A sound wave has a frequency of 350 Hz and a wavelength of 2.25m. At what speed will the
sound wave move? Show your work.

Answer:

About 5000 years.

Explanation:

Answer:

Explanation:

it will take more than half a million years for the photon to escape the sun. If you think it's about a centimeter, then it will take about 5,000 years for the photon to get outside the su

Hi there!

We can use Ampère's Law:

[tex]\oint B \cdot dl = \mu_0 i_{encl}[/tex]

B = Magnetic field strength (B)
dl = differential length element (m)
μ₀ = Permeability of free space (T/Am)

Since this is a closed-loop integral, we must integrate over a closed loop. We can integrate over a rectangular-enclosed area of the rim of the solenoid - ABCD - where AD and BC are perpendicular to the solenoid.

Thus, the magnetic field is equivalent to:
[tex]\oint B \cdot dl = \int\limits^A_B {B} \, dl + \int\limits^B_C {B} \, dl + \int\limits^C_D {B} \, dl + \int\limits^D_A {B} \, dl[/tex]

Since AD and BC are perpendicular, and since:
[tex]\oint B \cdot dl = B \cdot L = BLcos\phi[/tex]

[tex]BLcos(90) = 0[/tex]

If perpendicular to the field, the equation equals 0.

Additionally, since AB is outside of the solenoid, there is no magnetic field present, so B = 0. The only integral we integrate now is:
[tex]\oint B \cdot dl = \int\limits^C_D {B} \, dl[/tex]

Which is horizontal and inside the solenoid. Let the distance between C and D be 'L', and the enclosed current is equivalent to the number of loops multiplied by the current:

[tex]B L = \mu_0 Ni[/tex]

N = # of loops per length multiplied by the length, so:
[tex]BL = \mu_0 nL i \\\\B = \mu_0ni[/tex]

Plug in the given values and solve. Remember to convert # of loops to # of loops per unit length.

[tex]B = \mu_0 (100/0.1)(2) = (4\pi *10^{-7})(1000)(2) = \boxed{0.00251 T}[/tex]

Answer:

Work is best defined as a force exerted on a body to cause it move over a certain distance.

work=force×displacement×cosθ

Explanation:

Answer:

125J

Explanation:

[tex]work \: = force \: \times distance \\ = 25 \times 5 \\ = 125joules[/tex]

Answer:

Explanation:

The number of mole of the hydrogen sample at 27 C and 750 mmHg pressure is:

[tex]n=\frac{PV}{RT}=\frac{(0.9898\times 10^{5})(5\times 10^{-3})}{(8.314)(27+273)}\approx 0.0198[/tex] mol

This number of mole will be the same although we change the pressure and temperature. For STP (Temperature = 0 C = 273 K, and Pressure = 760 mmHg = 76 cmHg = 1 Bar = 1.01 x [tex]10^{5}[/tex] Pa):

[tex]P'V'=nRT' \rightarrow V'=\frac{nRT'}{P'}=\frac{0.0198(8.314)(273)}{1.01\times 10^{5}}\approx 44.49 \times 10^{-5}[/tex] in [tex]m^{3}[/tex]. But in L, we find that V' = 0.4449 L

number of protons

the atomic number of an atom is equal to the number of protons

Explanation:

no. of seconds in a year = 60 × 60 × 24 × 365

= 31,536,000 seconds

so 2.05 × 10⁵ into years = 0.0065 years (approximately)

Answer:

234 90 +energy

Explanation:

because alpha decrease by 4 2helium

Answer:

Explanation:

The velocity of sound from the plucked string in the tennis racquet is:

[tex]v=\sqrt{\frac{FL}{m}}=\sqrt{\frac{274(0.74)}{28}}\approx 2.69 ms^{-1}[/tex]

Then the frequency will be:

[tex]f=\frac{v}{2L}=\frac{2.69}{2(0.74)}\approx 1.81 Hz[/tex]

Answer:

Explanation:

You can calculate the total electric charge that passes through the conductor as [tex]q=It=(180\times 10^{-3})(5\times 60)= 54 C[/tex]. It means that the number of electron that passes through the conductor is:

[tex]n=\frac{q}{e}=\frac{54}{1.6\times 10^{-19}}=33.75\times 10^{19}[/tex]

Answer:

Explanation:

The fennec may be the world's smallest fox, but it sports outlandishly large ears! In fact, relative to body size, it has the largest ears of any member of the canid family. It uses those big ears to listen for sounds of prey in the sand. The ears also help dispel body heat to keep the fox cool.

