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alex41 [277]
3 years ago
12

Two fire trucks have sirens that emit waves of the same frequency. As the fire trucks approach a person, the person hears a high

er frequency from truck X than from truck Y.
1. Which of the following statements about truck X can be correctly inferred from this information?
A) It is traveling faster than truck Y.
B) It is closer to the person than truck Y.
C) It is speeding up, and truck Y is slowing down.
Physics
1 answer:
Mumz [18]3 years ago
5 0

Answer:

Only option A is correct

Explanation:

From the concept of Doppler effect, only speed matters. Thus, the faster a vehicle is moving, the closer together the sound waves get compressed and the higher the frequency. For example, for a very fast vehicle traveling at the speed of sound; the compressions are all right on top of each other. Thus, faster speed means closer compressions and higher frequencies. Hence, option only option A must be true because X is a higher frequency and so it must be going faster. The distance to the person will affect the volume but will not the pitch so Option B is not correct. Option C too is not correct because It doesn’t matter whether you are speeding up or slowing down, it only matters who is going faster. For example, from option c concept, if truck X was going 10 m/h and speeding up while truck Y was going 50 mph and slowing down, it would not meet the requirement that X has a higher frequency because only actual speed matters, not what is happening to that speed. Thus only option A is the correct answer

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An 30-turn coil has square loops measuring 0.341 m along a side and a resistance of 3.61 Ω. It is placed in a magnetic field tha
Verdich [7]

Answer:

Thus, the induced current in the coil at t =1.73s is 9.98 A.

Explanation:

Faraday's law says

$\varepsilon = N \frac{d \Phi_B}{dt} $

where N is the number of turns and \Phi_B is the magnetic flux through the square coil:

Now,

N = 30

\theta = 37.5^o

A = (0.341m)^2= 0.11623m^2

B = 1.45t^3;

therefore,

$\varepsilon = N \frac{d \Phi_B}{dt}  = N\frac{d ( BA\:cos(\theta))}{dt}  = 30*\frac{d ( (1.45t^2)(0.1163)\:cos(37.5^o))}{dt}$

=30*(0.1163)\:cos(37.5^o)*1.45*\dfrac{d ( t^3)}{dt}  = 12.04t^2

\boxed{\varepsilon = 12.04t^2}

is the emf induced in the coil.

Now, the loop is connected to R = 3.61\Omega resistance; therefore, at t = 1.73s

\varepsilon = RI = 12.04t^2

RI = 12.04(1.73)^2

RI = 36.03

I = \dfrac{36.03}{3.61\Omega }

\boxed{I = 9.98A }

Thus, the current in the coil at t =1.73s is 9.98 A.

8 0
4 years ago
What does the term “retroperitoneal” mean in the terms of the location of the kidneys ?
lakkis [162]

Answer:

"Retroperitoneal" refers to the back of the peritoneum, the membrane that lines the anatomical space in the abdominal cavity. Kidney stones may cause pain to the organs within the retroperitoneal space. A diagram of the aorta, a retroperitoneal structure.

Explanation:

5 0
3 years ago
Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds.
s344n2d4d5 [400]

This question is incomplete, the complete question is;

A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t² + 8√t, where t is the time in seconds and s(t) is the distance in feet from the top of the building.

(a) Write a formula for the average velocity of the ball for t near 4.

(b) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds

(c) What is your estimate for the instantaneous velocity of the ball at t = 4

Answer:

a)

Average velocity, (Vavg)  of the ball for t near 4.

Vavg = [s(4) - s(0)] / (4 - 0)

Where s(4) = -16 × 4² + 8 × √4= - 240 m

s(0) = -16 × 0 + 8 * 0 = 0

b)

duration = 1 sec

Vavg = [s(5) - s(4)] / (5 - 4)

s(5) = -16 × 52 + 8 × √5 = - 382 m

s(4) = -16 × 42 + 8  √4 = - 240 m

Vavg = (-382 - (-240)) / (5 - 4)

Vavg = - 142.1 m/s

duration = 0.5 sec

Vavg = [s(4.5) - s(4)] / (4.5 - 4)

s(4.5) = -16 × 4.52 + 8 × √4.5 = - 307 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-307 - (-240)) / (4.5 - 4)

Vavg= - 134.1 m/s

duration = 0.05 sec

Vavg = [s(4.05) - s(4)] / (4.05 - 4)

s(4.05) = -16 × 4.052 + 8 × √4.05 = - 246 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-246 - (-240)) / (4.05 - 4)

Vavg= - 126.8 m/s

c)

Instantaneous velocity, v = ds/dt

= - 16 × 2 × t + 8 ×× (0.5 / √t )

= - 32 × t + 4/√t

ds/dt at t = 4 is,

v = - 32 × 4 + 4 / √4

= - 126 m/s

5 0
3 years ago
In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

  • k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

7 0
3 years ago
A hot-air balloon and its basket are accelerating upward at 0.265 m/s2, propelled by a net upward force of 688 N. A rope of negl
Sergeu [11.5K]

Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>

Answer:

The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon m_1 and the mass of the basket m_2, then according to newton's second law:

(1). \:F = (m_1+m_2)a,

where a =0.265m/s^2 is the upward acceleration, and F = 688N is the net propelling force (counts the gravitational force).

Also, the tension T in the rope is 79.8 N more than the basket's weight; therefore,

(2). \:T = m_2g+79.8

and this tension must equal

T -m_2g =m_2a

(3). \:T = m_2g +m_2a

Combining equations (2) and (3) we get:

m_2a = 79.8

since a =0.265m/s^2, we have

\boxed{m_2 = 301.13kg}

Putting this into equation (1) and substituting the numerical values of F and a, we get:

688N = (m_1+301.13kg)(0.265m/s^2)

\boxed{m_1 = 2295 kg}

Thus, the mass of the balloon and the basket is  2295 kg and 301 kg respectively.

8 0
3 years ago
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