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n200080 [17]
2 years ago
16

A student hears a police siren. What would change the frequency that the student hears? Check all that apply.

Physics
2 answers:
BlackZzzverrR [31]2 years ago
9 0

Answer:

1

2

3

4

Explanation:

It's right on Edge

ser-zykov [4K]2 years ago
4 0
<span>A student hears a police siren.

The arithmetic of the Doppler Effect shows that if the distance between
the source and observer is changing, then the observer hears a different
frequency compared to the frequency actually radiating from the source. 

Thus the first four choices would cause the student to hear a different
frequency:

-- if the student walked toward the police car
-- if the student walked away from the police car
-- if the police car moved toward the student
-- if the police car moved away from the student

The last two choices wouldn't affect the frequency heard by the student,
since the perceived frequency of a sound doesn't depend on its intensity.

-- if the intensity of the siren increased
-- if the intensity of the siren decreased.</span>
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Nana76 [90]
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8 0
2 years ago
If 1.00 mol of argon is placed in a 0.500-L container at 28.0 ∘C , what is the difference between the ideal pressure (as predict
Rudik [331]

Answer:

1.98 atm

Explanation:

Given that:

Temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28 + 273.15) K = 301.15 K

n = 1

V = 0.500 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

P × 0.500 L = 1 ×0.0821 L atm/ K mol  × 301.15 K

⇒P (ideal) = 49.45 atm

Using Van der Waal's equation

\left(P+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT

R = 0.0821 L atm/ K mol  

Where, a and b are constants.

For Ar, given that:

So, a = 1.345 atm L² / mol²

b =  0.03219 L / mol

So,  

\left(P+\frac{1.345\times \:1^2}{0.500^2}\right)\left(0.500-1\times 0.03219\right)=1\times 0.0821\times 301.15

P+\frac{1.345}{0.25}=\frac{24.724415}{0.46781}

P=\frac{24.724415}{0.46781}-\frac{1.345}{0.25}

⇒P  (real) = 47.47 atm

Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm

4 0
2 years ago
An ocean liner is cruising at 10 meters/second and is about to approach a stationary ferryboat. A parcel is released from the oc
Afina-wow [57]
The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.

First, we determine how long the parcel will fall using:

s = ut + 1/2 at²

where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity. 

5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds

Now, we may use this time to determine the horizontal distance covered by the parcel by using:
distance = velocity * time

The horizontal velocity of the parcel will be equal to the horizontal velocity of the cruise liner.

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4 0
2 years ago
Sally finds herself stranded on a frozen pond so slippery that she can't stand up or walk on it. To save herself, she throws one
8_murik_8 [283]

Answer:

a) 2.5 m/s. (In the opposite direction to the direction in which she threw the boot).

b) The centre of mass is still at the starting point for both bodies.

c) It'll take Sally 12 s to reach the shore which is 30 m from her starting point.

Explanation:

Linear momentum is conserved.

(mass of boot) × (velocity of boot) + (mass of sally) × (velocity of Sally) = 0

5×30 + 60 × v = 0

v = (-150/60) = -2.5 m/s. (Minus inicates that motion is in the opposite direction to the direction in which she threw the boot).

b) At time t = 10 s,

Sally has travelled 25 m and the boot has travelled 300 m.

Taking the starting point for both bodies as the origin, and Sally's direction as the positive direction.

Centre of mass = [(60)(25) + (5)(-300)]/(60+5)

= 0 m.

The centre of mass is still at the starting point for both bodies.

c) The shore is 30 m away.

Speed = (Distance)/(time)

Time = (Distance)/(speed) = (30/2.5)

Time = 12 s

Hope this Helps!!!

7 0
2 years ago
Read 2 more answers
The Orion nebula is one of the brightest diffuse nebulae in the sky (look for it in the winter, just below the three bright star
Oxana [17]

Answer:

T=183.21K

Explanation:

We have to take into account that the system is a ideal gas. Hence, we have the expression

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Thus, it is necessary to calculate n and V

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V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (5.9*10^{15}m)^3=8.602*10^{47}m^3

V=8.86*10^{50}L

and for n

n=\frac{(4000M_s)/(2*mH)}{6.022*10^{23}mol^{-1}}=3.95*10^{36}mol

Hence, we have (1 Pa = 9.85*10^{-9}atm)

T=\frac{PV}{nR}=\frac{(6.8*10^{-9}*9.85*10^{-6}atm)(8.86*10^{50}L)}{(0.0820\frac{atm*L}{mol*K})(3.95*10^{36}mol)}\\\\T=183.21K

hope this helps!!

8 0
2 years ago
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