Standing at a crosswalk, you hear a frequency of 550 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of the siren is 475 Hz. Determine the ambulance's speed from these observations. (Take the speed of sound to be 343 m/s.)

Answer:

They take the form of heat, I think thats it but I cant see if you put down any answers

Explanation:

Answer:a

Explanation:

they form hear

The sun emits electromagnetic waves with a power of  

4.0 ∗ 10  (26)  W.

Answer:

Explanation:

The answer:

https://www.chegg.com/homework-help/questions-and-answers/race-car-starting-rest-travels-around-circular-turn-radius-247-m-certain-instant-car-still-q402991

Total acceleration of car is 148.31 m/s².

Centripetal acceleration is a characteristic of an object's motion along a circular path. Centripetal acceleration applies to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.

Given parameters:

Radius of the circular path: r = 22.5 m.

Angular speed: ω = 0.541 rad/s.

So, centripetal acceleration; α = ω²r = (0.541)² × 22.5 m/s² = 6.58 m/s².

Tangential acceleration: [tex]\alpha_t[/tex] = αr = 6.58 × 22.5 = 148.16 m/s².

Hence, total acceleration = √(α² + [tex]\alpha_t[/tex]²) = √(6.58² +148.16²) = 148.31 m/s².

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Answer:

Energy consumed is 0.00033 Joules.

Explanation:

the formula of Energy is:

Energy = power/ time.

Answer:

C

Explanation:

Pls brainliest

Hi there!

We can use the following equation to find the frequency of each harmonic:

[tex]f_n = \frac{n}{2L} \sqrt{\frac{T}{\lambda}}[/tex]

n = nth harmonic

L = length of string (m)

T = Tension of string (N)

λ = linear density (kg/m)

Begin by converting the linear mass density to kg:

2.00g /m · 1 kg / 1000g = 0.002 kg/m

Now, we can use the equation to find the first three harmonics.

First harmonic:

[tex]f_1 = \frac{1}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{131.76 Hz}[/tex]

Second harmonic:

[tex]f_2 = \frac{2}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{263.52Hz}[/tex]

Third harmonic:

[tex]f_3 = \frac{3}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{395.28Hz}[/tex]

Answer:

she will move in the same direction at the same speed forever.

Explanation:

If there are no outside forces like gravity the net force will never change, she will just keep flying for forever and ever! poor lady

From Young's single slit experiment, the distance away from the wall will be 1.068 m

Given that 587.9 nm of wavelength of light passes through a single slit 0.73 mm wide, it creates a diffraction pattern.

From the question, the following parameters are given:

The wavelength of the light λ  =  587.9 nm

The width of the slit a = 0.73 mm

Fringe width X = 0.86 mm

The distance away from the wall D = ?

The fringe width is related to the wavelength  of the light source by the equation:

X = Dλ ÷ a

Substitute all the parameters into the formula

0.83 × [tex]10^{-3}[/tex] = 587.9 × [tex]10^{-9}[/tex] D ÷ 0.73 ×

Cross multiply

587.9 × [tex]10^{-9}[/tex] D = 6.278 × [tex]10^{-7}[/tex]

make D the subject of the formula

D = 6.278 × [tex]10^{-7}[/tex] ÷  587.9 × [tex]10^{-9}[/tex]

D = 1.068 m

Therefore, the distance away from the wall is 1.068 m

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If the first minimum is [tex]0.86 mm[/tex] from the central maximum, the distance away is 1.07 meters.

Given the data in the question;

From Thomas Young's single slit experiment:

[tex]\frac{a*y}{L} = m * \lambda[/tex]    

Where a is the width of the slit, y is first minimum, L is the distance, m is the order number and λ is the wavelength.

We substitute our values into the equation

[tex]\frac{0.00073m\ *\ 0.00086m}{L} = 1\ *\ ( 5.879*10^{-7}m)\\\\\frac{0.0000006278m^2}{L} = 5.879*10^{-7}m\\\\L = \frac{0.0000006278m^2}{5.879*10^{-7}m} \\\\L = 1.07m[/tex]

Therefore, if the first minimum is [tex]0.86 mm[/tex] from the central maximum, the distance away is 1.07 meters.

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Answer:

F = - K x for spring     (note that that F here is given in grams, F = m g is correct)

K here is 100 g / cm      for the spring constant

x = -420 g / 100 g/cm = -4.2 cm

The spring would compress 4.2 cm for a total length of 20 - 4.2 = 15.8

d)  to compress the spring 6.8 cm one can see that the load would be 680 g

Answer:

Explanation:

Initial vertical velocity

vy₀ = 50.0sin60 = 43.3 m/s

This initial velocity is reduced to zero by gravity in a time of

t = v/a = 43.3/9.81 = 4.41 s

h(max) = ½gt² = ½(9.81)4.41² = 95.6 m

The ball will return to earth in the same amount of time

t(max) = 2(4.41) = 8.82 s

The horizontal velocity is

vx = 50.0cos60 = 25.0 m/s

d = vt = 25.0(8.82) = 221 m

That 's one heck of a kick! No air resistance of course.

