the g string on a guitar is a 0.44-mm-diameter steel string with a linear density of 1.1 g/m . when the string is properly tuned to 196 hz, the wave speed on the string is 250 m/s. tuning is done by turning the tuning screw, which slowly tightens-and stretches-the string.

The time interval taken by the ball to strike the ground is 1.75 s.

First calculate the vertical velocity by using the kinematics equation of motion for constant acceleration.

Finally use the equation of motion which relates the vertical velocity with time and calculate the time.

The equation of motion for constant acceleration is,  

vf^2 = vi^2 + 2aΔy

vf is final velocity, vi is initial velocity, y  is vertical distance

vi = 7.60  vf = ? a = 9.8m/s^2  y = 29.3m

y = [tex]\sqrt{8.70^2-19.6-30.2[/tex]   , since height is vertical so we took -9.8 accelaration

y= -25.83

so now

The equation of motion which relates velocity with time is

vf = vi + at

t = [tex]\frac{ vf - vi }{ a}[/tex]

t = [tex]\frac{-25.83- (-8.70}{ -9.80}[/tex]

t = 1.748 s

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Therefore, the missile's individual speed would be 14c + 14c = 28c.

The missile is traveling in the opposite direction from the Earth, which has an individual speed of zero.

Both the rocket and the missile are traveling in the same direction relative to one another. The moving item whose speed is being measured and the frame of reference for the point of observation must both be in relative motion.

The difference between their respective speeds is hence the relative speed.

Therefore, the missile's individual speed would be 14c + 14c = 28c.

The missile is traveling in the opposite direction from the Earth, which has an individual speed of zero.

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The horizontal pulling force on the block is 22.5 N.

When a block is pulled by applying a force, friction also acts on it.

The net force is calculated by taking the net value of the two forces.

The free-body diagram(FBD) of the block,  is shown in the adjoining diagram.

Thus,

F - f = ma

Here F is the applied horizontal force and f is the frictional force.

Putting the given values in the above equation

F - 10 N = 5 x 2.5

F = 22.5 N

Thus, the horizontal pulling force on the block is 22.5 N.

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The horizontal pulling force on the block is 22.5 N.

When a block is pulled by applying a force, friction also acts on it.

The net force is calculated by taking the net value of the two forces.

The free-body diagram(FBD) of the block,  is shown in the adjoining diagram.

Thus,

F - f = ma

Here F is the applied horizontal force and f is the frictional force.

Putting the given values in the above equation

F - 10 N = 5 x 2.5

F = 22.5 N

Thus, the horizontal pulling force on the block is 22.5 N.

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The correct answer is 1.5 seconds.

We are aware that the differentiation is used to calculate the rate at which a quantity changes over time. We are aware that the velocity is the name given to the derivative of the displacement. Acceleration is what we get when we differentiate velocity.

x = t^4 - 6t^2 + 9t

To calculate the velocity, take the derivative.

v = 4t^2 - 12t + 9

To acquire acceleration, repeat the derivative.

a = 8t - 12

force:

F = ma

If a = 0, F = 0. As a result, when a = 0, the net force on the locomotive is equal to zero.

a = 0 = 8t -12

8t -12 = 0

8t = 12

t = 12/8 = 1.5 seconds

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Answer:

Rolling Friction

Explanation:

Friction always acts in the opposite direction to that of the motion of the other object. Meaning, friction slows down the motion of an object by exerting opposing forces on it

Force necessary to overcome static friction= (μs mg ) / (cos Ф + μs sinФ)

We have to find Force(F) in terms of m, g,  Ф ,and μs

To get the value of F first we have to find the value of N by using the formula of friction,

F sin Ф + N - mg =0

N = mg - F sinФ

As the body does not move so the net force is zero, therefore we have,

FcosФ -μs N = 0

Now we have to put the value of N in the equation. To know the force necessary to overcome the static friction we have,

F cosФ - μs ( mg - FsinФ) = 0

F cosФ - μs mg + μs FsinФ = 0

F( cos Ф + μs sinФ ) = μs mg

F = μs mg/ (cosФ +μs sin Ф)

Therefore we get the value of F i.e., force in terms of m, g, Ф, and μs.

