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LuckyWell [14K]
3 years ago
5

Two trains on separate tracks move toward each other. Train 1 has a speed of 109 km/h; train 2, a speed of 99.0 km/h. Train 2 bl

ows its horn, emitting a frequency of 500 Hz. What is the frequency heard by the engineer on train 1?
Physics
1 answer:
Rufina [12.5K]3 years ago
4 0

Answer:

f_o=592.36 Hz

Explanation:

Given that

Train 1 (observer):

Speed = 109 km/h

Train 2 (source):

Speed = 99 km/h

Train 2 emitting frequency = 500 Hz

We know that observer and source are moving toward each other, then frequency heard by observer can be given as follows

f_o=\left(\dfrac{C+V_o}{C-V_s}\right)f_s

Where C is the velocity of sound  (C=1225 Km/h)

Now by putting the values

f_o=\left(\dfrac{C+V_o}{C-V_s}\right)f_s

f_o=\left(\dfrac{1225+109}{1225-99}\right)\times 500

f_o=592.36 Hz

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A 1.25 x 10-4 C charge is moving5200 m/s at 37.0° to a magnetic fieldof 8.49 x 10-4 T. What is the magneticforce on the charge?
juin [17]

0.0003321 Newtons

Explanation

We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equation to find the force

\begin{gathered} F=qvBsinθ \\ where\text{ F is the magnetic force} \\ q\text{ is the charge} \\ v\text{ is the velocity of the charge} \\ Bis\text{ the magnetic field} \\ \theta\text{ is the angle} \end{gathered}

so

Step 1

Let

\begin{gathered} q=1.25*10^{-4\text{ }}C \\ v=5200\frac{m}{s} \\ \theta=37\text{ \degree} \\ B=8.49*10^{-4}T \end{gathered}

now, replace

\begin{gathered} F=qvBs\imaginaryI n\theta \\ F=1.25*10^{-4}\text{ C*5200 }\frac{m}{s}*8.49*10^{-4}Tsin(37) \\ F=0.0003321\text{ Newtons} \\  \end{gathered}

so, the answer is

0.0003321 Newtons

I hope this helps you

7 0
1 year ago
A nonconducting sphere has radius R = 1.81 cm and uniformly distributed charge q = +2.08 fC. Take the electric potential at the
Sloan [31]

Answer:

a) V = -0.227 mV

b) V = -0.5169 mV

Explanation:

a)

Inside a sphere with a uniformly distributed charge density, electric field is radial and has a magnitude

E = (qr) / (4πε₀R³)

As we know that

V = -\int\limits^r_0 {E} \, dr

By solving above equation, we get

V = (-qr²) / (8πε₀R³)

When

R = 1.81 cm

r = 1.2 cm

q = +2.80 fC

ε₀ = 8.85 × 10⁻¹²

V = (-2.80 × 10⁻¹⁵ × (1.2 × 10⁻²)²) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²)³)

V = -2.27 × 10⁻⁴ V

V = -0.227 mV

b)

When

r = R

R = 1.81 cm

q = +2.80 fC

ε₀ = 8.85 × 10⁻¹²

V = (-qR²) / (8πε₀R³)

V = (-q) / (8πε₀R)

V = (-2.80 × 10⁻¹⁵) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²))

V = -5.169 × 10⁻⁴ V

V = -0.5169 mV

4 0
4 years ago
What is the measure of how much a material resists the formation of an electric field?
shusha [124]
The answer is c capacitance
5 0
4 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!
Varvara68 [4.7K]

The answer to your question is,

4 kilometers north

-Mabel <3

6 0
3 years ago
An eartly method of measuring the speed of light makes use of a rotating slotted wheel.
Pavel [41]

Answer:

Explanation:

Distance traveled by light = 2 x 550 m = 1100 m

time taken by light to travel this distance = 1100 / 3 x 10⁸

=  366.67 x 10⁻⁸ s

angle between two consecutive slots = 2π / 500 rad

= .004π

angular velocity of wheel = angle moved / time taken

= .004π / 366.67 x 10⁻⁸

= 1091π radian / s

b )  linear speed of a point on the edge of the wheel

= ω R , r is radius of wheel , ω is angular velocity.

= 1091π x 5 x 10⁻²

= 1712.8 m /s

4 0
4 years ago
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