An ultrasonic tape measure uses frequencies above 20 MHz todetermine dimensions of structures such as buildings. It does so byemitting a pulse of ultrasound into air and then measuring the timeinterval for an echo to return from a reflecting surface whosedistance away is to be measured. The distance is displayed as adigital read-out. A tape measure emits a pulse of ultrasound with afrequency of 25.0 MHz.
(a) What is the distance to an object fromwhich the echo pulse returns after 24ms when the air temperature is 26°C?
(b) What should be the duration of the emitted pulse if it is toinclude 10 cycles of the ultrasonic wave?
(c) What is the spatial length of such a pulse?

Answer:

0.02 kg/cm³

Explanation:

Answer:

I'm pretty sure its helium

Answer:

surface

Explanation:

You need friction created by your tires and the road surface

to control your speed and direction.

Answer:

C

Explanation:

Primacy means being first or important so thats not an important facial display as the others.

Answer:

a) v₂ = 30 m/s

b) m₁ = 12600 kg

c) m₂ = 12600 kg

Explanation:

a)

Using the continuity equation:

[tex]A_1v_1 = A_2v_2[/tex]

where,

A₁ = Area of inlet = π(0.15 m)² = 0.07 m²

A₂ = Area of outlet = π(0.05 m)² = 0.007 m²

v₁ = speed at inlet = 3 m/s

v₂ = speed at outlet = ?

Therefore,

[tex](0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}[/tex]

v₂ = 30 m/s

b)

[tex]m_1 = \rho A_1v_1t[/tex]

where,

m₁ = mass of water flowing in = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

[tex]m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\[/tex]

m₁ = 12600 kg

c)

[tex]m_1 = \rho A_1v_1t[/tex]

where,

m₂ = mass of water flowing out = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

[tex]m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\[/tex]

m₂ = 12600 kg

Answer:

[tex]523269.9\ \text{N/m}[/tex]

Explanation:

q = Charge

r = Distance

[tex]q_1=25\ \text{C}[/tex]

[tex]r_1=3000\ \text{m}[/tex]

[tex]q_2=40\ \text{C}[/tex]

[tex]r_2=850\ \text{m}[/tex]

The electric field is given by

[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]

The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁  + 2.43 x 0  =  v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂²  = 0.07032

v₂²  = 0.07032  / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Answer:

0.337 lb-in

Explanation:

From the law of conservation of angular momentum,

L' = L" where L = initial angular momentum of system and L" = final angular momentum of system

Now L = Iω + Mt where Iω = angular momentum of shaft + wheel and Mt = impulse on system due to couple M.

L' = Iω' + (-Mt) (since the moment about the shaft is negative-anticlockwise)

L' = Iω' - Mt where Iω' = angular momentum of shaft at t' = 0 + wheel and Mt = impulse on system due to couple M in time interval t = 70 s.

L" = Iω"  where Iω" = angular momentum of shaft at t" = 70 s.

Now I = moment of inertia of system = mk² where m = mass of system = W/g where W = weight of system = 6 lb and g = acceleration due to gravity = 32 ft/s². So, m = W/g = 6lb/32 ft/s² = 0.1875 lb-s²/ft and k = radius of gyration = 2 in = 2/12 ft = 1/6 ft.

So, I = mk² = (0.1875 lb-s²/ft) × (1/6 ft)² = ‭0.00521‬ lb-ft-s², ω' = initial angular speed of system = 3600 rpm = 3600 × 2π/60 = 120π  rad/s = 377 rad/s,  ω" = final angular speed of system = 0 rad/s (since it stops), t' = 0 s, f" = 70 s and M = couple on system

So,

Iω' - Mt" = Iω"

Substituting the values of the variables into the equation, we have

Iω' - Mt" = Iω"

0.00521‬ lb-ft-s² ×  377 rad/s - M × 70 s = 0.00521‬‬ lb-ft-s² × 0 rad/s"

0.00521‬ lb-ft-s² ×  377 rad/s - 70M = ‭0

1.964‬ lb-ft-s = 70M

M = 1.964‬ lb-ft-s/70 s

M = 0.0281‬ lb-ft

M = 0.0281 lb × 12 in

M = 0.337 lb-in

Answer:

When the spring compresses, elastic potential energy increases.

