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atroni [7]
3 years ago
13

An 800-kHz radio signal is detected at a point 4.5 km distant from a transmitter tower. The electric field amplitude of the sign

al at that point is 0.63 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the magnetic field amplitude of the signal at that point
Physics
1 answer:
Hitman42 [59]3 years ago
3 0

Answer:

B_2=2.1nT

Explanation:

From the question we are told that:

Frequency F=800kHz

Distance d=4.5km

Electric field amplitude B_2=0.63V/m

Generally the equation for momentum is mathematically given by

 B=\frac{E}{C}

Therefore

 B_2=\frac{0.63}{3*10^8}

 B_2=0.21*10^{-8}

 B_2=2.1nT

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The largest tendon in the body, the Achilles tendon, connects the calf muscle to the heel bone of the foot. This tendon is typic
tia_tia [17]

Answer:

Tension on tendon = 1669800N

Explanation:

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7 0
3 years ago
A car travels 40 miles in 30 minutes.
lukranit [14]

Answer:

(a)Average velocity ,v =128.74 Km/hr

(b)Kinetic Energy , K=958546.875 Joule

(c)Distance, s=268.8m

(d)Acceleration, a= - 2.38 m/s^2

<u>Explanation</u>:

<u>Given</u>:

Distance travelled = 40 miles

Time taken = 30 minutes.

(A) The average velocity in kilometres/hour

Converting 40 miles into km ,

we know that,

1 mile = 1.60934

40 miles =  40 x 1.60934

so 40 miles  =  64.3738 Km

similarly converting 30 minutes into hours

1 minute = \frac{1}{60}hours

30 minute = \frac{30}{60}hours

30 minute = \frac{1}{2}hours

Now

Average velocity = \frac{Speed}{time}

Substituting Values,

Average velocity = \frac{64.3738}{\frac[1}{2}}

Average velocity = 64.3738 \times 2

Average velocity =128.74 Km/hr

(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

converting  velocity to m/s

128.74  \times \frac{5}{18}

=>35.75 m/s----------------------------------------------------------(1)

kinetic energy  K= \frac{1}{2}mv^2

Substituting the values,

K= \frac{1}{2}1500(35.75)^2

K= \frac{1}{2}1500(1278.06)

K= \frac{1500 \times (1278.06)}{2}

K= \frac{1917093.75}{2}

K=958546.875 Joule---------------------------------------------(2)

(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping

Lets use Distance formula,

S= ut+\frac{1}{2}at^2

Substituting the known values,

s= ut+\frac{1}{2}at^2

s= (37.75)(15)+\frac{1}{2}a(15)^2

s=566.25+\frac{1}{2}a(225)

s=566.25+\frac{(225a)}{2}-------------------------------------(3)

(D) What is the average acceleration of the car (in m/s2) during braking?

Using the formula

v=u +at

re arranging the formula we get,

a = \frac{v - u}{t}

Substituting the values

a = \frac{0 - 35.75}{15}

a = \frac{- 35.75}{15}

a= - 2.38 m/s^2----------------------------------------(4)

Now substituting 4 in 3 we get

s=566.25+\frac{(225( - 2.38)}{2}

s=566.25+\frac{-535.5}{2}

s=536.25-267.75

s=268.8m--------------------------------------------------------------(5)

4 0
3 years ago
Which of the following is a small star that has reached the end of its stellar evolution?
Akimi4 [234]
D.) White Dwarf

It is the smallest star whose mass is approximately equal or greater than 1.4M
Here, M = mass of the Sun.

Hope this helps!
3 0
3 years ago
Read 2 more answers
How many times farther is Proxima Centauri from the sun than is earth from the sun
Aleks04 [339]

Answer:

Of the three stars in the system, the dimmest - called Proxima Centauri - is actually the nearest star to the Sun. The two bright stars, called Alpha Centauri A and B form a close binary system; they are separated by only 23 times the Earth - Sun distance.

5 0
3 years ago
Read 2 more answers
A person throws a ball straight up. He releases the ball at a height of 1.75 m above the ground and with a velocity of 12.0 m/s.
Lyrx [107]

Answer:

a) 1.22 s

b) 9.089 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-12}{-9.81}\\\Rightarrow t=1.22\ s

Time taken by the ball to reach the highest point is 1.22 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=12\times 1.22+\frac{1}{2}\times -9.81\times 1.22^2\\\Rightarrow s=7.339\ m

The maximum height the ball will reach above the ground is 1.75+7.339 = 9.089 m

8 0
3 years ago
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