Answer:

3 pi R / s     or   9.42 R/s    or  540 degrees/sec

Explanation:

Ball covers 2 pi radians each rotation

2 pi * 3 R / 2  s =   3 pi R / s

the angular speed of the ball is 3 pi R / s     or   9.42 R/s    or  540 degrees/sec

Angular speed can be defined as the measures of speed of how fast the central angle of a rotating body changes with respect to time

Angular speed mainly implies how quickly the rotation of an object occur means it is described as the change in the angle of the object per unit of time.

To calculate the speed of a rotational motion the angular speed value need to be known, hence  The angular speeds formula is used to calculate the distance traveled by an object in terms of rotation and revolutions per unit of time.

The unit of angular speed of an object is radian per second and Both angular speed, angular velocity are determined by using the same formula, angular velocity is  vector quantity which describes both magnitude and direction.

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Answer:

when it has a fewer amount of electrons than protons

nothing

Explanation:

we can say that it was certainly bad for the shopkeeper obviously and we should not be making questions about the physics behind that accident and should call the cops or 911

Answer:

Energy = mcΔt

where m the mass ,  c specific heat capacity

Δt is the change in temperature

1200x10^3=75x4184xΔt

Δt=3.82°

Explanation:

Answer:

The answer is in the picture above.

During the phase change in which heat is being added but the temperature remains constant, the heat energy is used to break the bonds in the molecules.

A phase change is the conversion of one state of matter to another. e.g. solid to liquid or liquid to gas.

Substances can exists in three states namely;

In a change of phase, heat energy is needed to break the existing bonds between the molecules of the initial state. For example, heat energy is needed to break the bonds of ice to form water.

However, during this phase change, the temperature remains constant.

Therefore, during the phase change in which heat is being added but the temperature remains constant, the heat energy is used to break the bonds in the molecules.

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Answer:

See below

Explanation:

Assuming starting from zero velocity

vfinal = at

24   = a (5.5)

a = 4.36  m/s^2        I think part of your question is missing....what was the initial velocity?

Answer:

0.73kg

Explanation:

Potential gravitational energy is equal to [tex]Pe=mgh[/tex], where Pe is potential gravitational energy, m is mass in kilograms, g is gravity in m/s^2 and h is height in meters.

Substitute:

[tex]Pe=mgh\\23=m*9.8*3.2\\23=31.36m\\0.73=m[/tex]

Answer:

the answer is B I hope it helps out

The instant acceleration of the object is 10 m/s² and the displacement of the object is 228 m.

The instant acceleration of the object is determined by taking the derivative of the velocity as follows;

v = 8 + 2.5t²

a = dv/dt

a = 5t

a(2) = 5(2)

a(2) = 10 m/s²

The displacement of the object is determined by taking the integral of velocity as follows;

v =  8 + 2.5t²

x = ∫v

x = 8t + 2.5t³/3

x(0) = 0

x(6) = 8(6) + 2.5(6)³/3

x(6) = 228 m

Thus, the displacement of the object is 228 m.

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Answer:

0.52     8.9  

Explanation:

calc.

The tension in the rope just before the student releases it is 576 N.

Tension is defined in physics as the pulling force transmitted axially by a string, rope, chain, or similar object.

Just before the student releases the rope, his velocity is zero. At this point, all of his kinetic energy has been converted into potential energy. Therefore, we can write:

1/2 m [tex]v^2[/tex] = mgh

h = [tex]v^2[/tex] / (2g) = [tex](6.2 m/s)^2[/tex] / (2 x 9.8 [tex]m/s^2[/tex]) ≈ 1.95 m

m g = 56 kg x 9.8 [tex]m/s^2[/tex] ≈ 549 N

The tension required to keep the rope taut is equal to the horizontal component of the force due to the weight of the student, which is:

T = m g / cos(θ)

tan(θ) = h / L

tan(θ) = 1.95 m / 10.0 m ≈ 0.195

θ ≈ 10.9°

So, T = m g / cos(θ) ≈ 576 N

To find the maximum tension in the rope during the swing, we can use conservation of angular momentum.