Answer:

the answer is b

Explanation:

elastic potential energy

Answer:

0.01j

Explanation:

the energy equals the work done by the bulb.

Workdone=

[tex]workdone = \frac{power}{time} [/tex]

power=voltage×current

=9×0.2

=1.8 W.

THEREFORE,

time=3×60

= 180s

workdone=1.8/180

=0.01 j

Newton’s first law of motion states that there must be a cause—which is a net external force—for there to be any change in velocity, either a change in magnitude or direction. An object sliding across a table or floor slows down due to the net force of friction acting on the object.

got it off g lol..

Answer:

it state that everybody in the universe is state that" universe continues its state of rest or uniform motion in a straight path unless it is acted upon by external force."

Answer:

aim at prisma and will have all colors

Explanation:

Answer:

If the vision position is above the actual image location then the light travel from the object in such a way that the angle of incidence is less than the angle of reflected ray which means that the reflected beam is above the incident beam.

Explanation:

This question is incomplete, the complete question is;

A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.

Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.

Hint : ΔV = Ed

Answer:

the required potential difference, between the capacitor plates is 600 V

Explanation:

Given the data in the question;

B = 0.60 T

d = 2.0 mm = 0.002 m

v = 5.0 × 10⁵ m/s.

since particle pass straight through without deflection.

F[tex]_{net[/tex] = 0

so, F[tex]_E[/tex] = F[tex]_B[/tex]

qE = qvB

divide both sides by q

E = vB

we substitute

E = (5.0 × 10⁵) × 0.6

E = 300000 N/C

given that; potential difference ΔV = Ed

we substitute

ΔV = 300000 × 0.002

ΔV = 600 V

Therefore, the required potential difference, between the capacitor plates is 600 V

Answer:

68 meters moved in the next seconds

Explanation:

Given

[tex]u= 30m/s[/tex]

[tex]a = 4m/s^2[/tex]

Required

Distance covered by the car in the next second

At a point in time t, the current distance is calculated as:

[tex]s(t) = ut + \frac{1}{2}at^2[/tex]

Substitute values for a and u in the above equation.

[tex]s(t) =30 * t + \frac{1}{2} * 4 * t^2[/tex]

[tex]s(t) =30t + 2t^2[/tex]

Next, we generate the second degree Taylor polynomial as follows;

Calculate velocity (s'(t))

Differentiate s(t) to get velocity

[tex]s(t) =30t + 2t^2[/tex]

[tex]s'(t) =30 + 4t[/tex]

Calculate acceleration (s"(t))

Differentiate s'(t) to get acceleration

[tex]s'(t) =30 + 4t[/tex]

[tex]s"(t) =4[/tex]

When t = 0

We have:

[tex]s(0) = 30 * 0 + 2 * 0^2 = 0[/tex]

[tex]s'(0) =30 + 4*0 = 30[/tex]

[tex]s"(0) = 4[/tex]

So, the second degree tailor series is:

[tex]T_2(t) = s(t) * t^0 + s'(t) * \frac{t^1}{1!} + s"(t) * \frac{t^2}{2!}[/tex]

To see the distance moved in the next second, we set t to 1

So, we have:

[tex]T_2(1) = s(1) * 1^0 + s'(1) * \frac{1^1}{1!} + s"(2) * \frac{1^2}{2!}[/tex]

[tex]T_2(1) = s(1) * + s'(1) * \frac{1}{1} + s"(1) * \frac{1}{2}[/tex]

[tex]T_2(1) = s(1) * + s'(1) * 1 + s"(1) * \frac{1}{2}[/tex]

