u-shaped Tube of uniform cross-sectional area, the length of each branch is 33 cm and its area is 2cm² contains an amount of mercury of height 6.8cm. find the volume of the largest amount of water that can be poured in one of the two branches knowing that the density of water is 1 g/cm³, the density of mercury is 13.6 g/cm³

Answer:

tbh vector does not have any direction at all the answer is 0

To calculate the efficiency of a machine, divide the work output by the work input and multiply the result by 100%.

Correct question:

The work output of a machine is 80 Joules, and its work input is 120 Joules. What is the efficiency of the machine?

Work output = 80 Joules

Work input = 120 Joules.

Efficiency = Working output/Work input *100

Efficiency = 80/120 *100

Efficiency = 0.67 *100

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Answer:

the answer is equal to the microwave gamma rays visible light x-rays.

The circumference of a circle of radius 17 mm is 34π or 34 × 3.14 = 106.76 mm.

The length of a circular or other curved geometrical object is referred to as its circumference. It is the edge of a surface like this with two dimensions, calculated linear in one dimension. A circle's circumference, which is equivalent to the length of it's own edge measured in linear distance, serves as a measure of length. In order to measure the circumference of a circle in the metric system, a most common units are millimetre, centimeter, and meter, but in the imperial system, the most common units include inch, foot, and yard.

Given :

Radius of circle = 17mm

By using the formula of circumference -

circumference = 2 × π × r

circumference = 2 × π × 17

circumference = 34 π

or 34 × 3.14 = 106.76 mm

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A thin lens of focal length 68 cm forms a real image 4.6 times as high as the object. The object and image are approximately 299.13 cm apart.

To solve this problem, we can utilize the lens formula:

1/f = 1/v - 1/u

where f is the focal length of the lens, v is the image distance, and u is the object distance. Given that the focal length (f) is 68 cm, we can proceed to find the image and object distances.

Let's assume the height of the object is "h." Since the problem states that the real image formed is 4.6 times as high as the object, the height of the image is 4.6h.

Now, using the magnification formula:

magnification (m) = height of the image (h') / height of the object (h) = -v/u

Given that the magnification is 4.6, we can write:

4.6 = 4.6h / h = -v / u

Simplifying this equation, we get:

-4.6u = v

Now, substituting the values into the lens formula, we have:

1/68 = 1/(-4.6u) - 1/u

To solve for u, we can combine the fractions:

1/68 = (-1 + 4.6) / (-4.6u)

1/68 = 3.6 / (-4.6u)

Cross-multiplying, we find:

-4.6u = 3.6 * 68

Simplifying further:

-4.6u = 244.8

Dividing both sides by -4.6, we get:

u ≈ -53.39 cm

Since the object distance cannot be negative, we take the magnitude of the value:

u ≈ 53.39 cm

Now, we can substitute the value of u into the magnification equation to find v:

4.6 = -v / 53.39

Cross-multiplying and solving for v, we find:

v ≈ -245.74 cm

Taking the magnitude of v, we have:

v ≈ 245.74 cm

To find the distance between the object and image, we add the magnitudes of u and v:

53.39 + 245.74 ≈ 299.13 cm

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Answer:

The trains will collide at a distance 1660 m from the station

Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

∴ If train A has covered a distance \(x\) m by the time of collision, then train B would have traveled \((2881 - x)\) m.

Also,

At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

\(t_{A} = t_{B}\)

\(t_{A}\) is the time spent by train A

\(t_{B}\) is the time spent by train B

From,

\(Velocity = \frac{Distance }{Time }\\\)

\(Time = \frac{Distance}{Velocity}\)

Since the time spent by the two trains is equal,

Then,

\(\frac{Distance_{A} }{Velocity_{A} } = \frac{Distance_{B} }{Velocity_{B} }\)

\({Distance_{A} = x\) m

\({Distance_{B} = 2881 - x\) m

\({Velocity_{A} = 100\) m/s

\({Velocity_{B} = 136\) m/s

Hence,

\(\frac{x}{100} = \frac{2881 - x}{136}\)

\(136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\\)

\(x\)≅ 1,221 m

This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

Answer:

W = 882.9[J]

Explanation:

In order to be able to calculate the work, we must first calculate the force necessary to lift the box. Since the necessary force is equivalent to the weight of the box, we can determine the weight of the box by means of the product of mass by gravitational acceleration.