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The complete question is:

If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as Fcosθ−μsN=0 Fsinθ+N−mg=0 To find the magnitude of force F, we have to solve a system of two equations with both F and the normal force N unknown. Use the methods we have learned to find an expression for F in terms of m, g, θ, and μs (no N).

The relative velocity of the first stone concerning the second stone is 20 m/s.

The time interval at which two stones are dropped from the tower is t.

t = 2 seconds

Let the initial velocity of the first stone be u1.

Let the initial velocity of the second stone be u2.

Let the final velocity of the first stone be v1.

Let the final velocity of the second stone be v2.

The velocity of the first stone is,

[tex]v _{1} = u _{1} + at[/tex]

[tex]v _{1} = 0 + gt[/tex]

[tex]v _{1} = gt[/tex]

The velocity of the second stone is,

[tex]v _{2} = g(t - 2)[/tex]

The relative velocity of the first stone concerning the second stone is,

[tex]v = v _{1} - v _{2}[/tex]

= g - g ( t - 2)

= gt -gt - 2g

= 2 × g

= 2 × 10

= 20 m/s

Therefore, the relative velocity of the first stone concerning the second stone is 20 m/s.

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Answer:

why?

Explanation:

Do you need any help on anything else?

The potential difference between the plates will be 1552 Volts.

Voltage, or the difference in electric potential between two points, is defined as the amount of labor per unit of charge needed to move a test charge between the two points.

Given that a 3.7-f capacitor that stores sufficient energy to operate a 75.0-w

The potential difference will be calculated by the formula below:-

Q = I t            

Where I = charge / time

Q = V * C    

V C = I t

V = I t / C

V = 75 C/s x 60 sec / 2.9 faraday

V =  1552 Volts

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The speed of the electron as it emerges from the field is; 388.587 m/s

Initial speed; v₀x = 1.1 * 10⁶ m/s

Acceleration in horizontal direction = 0 m/s²

distance; s_x = 2 cm = 0.02 m

Thus, formula to find time here is;

t = s_x/v₀x

t = 0.02/(1.1 * 10⁶)

t = 1.82 * 10⁻⁶ s

Now for the vertical distance; v,y_o = 0 m/s

Thus, the equation of motion becomes;

s_y = ¹/₂at²

0.005 = ¹/₂a(1.82 * 10⁻⁶)²

Solving for a gives;

a = 3.02 * 10¹³ m/s²

Thus the speed of the electron as it emerges from the field is;

v² = u² + 2as

v = √(0² + 2(3.02 * 10¹³ * 0.005))

v = 388.587 m/s

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Complete Question is;

An electron is projected with an initial speed v0 = 1.1 * 10⁶ m/s into the uniform field between the parallel plates. The distance between the plates is 1 cm and the length of the plates is 2 cm. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. E = N/C

find the speed of the electron as it emerges from the field?.

Answer:

The car moving at velocity 15 m/s will take 2.67 s to travel 40 m.

What do you mean by velocity?

An object's velocity can be defined as the rate of change in the object's position with respect to a frame of reference and time.

A rocket launching into space and a car going north on a busy freeway may both be quantified using velocity.

Explanation:

Velocity of the car = 15 m/s

Distance = 40 m

Therefore, time taken to travel 40 m at a velocity of 15 m/s

= 40/15 = 2.666 ≈ 2.67 s

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a. Nyquist rate = 30 kHz is the Nyquist rate.

compact disc (CD) records audio signals digitally by means of a binary code. Assume an audio signal bandwidth of 15 kHz.

⇒ According to the Nyquist criteria, the sampling frequency must be at least 2 times the bandwidth.

⇒Nyquist rate = 2 x 15 kHz

⇒Nyquist rate = 30 kHz

So that means we need to take at least 30,000 sample/s to meet Nyquist criteria.  

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The basic harmonic motion of the mass-spring system lasts for 5.024 seconds.

We have a 16-pound mass that is fastened to a spring with a 25-pound-per-foot spring constant. This entire spring and mass system is moving harmonically.

Simple harmonic motion (sometimes referred to as SHM) is a particular type of periodic motion in mechanics and physics in which the restoring force on the moving object is inversely proportional to the magnitude of the object's displacement and acts in the direction of the object's equilibrium position.