[tex]T_2(1) = s(1) * + s'(1) + \frac{s"(1)}{2}[/tex]

Solving s(1), s'(1) and s"(1)

We have:

[tex]s(1) =30*1 + 2*1^2 = 32[/tex]

[tex]s'(1) =30 + 4*1 = 34[/tex]

[tex]s"(1) =4[/tex]

Hence:

[tex]T_2(1) = 32 + 34 + \frac{4}{2}[/tex]

[tex]T_2(1) = 32 + 34 + 2[/tex]

[tex]T_2(1) = 68[/tex]

Answer:

the required or need power is 115960.57 Watts  

Explanation:

First of all, we take down the data we can find from the question, to make it easier when substituting values into formulas.

mass of swordfish m = 650 kg

Cross - sectional Area A = 0.92 m²

drag coefficient C[tex]_D[/tex] = 0.0091

speed v = 30 m/s

density p = 1026 kg/m³

Now, we determine our Drag force F[tex]_D[/tex]

Drag force F[tex]_D[/tex]  = [tex]\frac{1}{2}[/tex] × C[tex]_D[/tex] × A × p × v²

Next, we substitute the values we have taken down, into the formula.

Drag force F[tex]_D[/tex]  = [tex]\frac{1}{2}[/tex] × 0.0091 × 0.92 × 1026 × (30)²

Drag force F[tex]_D[/tex]  = 4.294836 × 900

Drag force F[tex]_D[/tex]  = 3865.3524

Now, we determine the power needed P[tex]_w[/tex]

P[tex]_w[/tex] = F[tex]_D[/tex] × v

we substitute  

P[tex]_w[/tex] = 3865.3524 × 30

P[tex]_w[/tex] = 115960.57 Watts  

Therefore, the required or need power is 115960.57 Watts  

Answer:

1) The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object at any point of its motion is 16.892 joules.

Explanation:

1) An object under simple harmonic motion is conservative, since there is no dissipative forces acting during motion (i.e. friction, air viscosity). The amplitude of the motion can be found easily by Principle of Energy Conservation by the fact that maximum elastic potential energy ([tex]U_{e}[/tex]), in joules, is equal to maximum translational kinetic energy ([tex]K[/tex]), in joules:

[tex]U_{e} = K[/tex]

[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)

Where:

[tex]k[/tex] - Spring constant, in newtons per meter.

[tex]A[/tex] - Amplitude, in meters.

[tex]m[/tex] - Object mass, in kilograms.

[tex]v[/tex] - Speed of the object at equilibrium, in meters per second.

If we know that [tex]k = 450\,\frac{N}{m}[/tex], [tex]m = 0.25\,kg[/tex] and [tex]v = 0.3\,\frac{m}{s}[/tex], then the amplitude of the motion is:

[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]

[tex]k\cdot A^{2} = m\cdot v^{2}[/tex]

[tex]A = v\cdot \sqrt{\frac{m}{k} }[/tex]

[tex]A = \left(0.3\,\frac{m}{s} \right)\cdot \sqrt{\frac{0.25\,kg}{0.3\,\frac{m}{s} } }[/tex]

[tex]A \approx 0.274\,m[/tex]

The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object ([tex]E[/tex]), in joules, is found either by maximum elastic potential energy or by maximum translational kinetic energy, that is: ([tex]k = 450\,\frac{N}{m}[/tex], [tex]A \approx 0.274\,m[/tex])

[tex]E = U_{e}[/tex]

[tex]E = \frac{1}{2}\cdot k\cdot A^{2}[/tex]

[tex]E = \frac{1}{2}\cdot \left(450\,\frac{N}{m} \right) \cdot (0.274\,m)^{2}[/tex]

[tex]E = 16.892\,J[/tex]

The total energy of the object at any point of its motion is 16.892 joules.

Answer:

C.A single nucleus divides into two or more nuclei and gives off energy best describes what occurs in a fission reaction.

