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Naddik [55]
3 years ago
14

As an emergency vehicle approaches Bob and moves away from Jill, how does the actual frequency of the siren change? A) As an eme

rgency vehicle approaches Bob the frequency increases. B) As an emergency vehicle moves away from Jill the frequency decreases. C) The actual frequency of the siren does not change despite appearances. D) Bob does not notice any change until the vehicle passes him and the siren increases.
Physics
2 answers:
asambeis [7]3 years ago
8 0

its C on USAtestprep

Tamiku [17]3 years ago
7 0
The correct answer is: 
<span>C) The actual frequency of the siren does not change despite appearances.

In fact, Bob will observe an increase in the apparent frequency as the emergency vehicle approaches him, while Jill will observe a decrease in the apparent frequency as the emergency vehicle moves away from him, because of the Doppler effect (the relative velocity between the observer and the source of the sound is changing), but this effect involves the apparent frequency, while the real frequency of the siren will remain the same.</span>
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Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu
Luba_88 [7]

Answer:

V_1=8 V_2

Explanation:

Given that:

  • Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
  • separation distance of capacitor 2, d_2=d
  • separation distance of capacitor 1, d_1=2d
  • quantity of charge on capacitor 2, Q_2=Q
  • quantity of charge on capacitor 1, Q_1=4Q

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

C=\frac{k.\epsilon_0.A}{d}.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}

From eq. (1)

For capacitor 2:

C_2=\frac{k.\epsilon_0.A}{d}

For capacitor 1:

C_1=\frac{k.\epsilon_0.A}{2d}

C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]

We know, potential differences across a capacitor is given by:

V=\frac{Q}{C}..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}

V_2=\frac{Q.d}{k.\epsilon_0.A}

& for capacitor 1:

V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}

V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}

V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]

V_1=8 V_2

6 0
3 years ago
In 1.71 minutes, a ski lift raises four skiers at constant speed to a height of 148 m. The average mass of each skier is 62.9 kg
Vitek1552 [10]

Answer:

3560.36 Watts

Explanation:

Power, P=\frac {nΔW}{Δt) where P is power, n is the number of skiers, t is time in seconds and Δt is change in time, ΔW is given by mgh where m is mass, g is gravitational constant, h is height

Substituting n for 4 skiers, m for 62.9 Kg, g for 9.81, h for 148 m and t for 1.71*60=102.6 seconds

P=\frac {4*62.9*9.81*148}{1.71*60}=3560.360702  Watts

Average power is approximately 3560.36 Watts

6 0
4 years ago
At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf. What is the local acceleration of g
Strike441 [17]

Answer:

31.905 ft/s²

Explanation:

Given that

Mass of the pilot, m = 120 lb

Weight of the pilot, w = 119 lbf

Acceleration due to gravity, g = 32.05 ft/s²

Local acceleration of gravity of found by using the relation

Weight in lbf = Mass in lb * (local acceleration/32.174 lbft/s²)

119 = 120 * a/32. 174

119 * 32.174 = 120a

a = 3828.706 / 120

a = 31.905 ft/s²

Therefore, the local acceleration due to gravity at that elevation is 31.905 ft/s²

3 0
3 years ago
A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-dire
oee [108]

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v).  From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

(c) The collision is elastic

6 0
3 years ago
What is a force that attracts all matter to each other?
Temka [501]
The answer is gravity. I hope this helps. 
5 0
3 years ago
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