizontal surface with negligible friction between the block and the surface will be 4 m/s² when it reads 32 n.

Acceleration is physical quantity used to express the rate of change in velocity of a moving massive body. Acceleration of an object is directly proportional to the force acting and inversely proportional to the mass.

The equation relating force and acceleration is F = ma. Thus as the mass of the body increases, acceleration decreases and this is called deceleration.

Given that the weight of the body is 80 n. It is the product of mass and gravity of 10 m/s². Thus mass m = 80/10 = 8 kg.

The scale reads a force of 32 Newton for the body weighing 8 Kg. Hence the acceleration can be calculated as follows:

a = F/m

   = 32 N/ 8 Kg

   = 4 m/s².

Therefore, when the acceleration of the body of 8 kg when the scale reading is 32 N is 4 m/s².

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Answer:

7.81 moles of Neon.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This simply means that 1 mole of Neon also contains 6.02×10²³ atoms.

From the above illustration, we can convert 4.7×10²⁴ atoms of Neon to moles. this is illustrated below:

6.02×10²³ atoms = 1 mole

Therefore,

4.7×10²⁴ atoms = (4.7×10²⁴ × 1)/6.02×10²³

4.7×10²⁴ atoms = 7.81 moles

Therefore, 4.7×10²⁴ atoms of Neon is equivalent to 7.81 moles of Neon.

A race car is moving at a constant speed around a track. The race car is changing its velocity as the direction of motion changes.

The primary indicator of an object's position and speed is its velocity. It is the distance that an object travels in one unit of time. The displacement of the item in one unit of time is the definition of velocity.

The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.

As the race car is moving at a constant speed around a track, the magnitude of velocity remains same but during race it may changes its direction of motion, that is why, velocity of it, which depends on both magnitude and direction, may changes.

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D. The average speed is 2.5 m/s and the average velocity of the bicycle is 0 m/s east.

The average speed of an object is the ratio of total distance to total time of motion of the object.

average speed = total distance traveled / total time taken

average speed = (150 m + 150 m) / (2 mins x 60s/min)

average speed = (300 m) / (120 s)

average speed = 2.5 m/s

The average velocity of an object is the ratio of change in displacement to change in time of motion.

v = Δx/t

v = (150 m - 150 m) / (120 s)

v = 0 m/s

Thus, the average speed is 2.5 m/s and the average velocity of the bicycle is 0 m/s east.

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Answer:

Its angular acceleration is zero.

Explanation:

If the angular velocity of an object (in this case the toy) is constant, then its angular acceleration will be zero. Why? Because angular acceleration is the time rate change of angular velocity. Since there is no change, this brings the answer to zero.