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3 years ago

14

Do i become a professional astrophysicist after getting PhD in Astrophysics ? Or will i be a postdoc research student i mean wil

l i be practicing under another researcher ? Will i get the correct salary just after finishing PhD ?

1 answer:

DaniilM [7]3 years ago

4 0

Most jobs in Astrophysics require a PhD. That's why that's what I'll be going for when I start my degree in it.

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Your friend just challenged you to a race through an obstacle course. You know in order to beat him, you must run 30 meters with

seropon [69]

Answer:

Velocity = 0.5 m/s South (A)

Explanation:

You need to determine the average rate of velocity.

The equation you will use is velocity = displacement/time

The displacement is 30m South.

The time is 60 seconds.

Plug into the equation Velocity = 30m South/60 s

Velocity = 0.5 m/s South

3 0

3 years ago

A pendulum clock gives correct time at 20°c how many sec/day will it gain or loose when the temperature fall to 5°c? cofficient

rewona [7]

Answer:

129.6 seconds

Explanation:

Given that :

α = 0.0002°c-1

θ1 = 20°C

θ2 = 5°C

Time t = one day ; Converting to seconds ; number of seconds in a day ; (24 * 60 * 60) = 86400 seconds

Let dT= change in time

Using the relation :

dT = 0.5* α * dθ * t

dθ = (20 - 5) = 15°C

dT = 0.5 * 0.0002 * 15 * 86400

dT = 129.6 seconds

5 0

2 years ago

What is the gravitational potential energy of a 65.7 kg person standing on the roof of a 135 meter building?

horrorfan [7]

For the purpose we will use the following equation for potential energy:
U = m * g * h
In the above equation, m represents the mass of the object, h represents the height of the object and g represents the gravitational field strength (9.8 N/kg on Earth).
When we plug values into the equation, we get following:
U= 65.7kg * 9.8 N/kg *135m = 86921.1 J = 86.92 kJ

8 0

3 years ago

If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 64 ft/sec, its height after t seconds

stepan [7]

Answer:

a) $s_{max} = 96\,ft$ , b) $v(4.449\,s) = -78.368\,\frac{ft}{s}$

Explanation:

a) The maximum height is obtained with the help of the First and Second Derivative Tests:

First Derivative

$v(t) = 64 - 32\cdot t$

$64 - 32\cdot t = 0$

$t = 2\,s$

Second Derivative

$a(t) = -32$ (absolute maximum)

The maximum height reached by the ball is:

$s (2\,s) = 32 + 64\cdot (2\,s) - 16\cdot (2\,s)^{2}$

$s_{max} = 96\,ft$

b) The time required by the ball to hit the ground is:

$32+64\cdot t - 16\cdot t^{2} = 0$

$-16\cdot (t^{2}-4\cdot t - 2) = 0$

$t^{2}-4\cdot t - 2 = 0$

$(t -4.449)\cdot (t+0.449)\approx 0$

Just one root offers a solution that is physically reasonable:

$t = 4.449\,s$

The velocity of the ball when it hits the ground is:

$v(4.449\,s) = 64 - 32\cdot (4.449\,s)$

$v(4.449\,s) = -78.368\,\frac{ft}{s}$

6 0

3 years ago

Read 2 more answers

A steel alloy specimen having a rectangular cross section of dimensions 18.1 mm × 3.0 mm ( 0.7126 in. × 0.1181 in.) has the stre

Natalka [10]

Answer:

Your search - A steel alloy specimen having a rectangular cross section of dimensions 18.1 mm × 3.0 mm ... - did not match any documents.

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Make sure that all words are spelled correctly.

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Explanation:

i hope it help please mark me as a brainless

4 0

3 years ago