\(w = m*g\\\)

where:

w = weight [N]

m = mass = 30 [kg]

g = gravity acceleration = 9.81 [m/s²]

\(w = 30*9.81\\w = 294.3 [N]\)

Now, the work can be calculated multiplying the force (weight) by the distance [m]

\(W = w*d\\W = 294.3*3\\W=882.9[J]\)

A third positive charge \(x = \frac{2\sqrt{2}a }{1+\sqrt{2} }\) can be placed on the x-axis so that the net electric force on it is zero.

whilst -factor charges are in touch, there may be

\(F = k \frac{q_{1}q_{2} }{ r^{2} }\)

Where

\(k = 9 * 10^{9} Nm^{2}C^{-1}\) = the coulomb constant.

The charges' magnitudes are \(q_{1}\) and \(q_{2}\).

r is the separation of the charges.

Let,

The third charge equals Q at, x equals x from the source.

because there is no net force acting on the third charge.

Hence,

\(f_{1} + f_{2} = 0\)

⇒ \(k\frac{(2q)Q}{x^{2} }-k\frac{qQ}{(2a-x)^{2} } = 0\)

⇒ \(\frac{2}{x^{2} } - \frac{1}{(2a-x)^{2} } = 0\)

⇒ \(\frac{2}{x^{2} } = \frac{1}{(2a-x)^{2} }\)

⇒ \(\frac{\sqrt{2} }{x} = \frac{1}{2a-x}\)

⇒ \(2\sqrt{2}a - \sqrt{2} = x\)

⇒ \(x = \frac{2\sqrt{2}a }{1+\sqrt{2} }\)

handiest when forces resulting from previous prices cancel every different out will the pressure on a third fee be 0.

at the same time as the pressure between  identical costs is repulsive and is interpreted as high-quality, the pressure between two opposing expenses is attracted and is interpreted as bad.

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Answer:

1kW = 1000W

600W = 0.6kW

Cost of electric bill = 0.6kWh × 24 × 30 × $0.05

                               = $21.60

The capacitance characteristics allow to find that the response to increase the separation of the plates are:

A capacitor is a system formed by two parallel plates charged with opposite charge that stores energy, the capacitance is

            C =     \(\frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}\)  

Where C is the capacitance, Q the stored charge, ΔV the voltage difference, A the area of ​​the plates, d the separation between them and ε₀ the dielectric permittivity (ε₀ = 8.85 10⁻¹² \(\frac{C^2}{N \ m^2 }\))

In this case the capacitor is charged and then the separation of the plates increases, the different amounts are affected:

In conclusion of the characteristics of capacitance, let us find that the response for increased separation of the plates are:

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The refracting telescope uses large lenses. The larger lenses increase the weight of the telescope. The weight of the refracting telescopes makes it difficult for the observer to keep them aligned.

Thus the correct answer is option b, refracting telescopes.

Answer:

Is Reflecting telescope

Explanation:

I took the test

The power, in terms of P₀, dissipated by this circuit is equal to (1/2)P₀.

Power dissipation can be defined as the generation of the rate of heat release from electrical devices.

A single resistor is wired to a battery which is shown in the attached diagram below:

The voltage is V₁ = V, resistance is R₁ = R

A second identical resistor is wired in series with the first resistor as shown in the attached diagram:

The voltage is V₂ = V, resistance is R₁ = R₂ = R therefore, total resistance, Rₙ = R + R = 2 R

We know that the Power, P = V²/R

In the first case, P₀= V₁²/R₁ = V²/R

In 2nd case, P = V₂²/Rₙ = V²/2R

P = (1/2)P₀

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Answer:

A

Explanation: The hotter things get the more thermal energy they have

Also don't ever take pictures like this again

Answer:A

Explanation:

1  kg of boiling water is hot enough to melt a block of ice more quickly

Answer:

kg

Explanation:

The linear velocity of the solid sphere at the bottom of the incline is approximately 6.69 m/s.

To solve this problem, we can analyze the forces acting on the solid sphere as it rolls down the incline.

Diameter of the sphere (d) = 7.80 cm = 0.078 m

Mass of the sphere (m) = 320 g = 0.320 kg

Length of the incline (h) = 1.70 m

Incline angle (θ) = 20°

First, let's calculate the gravitational potential energy (PE) of the sphere at the top of the incline:

PE = m * g * h

where g is the acceleration due to gravity.

PE = 0.320 kg * 9.8 m/s² * 1.70 m

Next, let's determine the rotational kinetic energy (KE_rot) of the rolling sphere at the bottom of the incline:

KE_rot = (1/2) * I * ω²

where I is the moment of inertia and ω is the angular velocity.

