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3 years ago

14

A boat in a race accelerates uniformly at 0.2 m/s² for 18 seconds to reach a final speed of 6 m/s. Work out the initial speed of

the boat.

1 answer:

Xelga [282]3 years ago

4 0

Answer:

initial velocity=2.4m/s

Explanation:

Using the formula:

v=u+at

v= final velocity (6m/s)

u=initial velocity (???)

a=acceleration (0.2 m/s²)

t=time(18 seconds)

v=u+at

6=u+(0.2*18)

6=u+3.6

u=6-3.6

u (initial velocity)=2.4m/s

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a square loop whose sides are long is made with copper wire of radius , assuming resistivity of copper is . if a magnetic field

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A square loop whose sides are long is made of copper wire of radius , given the resistivity of copper is . if the magnetic field perpendicular to the loop changes at a constant rate of I = 14.029 mA.

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The complete question is :

A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?

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1 year ago

It appears as though the moon disappears and the sun comes out in the daytime. Based on what you know, explain why this is happe

zimovet [89]

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The Earth is toward the sun

Explanation:

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3 years ago

Read 2 more answers

Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion

stellarik [79]

Answer:

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

(c) The entropy change of the gas is equal to zero.

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = 240 K

= initial volume of gas =

= final volume of gas =

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Explanation:

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2 years ago

Convert the following quantities <br><img src="https://tex.z-dn.net/?f=25m%20%7B%7D%5E%7B2%7D%20%20%5C%3A%20into%20%5C%3A%20cm%2

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Answer:

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Explanation:

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2 years ago

A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite

sladkih [1.3K]

Answer:

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Explanation:

From the question we are told that

The radius of the orbit is $r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m$

The gravitational force is $F_g = 6600 \ N$

The kinetic energy of the satellite is mathematically represented as

$KE = \frac{1}{2} * mv^2$

where v is the speed of the satellite which is mathematically represented as

$v = \sqrt{\frac{G M}{r^2} }$

=> $v^2 = \frac{GM }{r}$

substituting this into the equation

$KE = \frac{ 1}{2} *\frac{GMm}{r}$

Now the gravitational force of the planet is mathematically represented as

$F_g = \frac{GMm}{r^2}$

Where M is the mass of the planet and m is the mass of the satellite

Now looking at the formula for KE we see that we can represent it as

$KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

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3 years ago