Answer:

im pretty sure the answer is C

Explanation:

F = Icurrent×length×Bfieldstrength×sin(angle field to wire)

in our case

Icurrent = 10 A

length = 0.02km = 20 meters

B = 10^-6 T

angle = 30 degrees.

F = (20 A)(20m)(10^-6 T)×sin(30) = 400× 10^‐6 ×0.5 N =

= 200 × 10^-6 = 2 × 10^‐4 N

Answer:

An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.

Explanation:

Answer:

An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.

Explanation:

Answer:

d. The truck that left later was travelling faster

Explanation:

Both trucks travelled from the same place to the same place, meaning they both travelled the same distance;

They both arrive at the same time, but the second truck left later so it took less time to travel the distance than the first truck;

The only variable that can account for this difference is speed;

The one that left later, therefore, must have been going faster.

Answer:

C number is write i think

Hi there!

For this problem, we must use the conservation of angular momentum. This is an example of an inelastic "collision", so:

I₁w₁ + I₂w₂ = (I₁ + I₂)wf

We know that the moment of inertia of a disk is 1/2mR², so we can calculate the moments of inertia for both disks:

Disk 1: 1/2(2)(0.40²) = .16 kgm²/s

Disk 2: 1/2(2)(0.20²) = .04 kgm²/s

Plug in the values. Let counterclockwise be positive.

.16(-50) + .04(50) = (.16 + .04)wf

Solve:

wf = -30 rev/s

Answer:

Explanation:

Summing moments about the CG to zero will show that the two normal forces are equal and have a value of FL = FR = 684/2 = 342 N

Pressure on the left

PL = 342 / 6.20e-4 = 551,612.9032... = 5.5e5 Pa

Pressure on the right

PR = 342 / 9.20e-4 = 371,739.1304... = 3.7e5 Pa

A) The force from the left tire is; FL = 342 N

B) The pressure from the left tire is; PL = 551613 N/m²

C) The force from the right tire is; FR = 342 N

D) The pressure from the right tire is; PR = 371739 N/m²

We see that;

FL and FR are upward forces

W is the downward force.

We know that in equilibrium;

Sum of upward forces = sum of downward forces

Thus;

FL + FR = W

We are given W = 684 N

Since W is at the center, it means that FL = FR. Thus;

FL = FR = 684/2

FL = FR = 342 N

We are given;

Contact area of left tire; AL = 6.2 × 10⁻⁴ m²

Contact area of right tire; AR = 9.2 × 10⁻⁴ m²

Formula for pressure is;

Pressure = Force/Area

Pressure on the left tire;

PL = FL/AL

PL = 342/(6.2 × 10⁻⁴)

PL = 551613 N/m²

Pressure on the t right tire;

PR = FR/AR

PR = 342/(9.2 × 10⁻⁴)

PR = 371739 N/m²

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Answer:

Explanation:

Let d be the distance to the center of mass from the front wheels

Sum moments about the front wheel contact point to zero

1240(9.8)[d] - 1240(9.8)(1 - 0.67)[3.2] = 0

1240(9.8)[d] = 1240(9.8)(1 - 0.67)[3.2]

                d = (1 - 0.67)[3.2]

                d = 1.056 m

This question involves the concepts of work done and the frictional force.

a. Work done by the person is "692.82 N".

b. Work done by the frictional force is "490.5 N".

a.

Work done by the person can be given by the following formula:

[tex]W=FdCos\theta[/tex]

where,

W = work done by the person = ?

F = Force applied by the person = 80 N

d = distance traveled = 10 m

θ = angle between force and motion = 30°

Therefore,

[tex]W=(80\ N)(10\ m)Cos30^o[/tex]

W = 692.82 N

b.

Work done by the frictional force is given by the following formula:

[tex]W_f=fd\\W_f=\mu mgd[/tex]

where,

[tex]W_f[/tex] = work done by the frictional force = ?

μ = coefficient of friction = 0.5

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]W_f=(0.5)(10\ kg)(9.81\ m/s^2)(10\ m)[/tex]

[tex]W_f=490.5\ N[/tex]

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Answer:

Explanation:

20 n is an unknown amount

If that is supposed to be 20 N(ewtons)

then W = Fd = 20(15) = 300 J

Answer: it will be 300 newton meters

Explanation:

Answer:

In space we feel weightlessness because the earth's gravity has less effect on us. The Earth's gravitational attraction at those altitudes is only about 11% less than it is at the Earth's surface. If you had a ladder that could reach as high as the shuttle's orbit, your weight would be 11% less at the top.