For a solid sphere rolling without slipping, the moment of inertia is given by I = (2/5) * m * r², where r is the radius of the sphere.

r = d/2 = 0.078 m / 2 = 0.039 m

I = (2/5) * 0.320 kg * (0.039 m)²

Next, we need to find the linear velocity (v) of the sphere at the bottom of the incline. Since the sphere rolls without slipping, the linear velocity is related to the angular velocity by v = ω * r.

v = ω * r

To find ω, we can relate it to the linear velocity using the equation ω = v / r.

Finally, we can equate the gravitational potential energy at the top to the rotational kinetic energy at the bottom:

PE = KE_rot

m * g * h = (1/2) * I * (v/r)²

Substituting the expressions for I and v, we can solve for v:

m * g * h = (1/2) * (2/5) * m * r² * (v/r)²

Simplifying and canceling out the mass and r terms:

g * h = (1/5) * r * (v/r)²

g * h = (1/5) * v²

Now, we can solve for v:

v = √(5 * g * h)

Substituting the known values:

v = √(5 * 9.8 m/s² * 1.70 m)

v ≈ 6.69 m/s

Therefore, the linear velocity of the solid sphere at the bottom of the incline is approximately 6.69 m/s.

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Answer:

With regards to NASA's ambition to send humans to Mars, is an encouraging idea.  Also, the aim of sending food supply by unmanned spacecraft before the 2030 project though noble is filled with constraints and design problems.

For the constraints,

1. The spacecraft could be lost on the space before it reaches its destination being Mars. This is because, lack of communication between the control unit on earth and the spacecraft signify it being lost forever.

2. The food supply is bound to spoil on Mars since it would be stored unrefrigerated. This is as a result of the actions of the micro organisms acting on them.

For the criteria for the engineering design problem:

1. The distance between earth and Mars is very far

2. Not factoring in the preservation method into the unmanned aircraft.

Explanation:

Answer:

B. seafloor spreading

Explanation:

divergent motion between the Eurasian and North American, and African and South American Plates. ... In this way, as the plates move further apart new ocean lithosphere is formed at the ridge and the ocean basin gets wider

geological society

Increasing resistance in a circuit results in higher voltage drops, reduced current, increased power dissipation, and potentially increased heating effects. Refining resistance values within a range from 100 to 1000 allows identification of resistances where the circuit behaves primarily resistively with minimal influence from capacitance or inductance.

As the value of resistance (r) increases in the circuit, it has several effects on the system. Let's examine the effects of increasing resistance on the circuit's behavior.

1. Voltage Drop: According to Ohm's Law (V = IR), as resistance increases, the voltage drop across the resistor increases for a given current. This means that a higher proportion of the applied voltage is consumed by the resistor.

2. Current: Following Ohm's Law, as resistance increases, the current in the circuit decreases for a constant voltage. This is because a higher resistance restricts the flow of electric current.

3. Power Dissipation: The power dissipated in a resistor is given by P = \(I^2\) * R. As resistance increases, with a constant current, the power dissipated by the resistor also increases.

4. Heating Effect: With a higher resistance, more power is dissipated as heat in the resistor. This can lead to an increase in the temperature of the resistor and potentially affect its performance or longevity.

To find a wide range of resistances that yield purely resistive behavior, it is necessary to refine the values of resistance within the vector range from 100 to 1000. By incrementing the resistance values and observing the behavior of the circuit (such as current, voltage drop, and power dissipation), a range can be determined where the effects of capacitance or inductance are negligible, and the circuit behaves primarily as a resistive load.

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Answer:

The time between the emission of the wave and it's reception was 0.4 seconds.

Explanation:

The sound wave emitted by the sonar traveled from the boat to the fishes and came back, therefore the distance of it's total route is twice the distance from the boat to the fishes. We can apply the average speed formula in order to solve this problem, this is done below:

\(\text{speed} = \frac{\text{distance}}{\text{time}}\)

We were given the distance traveled by the wave and it's speed, therefore we can solve for the time:

\(\text{time} = \frac{\text{distance}}{\text{speed}}\\\\\text{time} = \frac{(2*300)}{1500}\\\\\text{time} = \frac{600}{1500} = 0.4 \text{ s}\)

The time between the emission of the wave and it's reception was 0.4 seconds.

Answer:

x = 3.6 10⁷ m

Explanation:

The speed of an electromagnetic wave is constant, in a vacuum its value is    c = 3 10⁸ m / s, this speed decreases when entering a material medium,

         v = c / n

where n is the refractive index, but in this case the refractive index of air is n = 1.00002. The refractive index of a vacuum is n = 1, so we can assume that the change in velocity is negligible.