The particle's acceleration in a simple harmonic motion is geared toward its mean location and directly proportionate to its displacement. Simple harmonic motion results in the conservation of the particle's total energy. SHM is a cyclical motion.

Simple harmonic motion is demonstrated by bungee jumping. Due to the flexibility of the bungee cord, the jumper is experiencing SHM as it oscillates up and down.

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The calculated greatest speed is v=0.725 vs.

The ratio of a moving object's speed to the fundamental speed of sound is known as the Mach number. Therefore, an object travelling at twice the speed of sound has a Mach number of 2. speed of sound, the rate at which sound waves move through various substances. Modern estimates of the speed of sound, in particular, for dry air at a temperature of 0 °C (32 °F), are 331.29 meters (1,086.9 feet) per second.

The sound speed is vs=331 m/s.

The duration is t=4.9s.

A=5g is the acceleration.

u=0 is the beginning speed.

Let v be the fastest possible speed without a blackout.

The initial equation of motion is given by v=u+at

v=0+5gt

v=5(9.81m/s2)(4.9s)

v=240m/s

v=240m/svs/331m/s

v=0.725 vs.

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Because of the cosmological principle, we can predict gravity works similarly on Earth as on Mars.

The cosmological principle says that an observer's view of the universe is independent of both the direction he looks and his location. This principle only applies to the universe's large-scale properties, but it does suggest that the world has no edge, implying that the big-bang origin took place not at a specific point in space, but instead throughout space at the exact time.

These two assumptions allow us to determine the history of the universe after a specific epoch known as the Planck time. Scientists are yet to discover what existed prior to Planck time.

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Initially, an automobile is moving north at a speed of 23 m/s. If the automobile accelerates at [tex]1.1 m/s^{2}[/tex] in the direction of travel, its speed after 5.3 seconds will be 28.83 m/s.

The initial speed of the car u = 23 m/s

Time is taken t = 5.3 s

Acceleration, a = [tex]1.1 m/s^{2}[/tex]

the first movement equation (Vf =Vi + at) offers with time, acceleration, very last pace, and starting pace. the first equation of motion can be used to calculate time, acceleration, beginning pace, and final pace.

Applying 1st equation of motion,

v = u + at

v = 23 + 1.1 × 5.3       (Since both beginning speed and acceleration are in the same direction)

⟹  v = 28.83 m/s      (due North)

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The larger magnification will be based on Focal length

The narrower the focal length, the smaller the magnification, and the larger the focal length, the larger the magnification.

it is obvious that the angle of the photograph from the objective lens or reflect to the eyepiece could be narrower the higher the focal duration. therefore, the photo may be considered via the eyepiece deeper with out dropping attention the narrower the angle (higher focal period).

The intensity of consciousness is shorter than the focal period. It approach that in comparison to large focal lengths, decrease focal lengths provide you with much less focuser play.

although you could nevertheless acquire the equal magnification with either focal period, the shorter focal duration will lose awareness a great deal earlier than the bigger focal period.

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The billiard ball involves relaxation 80 cm

Displacement refers to the space among the very last and the preliminary role. Hence the displacement of the ball can be the distinction among the preliminary and the very last displacement.

Let the preliminary role be 0

Final role = 8 cm

So the distinction among preliminary role and very last role = 0 - 88 cm

So the billiard ball involves relaxation 80 cm at the back of its orbital role.

The displacement thickness for the boundary layer is described as the gap the floor could need to pass withinside the y-course to lessen the glide passing via way of means of a extent equal to the actual impact of the boundary layer.

The distance from the boundary floor where the rate reaches 99% of the mainstream speed is used as an arbitrarily defined measure of the thickness of the boundary layer indicated by the symbol.

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The position when time, t =35s is  87.59 m.

The acceleration represented by the graph is 0.143 m/s².

a = Δv/Δt

where;

a = (6 - 5) / (7 - 0)

a = 0.143 m/s²

The position of the object is determined from the product of velocity and time of motion.

s = vt + ¹/₂at²

where;

s = 0 + ¹/₂(0.143)(35)²

s = 87.59 m

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The corresponding wavelengths for these jumps are as follows:

[tex]\lambda_{7-2}[/tex] = 397.7 nm

[tex]\lambda_{6-2}[/tex] = 411 nm

[tex]\lambda_{5-2}[/tex] = 434.9 nm

[tex]\lambda_{4-2}[/tex] = 487.1 nm

[tex]\lambda_{3-2}[/tex] = 657.5 nm

The distance that is present between ant two successive crests of a wave, mostly points present between a sound wave or electromagnetic wave. Wavelength is also called as the distance between identical points (adjacent crests) in the adjacent cycles of any waveform signal that is propagated in space or along a wire.