L = I ω

I = (1/3) m [tex]L^2[/tex]

I = (1/3) x 10.0 kg x [tex](10.0 m)^2[/tex] ≈ 3333.3 kg[tex]m^2[/tex]

The angular velocity of the rope just after the student releases it is:

ω = v / L

ω = 6.2 m/s / 10.0 m ≈ 0.62 rad/s

L = I ω ≈ 2070 N m s

Thus, the angular momentum of the system just after the student releases the rope is 2070 N m s.

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A likely source of bias in the survey is undercoverage bias because some parts of the community may not be represented in the survey results.

The bias present in the survey may lead to error in the result obtained from the random sampling survey.

One of the likely source of bias in the survey is undercoverage.

It is not possible to ascertain the accurate opinion of the entire community from the 50 residents chosen for the survey.

Thus, a likely source of bias present in survey is undercoverage bias because some parts of the community may not be represented in the survey results.

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Answer:

B. Question wording bias may be present because the question leads those responding toward a certain response.

Explanation:

Answer:

2.67 ohm

Explanation:

V=IR

R=V/I

In which V is voltage and I is current and R is resistance

Hi there!

A)

At this point, the pilot is experiencing a DOWNWARD NET centripetal force (ALWAYS points towards the center of the circle) as well as a downward force due to gravity. Additionally, we can say that the normal force exerted by the seat on the pilot is its apparent weight.

[tex]\Sigma F = F_g + F_N[/tex]

**The net force is the centripetal force.

We can assign both to be positive since they work in the same direction. (Assigning down as positive in THIS situation will make things easier).

We know that:
[tex]F_c = \frac{mv^2}{r}[/tex]

Fc = Centripetal force (N)
m = mass (kg)
v = velocity (m/s)

r = radius (m)

[tex]F_g = mg[/tex]

Fg = Force due to gravity (N)
m = mass (kg)
g = acceleration due to gravity (9.81 m/s²)

Solve for mass. (Given weight)

[tex]m = \frac{F_g}{g}\\\\m = \frac{719}{9.8} = 73.367 kg[/tex]
Now, we have everything we need to solve.

[tex]\frac{mv^2}{r} = F_g + F_N \\\\\frac{mv^2}{r} - F_g = F_N\\\\F_N = \frac{(73.367)(225^2)}{3110} - 719 = \boxed{475.278 N}[/tex]

B)
Weightlessness is achieved when the pilot is in free fall. In this instance, FN (Normal force by the seat) = 0 N.

We can use the same setup as above.

[tex]\frac{mv^2}{r} = F_g + 0 (F_N)[/tex]

[tex]\frac{mv^2}{r} = mg[/tex]

Solve for 'v'.

[tex]\frac{v^2}{r} = g\\\\v = \sqrt{gr} = \sqrt{(9.81)(3110)} = \boxed{174.669 \frac{m}{s}}[/tex]

C)
At the BOTTOM, we have the force of gravity working DOWNWARD (Negative in this instance), while the centripetal force (NET FORCE) and normal force are upward (Positive). Do another summation of forces:
[tex]\Sigma F = F_N - F_g\\\\\\frac{mv^2}{r} + F_g = F_N \\\\F_N = \frac{(73.367)(225^2)}{3110} + 719 = \boxed{1913.278 N}[/tex]

For two horses pull horizontally on ropes attached to a stump, the magnitude of F and its direction is mathematically given as

F=1711 N

dF= 135 degrees

Generally, the equation for the components of F1   is mathematically given as

For The x axis

1210 - F2x = 0

F2x = 1210

For the y axis

F2y = R

F2y = 1210

In conclusion

F = sqrt(2)*1210

F=1711 N

Fopt the direction of F2

dF= 90 + 45

dF= 135 degrees

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Answer:

No

Explanation:

It is not possible for a machine to act as a force multiplier and speed multiplier simultaneously. This is because machines which are force multipliers cannot gain in speed and vice-versa.

It is not possible for a machine to act as a force multiplier and speed multiplier simultaneously. This is because machines which are force multipliers cannot gain in speed and vice-versa.

(1) The time taken for the drag racer to accelerate is 4.89 s

(2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.

(1) To calculate the time required to accelerate 18 ft/sec² from rest to a velocity of 60 mph, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

(2) To calculate the speed of the contestant, we use the formula below

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

(a) In feet = (9.43/0.3048) = 30.94 ft/s

(b)  in miles per hr = (9.43×2.24) = 21.12 miles per hr

Hence, (1) The time taken for the drag racer to accelerate is 4.89 s (2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.

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Answer:

the answer to this question is option c

there process which eventually takes place is called refraction

Explanation:

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