For the calculation we can use the relations of the uniform motion

        v = x / t

        x = v t

let's calculate

        x = 3 10⁸ 0.12

        x = 3.6 10⁷ m

216960m/s is the speed of this wave. Sound waves can travel through other mediums, such as water or solids, but they carry on somewhat in an unexpected way depending on the properties of the medium.

A sound wave could be a sort of pressure that is created by the vibration of an object, such as a guitar string or a drumhead. As the object vibrates, it makes changes in discussing pressure that moves through the air as a wave.

Frequency alludes to the number of total waves (cycles) that pass through a point in one second. It is more often than not measured in units of Hertz (Hz), which suggests cycles per second. Wavelength, on the other hand, is the removal between two comparing points on a wave, such as the peak of one wave to the peak of the following.

To calculate the speed of this sound wave we have to multiply the frequency by wavelength. wave, we can use the formula:

Frequency = 339m/s (given)

Wavelength = 640 hertz.(given)

Speed = frequency × wavelength

            = 339m/s × 640 hertz

            = 216960m/s.

Therefore, the speed of this sound wave is 216960m/s.

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Answer:

f = 1.5 Hz

Explanation:

Given that,

18 waves pass by a person in 12 seconds.

We need to find the frequency of these water waves.

We know that,

Frequency = no. of waves per second

So,

\(f=\dfrac{18}{12}\\\\f=1.5\ Hz\)

So, the frequency of these water waves is equal to 1.5 Hz.

One of the most significant observations that has provided the most accurate values of the age, dark matter, dark energy content, and the average density of the universe is the cosmic microwave background (CMB) radiation.

The CMB radiation is the afterglow of the Big Bang and is a remnant of the hot, dense early universe. The CMB radiation provides a snapshot of the universe when it was only 380,000 years old, and its properties can be analyzed to infer the universe's current state.

By analyzing the CMB radiation, cosmologists have determined that the universe is approximately 13.8 billion years old. Furthermore, they have found that dark matter constitutes around 27% of the universe's total energy density, and dark energy constitutes around 68%.

The CMB radiation has also provided insight into the universe's average density. By measuring tiny fluctuations in the CMB, scientists have determined that the average density of the universe is very close to the critical density required for a flat universe. This result is consistent with the inflationary Big Bang model and the concept of a flat universe.

Therefore, the observation of the cosmic microwave background radiation has been crucial in providing some of the most accurate values of the age, dark matter, and dark energy content, and the average density of the universe.

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The Doppler effect refers to the change in frequency or pitch of a wave due to the motion of the source or observer.

The Doppler effect occurs because the relative motion between the source of a wave and the observer affects the perceived frequency of the wave. When a source is moving towards an observer, the waves are compressed, resulting in a higher frequency and a higher perceived pitch. Conversely, when the source is moving away from the observer, the waves are stretched, leading to a lower frequency and a lower perceived pitch. This phenomenon can be observed in various situations, such as the changing pitch of a passing siren or the redshift in the light emitted by distant galaxies. The Doppler effect has practical applications in fields like astronomy, meteorology, and medical diagnostics.

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An object's mass rather than its weight is used to indicate the amount of matter it contains because weight is defined as the amount of force experienced by a body. The weight of a body can be different as it depends on where and under what conditions it is kept.

However mass is the measure of the amount of matter contained in a body and it is always constant.

Weight is defined as the amount of force exerted on a body by a moving gravitational field or any other field at the surface of the earth. It is measured in Newton (N)

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Answer:

hope it will help you

Explanation:

Answer:

\(\boxed {\boxed {\sf E_K= \ 5,040 \ J}}\)

Explanation:

Kinetic energy is the energy an object possesses due to motion. It is calculated using the following formula:

\(E_K= \frac{1}{2} mv^2\)

The mass of the athlete is 70.0 kilograms. The velocity is 12 meters per second.

Substitute the values into the formula.

\(E_K= \frac{1}{2} (70.0 \ kg)(12 \ m/s)^2\)

Solve the exponent.

\(E_K= \frac{1}{2} (70.0 \ kg)(144 \ m^2/s^2)\)

Multiply the numbers together.

\(E_K= \frac{1}{2} (10, 080 \ kg*m^2/s^2)\)

\(E_K= 5,040 \ kg*m^2/s^2\)

Convert the units. 1 kilogram meter squared per second squared is equal to 1 Joule, so our answer is equal to 5,040 Joules.

\(E_K= 5.040 \ J\)

The athlete has 5,040 Joules of kinetic energy.

Answer:

Explanation:

This is a product of a mass and velocity of a moving or a rest body.

it is expressed as P(momentum)=M(mass) × V(velocity)

Its SI unit is kgm/s

From our question.