Frequency, which is defined as the quantity of wave cycles per second, and wavelength have an inverse relationship. The wavelength of a signal decreases with increasing frequency.

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The mass flow rate of water is 0.08757 kg/s

mCp (T2 - T1) = Q =w

m x 1.015 (360-315) = 4000

m = 4000/ 1.015 x 10^3 x (360-315)

m = 0.08757 kg/s

The density of water at 68°F / 20°C has a flow rate in our example of about 998 kg/m3. However, as the density of water varies with temperature, salinity, and pressure, use our water density calculator if you want to be extremely accurate. 30.58 lbs/s of mass flow were shown by the gadget.

By dividing the volume flow rate by the fluid’s mass density, one can determine the mass flow rate. The cross-sectional vector area, A, is multiplied by the flow velocity of the mass elements, v, to determine the volume flow rate.

The mass of a substance that moves per unit of time is known as the mass flow rate in physics and engineering.

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The velocity vector is (v = 3ti + 4tj + k) and the position vector is (x = (3t²/2 + 1) i + 2t²j + tk)

We need to know about vector to solve this problem. The velocity is depend on position vector and acceleration. It can be written as

v = dx / dt

where v is velocity, dx is displacement and dt is time interval.

By differentiating the velocity function, we can also get the acceleration

a = dv / dt

where a is acceleration.

From the question above, we know that

a(t) = 3i + 4j

Hence, the velocity is

a = dv / dt

dv = a . dt

v = ∫a . dt

v = ∫(3i + 4j) . dt

v = (3i + 4j)t + v(0)

because v(0) is k, thus

v = 3ti + 4tj + k

Find the position vector

v = dx / dt

dx = v . dt

x = ∫v . dt

x =  ∫(3ti + 4tj + k) . dt

x = (3t²/2 i + 2t²j + tk) + x(0)

because x(0) is i, thus

x = 3t²/2 i + 2t²j + tk + i

x = (3t²/2 + 1) i + 2t²j + tk

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The tension in the cord is equal to centripetal force which is 32 newton.

We need to know about centripetal force to solve this problem. When an object moves in a circular motion, the object is experiencing centripetal force. The magnitude of centripetal force is

Fc = m . v²/R

where Fc is the centripetal force, m is mass, v is velocity and R is the radius.

From the question, we know that

m = 2 kg

v = 4 m/s

R = 1 m

The tension in the cord is equal to centripetal force

T = Fc

T = m . v²/R

T = 2 . 4² / 1

T = 32 N

Hence, the tension in the cord is 32 newton.

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The following are the values:-

(a)Water content (W%)  = 13.99%

(b) Total unit weight (Υ[tex]_{t}[/tex]) = 1778 kg/m³

(c) Dry unit weight (Y[tex]_{d}[/tex]) = 1559.75 kg/m³

(d) Void ratio (e)  = 0.718

(e) Porosity (n) = 0.417

Volume of mold (V[tex]_{t}[/tex]) = 0.4 m³

Dry mass of sand (W[tex]_{s}[/tex]) = 623.9 kg

Wet mass of sand (W[tex]_{t}[/tex]) = 711.2 kg

Specific Gravity (G[tex]_{s}[/tex]) = 2.68

Water content (W%) = [tex]\frac{W_{w} }{W_{s} }[/tex] × [tex]100[/tex]

Water content (W%) = [tex]\frac{711.2 - 623.9}{623.9}[/tex] × [tex]100[/tex]

Water content (W%) = 13.99%

Total unit weight (Υ[tex]_{t}[/tex]) = [tex]\frac{W_{t} }{V_{t} }=\frac{711.2}{0.4}[/tex]

Total unit weight (Υ[tex]_{t}[/tex]) = 1778 kg/m³

Dry unit weight (Y[tex]_{d}[/tex]) = [tex]\frac{W_{s} }{V_{t} }=\frac{623.9}{0.4}[/tex]

