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IgorLugansk [536]
3 years ago
5

a truck can travle 300 miles in the same time that it takes a car to travel 330 miles. If the truck's rate is 5 miles per hour s

lower than the car's find the average rate for each
Physics
1 answer:
yanalaym [24]3 years ago
5 0
<h2>The speed of car is 55 miles per hour  and speed of truck is 50 miles per hour .</h2>

Explanation:

Let the speed of the car is v miles/h , therefore the speed of truck is ( v - 5 ) miles per hour .

Because time = \frac{Distance}{Speed}

Thus in case of truck the time t₁ = \frac{300}{v-5}

In case of car the time t₂ = \frac{330}{v}

But time is the same in both cases , therefore

\frac{300}{v-5} = \frac{330}{v}

or  30 v = 1650  

or v = 55 miles per hour

Therefore speed of car is 55 miles per hour

Thus speed of truck is 50 miles per hour

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A force of 75 N is applied to a spring, causing it to stretch 0.3 m. What is the spring constant of the spring?
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This city is the capital of the state of Georgia since 1868 and was burned during the Civil War.
Sphinxa [80]

The correct answer is - Atlanta.

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3 0
3 years ago
Read 2 more answers
A metal wire breaks when its tension reaches 100 newton. If the radius and length of the wire were both doubled then it would br
serg [7]

Answer:

200 N

Explanation:

Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.

So,  E = σ/ε = FL/eA

Now, since at break extension = e.

So making e subject of the formula, we have

e = FL/EA = FL/Eπr² where r = radius of metal wire

Now, when the radius and length are doubled, we have our extension as e' = F'L'/Eπr'² where F' = new force on metal wire, L' = new length = 2L and r' = new radius = 2r

So, e' = F'(2L)/Eπ(2r)²

e' = 2F'L/4Eπr²

e' = F'L/2Eπr²

Since at breakage, both extensions are the same, e = e'

So,  FL/Eπr² = F'L/2Eπr²

F = F'/2

F' = 2F

Since F = 100 N,

F' = 2 × 100 N = 200 N

So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.

7 0
3 years ago
The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t
Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = 854\times 10^{3}\ Hz

Power, P = 50 kW = 50\times 10^{3}\ W

I_{p} = 1\ photon/s/m^{2}

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s

Area of the sphere, A = 4\pi r^{2}

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is 1\ photon/s/m^{2}

I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}

1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}

r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m

Now,

We know that:

1 ly = 9.4607\times 10^{15}\ m

Thus

r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly

5 0
3 years ago
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