Given

Momentum=3600kgm/s

Velocity=6m/s

RTF=Mass

Solution

P=M×V

M=P/V

= 3600kgm/s

6m/s

The criterion of significant figures can to find the result with reliable figures

          X = 9.2 10⁻⁵

now with the propagation of errors we obtain the result with its uncertainty

         X ± ΔX = (9.2 ± 0.5) 10⁻⁵

given Parameter

     * expression values ​​with their absolute errors

to find

     * the result with the correct significant figures

     * the absolute error of the expression

Significant figures are defined with the number of decimals that give information, the number of figures in a quantity gives information about the uncertainty of this quantity.

There are two criteria for applying significant figures:

     * Add and subtract the result of going with the number of decimal places of the figure that has the least

    * Product and division as a result of going with the least number of significant figures than the value that has the least.

Remember that the zero to the left do not form a pair of the significant figures

Let's apply this belief to the case presented, let's write the precaution

              \(x = \frac{a-b}{c}\)

where in this case they are worth

         a = 0.0336 ± 0.0002

         b = 0.010 ± 0.001

         c = 255.4 ± 0.4

We see that the significant figures of each parameterize (a, b, c) and their absolute errors are correct.

Let's apply the criteria to the operation

          a-b = 0.0336 - 0.010

          a- b = 0.0236

we apply the criterion of significant figures for the subtraction, the result must be with 3 decimal places

        a - b = 0.024

let's do the other operation

         X = \(\frac{a-b}{c}\)

         X = 0.024 / 255.4

         X = 9.24 10⁻⁵

We apply the criterion of significant figures for the division, in this case the result is left with two significant figures

         X = 9.2 10⁻⁵

The uncertainty or error of the measurements is of most importance as it determines how many significative figures are reliable at a given magnitude.

If the magnitudes are measured with some type of instrument, the absolute error is given by the appreciation of the instrument, if the magnitude is calculated using some equation, the errors must be propagated using the variations of each parameter in the worst case.

the uncertainty of the calculated quantity (X) is

        \(\Delta X = | \frac{dX}{da}| \Delta a + | \frac{dX}{db} | \Delta b + | \frac{dX}{dc}| \Delta c\)

let's perform the derivatives

        \(\frac{dX}{da} = \frac{1}{c}\)

        \(\frac{dX}{db} = - \frac{1}{c}\)

        \(\frac{dX}{dc} = - \frac{a-b}{c^2}\)

we substitute

remember that the bulk value guarantees that we tune the worst case. So all the mistakes add up

          ΔX = \(\frac{1}{c}\)  Δa +

          ΔX = (Δa + Δb) + Δc

we substitute

         ΔX = \(\frac{1}{255.4}\)  (0.0002 + 0.001) + \(\frac{0.0336-0.010}{255.4^2}\)  0.4

         ΔX = 4.698 10⁻⁶ + 1.45 10⁻⁷

         ΔX = 4.8 10⁻⁶

Absolute errors must be given with a single significant figure

         ΔX = 5 10⁻⁶

The result of the requested quantity using the criterion of significant figures and propagation of errors is

          X ± ΔX = (9.2 ± 0.5) 10⁻⁵

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The value of force is 1.7 × 10⁻³ N, with the direction opposite to that of the bat's motion.

When an object collides with another object, they exchange energy. For example, a baseball and bat collision or a car collision. When two objects collide, the force of the collision has to be equal on both sides of the collision according to Newton's Third Law. So, to find the value of force, we will apply the equation:

F = ΔP / ΔT

where F is the force, ΔP is the change in momentum, and ΔT is the time of collision. The equation represents the impulse momentum theorem.

Now, let's apply the given values to the above equation.

Final momentum (p2) = mass × final velocity (v2)

p2 = 0.15 kg × (-50 m/s)

p2 = -7.5 kg.m/s

Initial momentum (p1) = mass × initial velocity (v1)

p1 = 0.15 kg × (40 m/s)

p1 = 6 kg.m/s

Change in momentum (ΔP) = p2 - p1

ΔP = -7.5 kg.m/s - 6 kg.m/s

ΔP = -13.5 kg.m/s

Time of collision (ΔT) = 8.0 × 10³ s

Now, putting the values of ΔP and ΔT in the equation of impulse momentum theorem, we get:

F = ΔP / ΔT

F = -13.5 kg.m/s ÷ 8.0 × 10³ s

F = -1.7 × 10⁻³ N

Thus, the value of force is 1.7 × 10⁻³ N, with the direction opposite to that of the bat's motion.

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