Dry unit weight (Y[tex]_{d}[/tex]) = 1559.75 kg/m³

Solid unit weight (Y[tex]_{s}[/tex]) = [tex]\frac{W_{s} }{V_{s} }[/tex]

Or

G[tex]_{s}[/tex] = [tex]\frac{Y_{s} }{Y_{w} }[/tex] ⇒ Y[tex]_{s}[/tex] = G[tex]_{s}[/tex]Y[tex]_{w}[/tex]

Y[tex]_{s}[/tex] = 2.68 × [tex]\frac{10^{-3} }{(10^{-2})^{3} }\frac{kg}{m}[/tex] = 2680 kg/m³

Now,

Void Ratio (e) = [tex]\frac{Y_{s} }{Yd}-1[/tex]

Void Ratio (e) = [tex]\frac{2680}{1559.72}-1[/tex]

Void Ratio (e) = 0.718

Porosity (n) = [tex]\frac{e}{1+e}[/tex]

Porosity (n) = [tex]\frac{0.718}{1+0.718}[/tex]

Porosity (n) = 0.417

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A laboratory sample of sand is formed inside a mold of 0.4 cubic meters. It took 711.2 kg of wet sand (dry mass of sand = 623.9 kg) to fill the mold. Assuming the specific gravity of the solid is 2.68, compute the:

(a)Water content (W%)  

(b) Total unit weight (Υ[tex]_{t}[/tex])

(c) Dry unit weight (Y[tex]_{d}[/tex])

(d) Void ratio (e)

(e) Porosity (n)

The mass of the girl is 30.6 Kg.

Inertia, a fundamental property of all matter, can be quantified as mass. In essence, it's the resistance of a mass of matter to changing its direction or speed in reaction to the application of a force. The more mass a body possesses, the smaller the change that is brought about by an applied force. The dimensionless quantity mass is used to express the amount of matter contained in a particle or object (symbolized m).

Given,

The pressure exerted by the girl = 15000 [tex]\frac{N}{m^{2}}[/tex]

Total area of contact = 0.02[tex]m^{2}[/tex]

As we know,

Pressure  = [tex]\frac{Force}{area}[/tex]

Force = Area x Pressure

Force = 0.02 x 15000

Force = 300N

For the conversion of weight into mass, use the formula "w = m x g." The gravitational pull on an object is referred to as its weight. The equation for the statement is written as w = m x g, or w = mg, by scientists.

Weight of the girl = Mass x Gravitational constant

300N = Mass of the girl x 9.8

Mass of the girl = 300 / 9.8

Mass of the girl =  30.6 Kg

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Complete Question

Suppose you have a pocket calculator through which 1.25 C of charge passes in 3.5 h? What is the average current in milliamperes produced by the solar cells .

It is 0.126mA .

An item has a fine fee if it incorporates more protons than electrons. when a thing loses a few electrons, the difference between the number of protons and electrons in the item rises. As a result, the item acquires a positive rate.

tremendous and negative fees cancel each other out when there are an equal range of high-quality and poor charges, leaving the item neutral.

general rate that passes via the calculator may be denoted by q = 1.25c

total time t = 2.75=9900s

current required is = 1.25/9900

=1.26×10^-4 A

=1.26×10^1mA

=0.126mA

Thus the solar cell must produce an average of 0.126mA of current.

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Answer:

lol the SUN

Explanation:

Answer:

bro SUN just sun giving us heat

The correct answer is 223 days.

The relationship between the duration of revolution and the separation between the sun is shown by Kepler's third law. Using the notions of circular motion and the gravitational and centripetal forces, we may obtain this equation.

According to Kepler's third rule, the semi-major axis of an orbit is linked to the orbital period of a planet around the sun as follows:

p² = a³

where an is the semi-major axis/distance to the star and p is the orbital period in years.

It is said that a = 0.72 AU for Venus.

P= √(0.72 AU)^3 = 0.61 years.

365 days in a year = 222.9 ≈ 223 days.

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The body of mass is resting on a horizontal plane, then the normal force exerted onto it by stattic and kinetic frictional forces both have an equal magnitude. In addition, the net resultant force acting upon the body of mass will be zero.

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