Answer:

C.

A single nucleus divides into two or more nuclei and gives off energy.

hope it is helpful to you

Answer:

B)60km[W]

Explanation:

The boat travels 20km/h. So every hour the boat goes 20 miles. So if one hour equals 20km. Then 3 hours will be 3*20km which equals 60km. The boat is also going west. So you should consider putting that in your answer as well. So the answer would be B)60km[W].

Hope that helps!

Answer:

θ = cos^(-1) (-A/B)

Explanation:

The image of the reauktant forces A & B are missing, so i have attached it.

Now, from the attached image, we will see that;

Angle between A and B is θ

Also;

A = Bcos(180° − θ)

Now, in trigonometry, we know that;

cos(180° − θ) = -cosθ

Thus;

A = -Bcosθ

cosθ = -A/B

Thus;

θ = cos^(-1) (-A/B)

Answer:

7374.4

Explanation:

I took the test

(filler so I can post)

Answer:

a)   m_v = m_s (([tex]\frac{w_o}{w}[/tex])² - 1) ,  b)  m_v = 1.07 10⁻¹⁴ g

Explanation:

a) The angular velocity of a simple harmonic motion is

           w² = k / m

where k is the spring constant and m is the mass of the oscillator

let's apply this expression to our case,

silicon only

         w₉² = [tex]\frac{K}{m_s}[/tex]

         k = w₀² m_s

silicon with virus

         w² = [tex]\frac{k}{m_s + m_v}[/tex]

          k = w² (m_v + m_s)

in the two expressions the constant k is the same and q as the one property of the silicon bar, let us equal

           w₀²  m_s = w² (m_v + m_s)

           m_v = ([tex]\frac{w_o}{w}[/tex])²  m_s - m_s

           m_v = m_s (([tex]\frac{w_o}{w}[/tex])² - 1)

b) let's calculate

          m_v = 2.13 10⁻¹⁶ [([tex]\frac{20.4}{2.85}[/tex])² - 1)]

          m_v = 1.07 10⁻¹⁴ g

Answer:

The Gas Giants have more moons.

Explanation:

Mercury-0

Venus-0

Earth-1

Mars-2

Jupiter-66

Saturn-62

Uranus-27

Neptune-13

Answer:

Explanation:

energy of photon having wavelength of 400 nm = 1237.5/400 eV

= 3.1 eV.

Maximum kinetic energy of photoelectrons = 1.1 eV .

Threshold energy Ф = 3.1 - 1.1 = 2 eV .

energy of photons having wavelength of 310 nm = 1237.5 / 310 eV = 4 eV .

Maximum kinetic energy of photoelectrons = energy of photons - Threshold energy

= 4 - 2 = 2 eV .

Required kinetic energy K₀= 2 eV.

Answer:

a)  w = 2.52 10⁷ rad / s, b)  K / K₀ = 1.19 10⁴

Explanation:

a) We can solve this exercise using the conservation of angular momentum.

Initial instant. Before collapse

         L₀ = I₀ w₀

Final moment. After the collapse

         L_f = I w

angular momentum is conserved

        L₀ = L_f

         I₀ w₀ = I w                 (1)

The moment of inertia of a sphere is

        I = 2/5 m r²

we take from the table the mass and diameter of the star

        m = 1,991 10³⁰ kg

        r₀ = 6.96 10⁸ m

        r = 6.37 10⁶ m

to find the angular velocity let's use

       w = L / T

where the length of a circle is

      L = 2π r

      T = 24 days (24 h / 1 day) (3600 s / 1h) = 2.0710⁶ s

we substitute

      w = 2π r / T

      wo = 2π 6.96 10⁸ / 2.07 10⁶

      wo = 2.1126 10³ rad / s

we substitute in equation 1

      w = [tex]\frac{I_o}{I}[/tex]