Explanation:

Hope this helps:)

We have that for the Question "a)what is the resistivity in (ohm-cm) of the material? b) what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material"

Answer:

From the question we are told

This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.

A) Resistivity is given as,

[tex]p = \frac{RA}{l}[/tex]

where,

[tex]R = \frac{V}{I}[/tex]

Therefore,

[tex]p = \frac{VA}{Il}\\\\p = \frac{1*(0.1*0.5)}{0.019*1.5}\\\\p = 1.754 ohm-cm[/tex]

B) Conductivity is given as,

[tex]U = \frac{\rho}{pe}\\\\U = \frac{10^{17}}{10^{10}*1.6*10^{-19}}\\\\U = 6.25*10^{25} cm^3/V-s[/tex]

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From the diagram, the angular speed will increase clockwise, the sign of the angular acceleration will be negative and the direction of rotation will be clockwise direction and the sign of the angular acceleration is negative. The correct answer is option B

Given that two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Counterclockwise is considered the positive rotational direction.

If m1 is 24 kilograms, m2 is 12 kilograms, and mbar is 10 kilograms, The moment of object m1 will be equal to the moment of object m2 without the Mbar

Let assume that the length L of the seesaw is 9 cm.

Anticlockwise moment = 24 x 9/3 = 72Nm

Clockwise moment = 12 x 2(9/3) = 72 Nm

With the consideration of mass of the bar Mbar, this will add to clockwise moment of the seesaw.

Therefore, the direction of rotation will be clockwise direction.

Angular acceleration is positive when object is speeding up and negative when slowing down. Also, angular acceleration is positive when speed increases in an anticlockwise direction and negative when speed increases in the clockwise direction.

From the diagram, since the angular speed increase clockwise, the sign of the angular acceleration will be negative.

We can conclude that the direction of rotation will be clockwise direction and the sign of the angular acceleration is negative.

The correct answer is option B

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20 seconds

Explanation:

Let [tex]x_a[/tex] be the distance traveled by the accelerating car and [tex]x_c[/tex] be the distance traveled by the car moving with a constant velocity. When they cover the same distance, we can write

[tex]x_a = x_c \Rightarrow v_{0a}t + \frac{1}{2}at^2 = v_ct[/tex]

where [tex]v_c[/tex] is the velocity of car moving at a constant rate and a is the acceleration of the accelerating car. Since the accelerating car started from rest, then [tex]v_{0a}[/tex] is zero so our equation above simplifies to

[tex]\frac{1}{2}at^2 = v_ct[/tex]

Note that the variable t cancels out so solving for t, we get

[tex]\frac{1}{2}at = v_c \Rightarrow t = \dfrac{2v_c}{a}[/tex]

Plugging in the given values,

[tex]t = \dfrac{2(20\:\text{m/s})}{2\:\text{m/s}^2} = 20\:\text{s}[/tex]

Answer:

Heat Loss = 5 kg * 700 deg K * 220 J / (kg*deg K) = 7.70E5 J

Since there are 4.186 J/cal

Heat Loss = 7.70E5J / 4.186 J/cal = 1.84E5 cal

Heat Gain = 2000 g * 85 deg K / cal / (deg K g) + M * 540 cal/g

Heat Gain = 1.70E5 cal + M * 540 cal/g

M = (1.84 - 1.70) E5 g / 540

25.9 g

25.9 g or 25.9 cm^3 or .0259 L   of water will boil away

Explanation:

1 minute is 60 seconds so you multiply 60 * 15 and then multiply that answer * 2

Answer:

Steady-state theory, in cosmology, a view that the universe is always expanding but maintaining a constant average density, with matter being continuously created to form new stars and galaxies at the same rate that old ones become unobservable as a consequence of their increasing distance and velocity of recession.

Cosmic inflation is a faster-than-light expansion of the universe that spawned many others. ... Cosmic inflation solves these problems at a stroke. In its earliest instants, the universe expanded faster than light (light's speed limit only applies to things within the universe).

Hi there!

We can use the equation:

d = v₀t + 1/2at²

Where:

v₀ = initial velocity downward

a = acceleration due to gravity

t = time

Plug in given values:

d = 4(12) + 1/2(9.8)(12²)

d = 48 + 705.6 = 753.6 m

W = 25 J

Explanation:

Work done on an object is defined as

[tex]W = Fd = (10\:\text{N})(2.5\:\text{m}) = 25\:\text{J}[/tex]

The pressure dropped by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.

Using the attached map below, in the morning of November 9, the pressure situated at the storm's center = 1000 MB isobar located at the center.

Meanwhile, on November 11, 0000z the pressure situated at the storm's center = 976 MB isobar  

The difference in this pressure is regarded as the pressure drop in the storm's center and it is determined as follows;

= 1000 MB - 976 MB

= 24 MB

Therefore, we can conclude that the pressure drop by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.

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