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3 years ago

6

Who studies this branch of physics? What types of careers are there in this branch of physics? What particular subjects should s

omeone study if they want a job in this branch of physics?

1 answer:

kolezko [41]3 years ago

8 0

The answers to all three questions depend on WHICH branch

of Physics you're talking about . . . a piece of information you've

neglected to specify.

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Help! PROJECTILE PROBLEM: A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25

dsp73

Here,

height at failure, h1 = 525 m,
upward acceleration, a = 2.25 m/s^2,
velocity = v m/s,
<span>
SO, </span>
<span>
v^2 = 2*a*h = 2*2.25*525 = 2362.5 </span>
Now, acceleration, g = 9.8 m/s^2,
<span>
SO, </span>
<span>
heigt, h1 = v^2/2g = 2362.5 / 2*9.8 = 120.54 meters </span>
Hence,
<span>
a) </span>
Total height = 525+120.54 = 645.54 meters

b)
<span>time, for h1, t = v/g = sqrt(2362.5)/9.8 = 4.96 sec

---------------------------------

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4 0

3 years ago

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A(n) 83 kg fisherman jumps from a dock into a 139 kg rowboat at rest on the West side of the dock. If the velocity of the fisher

Anna71 [15]

here we can say that there is no external force on fisherman and dock

so here we will use momentum conservation theory

As per momentum conservation

initial momentum of fisherman + boat = final momentum of fisherman + boat

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

now we will have

$83(3.1) + 139(0) = 83 v + 139 v$

$257.3 = 222v$

$v = 1.16 m/s$

so the speed of boat and fisherman will be 1.16 m/s

3 0

3 years ago

The rocket now has a thruster that malfunctions and is now pushing the rocket in the wrong direction. What is the new net force

Lana71 [14]

Answer:

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Explanation:

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3 years ago

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The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velo

Lina20 [59]

I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J.
At point A, all this energy has converted into kinetic energy, which is:
$K= \frac{1}{2}mv^2$
And since K=7.35 J, we can find the velocity, v:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s$

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3 years ago

Whats the answer to 45 + 54 ÷ p - q

Lynna [10]

There’s nothing to answer to

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3 years ago