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mihalych1998 [28]

3 years ago

5

At what point in a projectile’s trajectory does it experience a minimum velocity? as soon as it is launched half way to its high

est point at its highest point just before it lands

2 answers:

andrew-mc [135]3 years ago

6 0

Answer:

at the highest point

Explanation:

velocity of a particle becomes minimum when the particle projected halfway or from ground reaches its maximum height.

VLD [36.1K]3 years ago

5 0

No matter what direction a projectile is traveling, its velocity
will be minimum at the highest point in its trajectory.

I just realized ... that's also true of an Earth satellite.

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lozanna [386]

Answer:

$2.655\times 10^{13}\ photons$

Explanation:

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$E=\frac{hc}{\lambda}\\\Rightarrow E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{570\times 10^{-9}}\\\Rightarrow E=3.487\times 10^{-19}\ J$

Number of photons

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3 years ago

Newton's law of cooling is , where is the temperature of an object, is in hours, is a constant ambient temperature, and is a pos

kirill [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The time taken is $t = 6.89 \ hrs$

Explanation:

From the question we are told that

The value of $k = 0.15hr^{-1}$

The the interior temperature is $u(t = 0) = 70 ^oF$

The external temperature is $T = 11 ^oF$

The required interior temperature is $u_{t} = 32 ^oF$

The newton cooling law is

$\frac{du}{dt} = -k (u -T)$

=> $\frac{du}{u-T} = - kdt$

Now integrate both sides we have

$\int\limits \frac{du}{u-T} =\int\limits - kdt$

$ln (u - T) = -kt + c$

Since c is a constant lnC = c will also give a constant so

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=> $u = T + Ce^{-kt}$

substituting value

$70 = 11 +Ce^{-0* 0.15}$

$C = 59$

Hence

$u_t = 11 + 59 e^{-0.15 *t}$

$32= 11 + 59 e^{-0.15 *t}$

=> $t = -\frac{ln(\frac{21}{59} )}{0.15}$

$t = 6.89 \ hrs$

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