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3 years ago

6 How many seconds will it take for a satellite to travel 450 km at a rate of 120 m/s?please help me !!!!!!!!!!

Physics

1 answer:

katrin2010 [14]3 years ago

6 0

Explanation:

Distance = rate × time

450,000 m = 120 m/s × t

t = 3750 s

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Edit question A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a t

Pachacha [2.7K]

Answer:

176.9N

Explanation:

The following data were given

wire length,L=37cm=0.37m

linear density=18g/m

tube length,=192cm=1.92m,

speed of sound,v=343m/s

Since it is an open-closed tube, the second harmonic frequency is expressed as

$f_{3}=3(\frac{v}{4l} )\\f_{3}=3(\frac{343}{4*1.92})\\f_{3}=133.98Hz$

The relationship between the tension, linear density and second harmonic frequency is expressed as

$f_{3}=\frac{1}{2l_{w}}\sqrt{\frac{T}{\alpha } } \\T=(f_{3}*2l_{w})^{2}\alpha \\T=(133.984*2*0.37)^{2}*18*10^{-3}\\T=176.9N$

7 0

3 years ago

A car travels eastwards at 60km/h for 2h, then travels northwards at 20km/h for 8h. Find,

sveta [45]

Answer:

$a) Average\ Speed=28\ km/h\\b) Average\ Velocity= 20\ km/h$

Explanation:

$We\ are\ given\ that,\\Velocity\ of\ the\ car\ Eastwards=60\ km/h\\Time\ taken\ by\ the\ car\ Eastwards=2\ h\\Velocity\ of\ the\ car\ Northwards=20\ km/h\\Time\ taken\ by\ the\ car\ Northwards=8\ h\\Hence,\\As\ we\ know\ that,\\Speed=\frac{Distance}{Time}\\Distance= Speed* Time\\Now,\ lets\ find\ the\ distance\ covered\ by\ the\ car\ in\ both\ the\ cases.$

$Hence,\\Distance\ Covered\ During\ its\ Eastward\ Journey=60*2=120\ km\\Distance\ Covered\ During\ its\ Northward\ Journey=20*8=160\ km\\Now,\\As\ we\ know\ that\ Average\ Speed=\frac{Total\ Distance}{Total\ Time} \\Here,\\Total\ distance\ of\ the\ car=Distance\ Covered\ During\ its\ Northward\ Journey+Distance\ Covered\ During\ its\ Eastward\ Journey\\Hence,\\Total\ distance\ of\ the\ car=120+160=280\ km\\Total\ time\ taken\ by\ the\ car=8+2=10\ hours\\Hence,\\$ $Average\ Speed\ Of\ the\ Car\ throughout\ its\ journey=\frac{280}{10}=28\ km/h$

$Now,\\For\ Average\ Velocity\ we\ need\ to\ consider\ displacement\ as:\\Average\ Velocity\ =\frac{Total\ Displacement}{Total\ Time} \\Now,\\As\ we\ already\ know\ that\ displacement\ is\ the\ shortest\ distance\\ from\ the\ initial\ to\ the\ final\ point.\\We\ observe\ that, \\The\ car\ forms\ a\ right\ triangle\ during\ its\ complete\ journey.\\Hence,\\As\ we\ already\ know\ that,\\Distance\ travelled\ Eastwards= 120\ km\\Distance\ travelled\ Northwards= 160\ km\\Hence,\\$ $We\ may\ apply\ Pythagoras\ Property\ to\ find\ the\ net\ displacement.\\Hence,\\a^2+b^2=c^2\\120^2+160^2=c^2\\14400+25600=c^2\\40000=c^2\\c=\sqrt{40000}\\c=200\\Hence,\\Total\ displacement=200\ km\\Total\ Time\ taken=2+8=10\ hours\\Hence,\\Average\ Velocity\ Of\ the\ Car=\frac{200}{10}=20\ km/h$

7 0

3 years ago

A light bulb radiates 110 nW of single-frequency sinusoidal electromagnetic waves uniformly in all directions. Calculate the ave

krek1111 [17]

Answer:

The intensity of the light from the bulb would be

3.501 x $10^{-6}$ W/ $m^{2}$

Explanation:

Given

The Power = 110 n W = 110 x $10^{-9}$ W

the distance r = 50 mm = 50 /1000 = 0.05 m

The intensity can be obtained with the relationship below;

I = Power/area ......1

The area of the sphere would be used in this case since the bulb is spherical; A=4π $r^{2}$

Putting it into equation 1, we have;

I = P/ 4π $r^{2}$

I = 110 x $10^{-9}$ / 4 x π x $0.05^{2}$

I = 3.501 x $10^{-6}$ W/ $m^{2}$

Therefore the intensity of the light from the bulb would be

3.501 x $10^{-6}$ W/ $m^{2}$

6 0

3 years ago

A 12.0 g bullet was fired horizontally into a 1 kg block of wood. The bullet initially had a speed of 250 m/s. The block of wood

LuckyWell [14K]

Answer:

The rise in height of combined block/bullet from its original position is 0.45m

Explanation:

Given;

mass of bullet, m₁ = 12 g = 0.012 kg

mass of block of wood, m₂ = 1 kg

initial speed of bullet, u₁ = 250 m/s.

initial speed of block of wood, u₂ = 0

From the principle of conservation of linear momentum, calculate the final speed of the combined block/bullet system.

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final speed of the combined block/bullet system.

0.012 x 250 + 0 = v (0.012 + 1)

3 = v (1.012)

v = 3/1.012

v = 2.96 m/s

From the principle of conservation of energy, calculate the rise in height of the block/bullet combined from its original position.

¹/₂mv² = mgh

¹/₂v² = gh

¹/₂ (2.96)² = (9.8)h

4.3808 = 9.8h

h = 4.3808/9.8

h = 0.45 m

Therefore, the rise in height of combined block/bullet from its original position is 0.45m

7 0

3 years ago

Read 2 more answers

A light string is wrapped around the edge of the smaller disk, and a 1.50 kg block is suspended from the free end of the string.

LenaWriter [7]

Answer: 7.41 m/s

Explanation: By using the law of of energy, kinetic energy of the brick as it falls equals the potential energy before falling.

Kinetic energy = mv²/2, potential energy = mgh

mv²/2 = mgh

v²/2 = gh

v² = 2gh

v = √2gh

Where g = 9.8 m/s², h = 2.80m

v = √2×9.8×2.8 = 7.41 m/s

7 0

3 years ago

6 How many seconds will it take for a satellite to travel 450 km at a rate of 120 m/s?please help me !!!!!!!!!!​

6 How many seconds will it take for a satellite to travel 450 km at a rate of 120 m/s?please help me !!!!!!!!!!