      w = 2/5 mr₀² / 2/5 m r² w₀

      w = ([tex]\frac{r_o}{r}[/tex]) ² wo

      w = (6.96 10⁸ / 6.37 10⁶) ² 2.1126 10³

      w = 2.52 10⁷ rad / s

b) the kinetic energy ratio

      K = ½ m w²

       K₀ = ½ m w₀²

       K = ½ m w²

       K / K₀ = (w / wo) ²

       K / K₀ = 2.52 10⁷ / 2.1126 10³

       K / K₀ = 1.19 10⁴

Answer:

conservation of linear momentum

We were told that two objects became stuck together hence we have to use the principle of conservation of  momentum to obtain the final velocities of the carts.

The principle of conservation of momentum lets us know that the momentum before collision is equal to the momentum after collision. As such we can write; m1u1 + m2u2 = m1v1 + m2v2.

We can use this thus principle to obtain the final speeds of the carts since the two objects that collided became stuck together.

Learn more about conservation of momentum: https://brainly.com/question/11256472

Answer:

the smallest separation between two objects is 0.8067 m

Explanation:

Given the data in the question;

Altitude h = 5.75 km = 5750 m

Diameter D = 4.0 mm = 0.004 m

λ = 460 nm = 4.6 × 10⁻⁷ m

Now, Using Rayleigh criterion for Airy disks resolution.

we know that, Minimum angular separation for resolving two points is;

θ = 1.22λ / D

so we substitute

θ = (1.22 × 4.6 × 10⁻⁷)  / 0.004

θ = 5.612 × 10⁻⁷ / 0.004

θ = 1.403 × 10⁻⁴ rad  

so minimum separation [tex]d_{min[/tex] =  θh

so we substitute

[tex]d_{min[/tex] = (1.403 × 10⁻⁴) × 5750 m

[tex]d_{min[/tex] = 0.8067 m

Therefore, the smallest separation between two objects is 0.8067 m

Answer:

 w = 0.319 rad / s

Explanation:

This is an angular momentum problem, let's form a system composed of the feeder and the squirrel, therefore the forces during the collision are internal and the angular momentum is conserved.

initial instant. Before the squirrel jumps

           L₀ = m v r

final instant. After the trough and the squirrel are together

          L_f = (I_fetter + I_ardilla) w

angular momentum is conserved

          L₀ = L_f

          m v r = (I_fetter + I_ardilla) w

          w = [tex]\frac{mvr}{I_{fetter} + I_{ardilla} }[/tex]

the moment inercial ofbody is

         I_thed = 2.00 kg m²

We approach the squirrel to a specific mass

          I_ardilla = m r²

we substitute

            w = m v r / ( I_[feefer  + m r²)

let's calculate

              w = 3 3.40 6.30 10⁻² / (2.00 + 3.00 (6.30 10-2)² )

              w = 0.6426 / 2.0119

               w = 0.319 rad / s

Answer:

d. the star is a member and also a part of an eclipsing binary star system.

Explanation:

If any star happens to be brighter for an extended period of time, however, at some times, it becomes dimmer, is due to the fact that the star is being overshadowed (hiding behind another star that is known as eclipse).

The above-mentioned eclipsing binary star system is essentially what has been defined. It occurs when two stars' orbit planes are so similar that one star will obscure (the light) of the other.

Thus, option D is correct.

Answer:

It doesn't interact with baryonic matter and it's completely invisible to light and other forms of electromagnetic radiation, making dark matter impossible to detect with current instruments. But scientists are confident it exists because of the gravitational effects it appears to have on galaxies and galaxy clusters.

Explanation:

Answer:

mass, weight, strength of gravity increases, weight decreases

Explanation:

got it on edge

Answer:

An object’s

✔ mass

will remain constant throughout the universe, but its

✔ weight

can change from planet to planet.

If you increase the mass of a planet, what happens to its gravity?

✔ strength of gravity increases

If the gravity on a planet decreases, what happens to the weight of an object on that planet?

✔ weight decreases

Explanation:

right on edge 22