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miskamm [114]
3 years ago
13

Why should you explore the potential roadblocks to your chosen career?

Physics
1 answer:
valina [46]3 years ago
6 0

Answer: D

Explanation: D is the most reasonable answer because it's always good to plan ahead for anything, so if you were to plan ahead for future obstacles, then you can overcome them.

You might be interested in
Which statements accurately describe the Doppler effect? Select three options.
Katena32 [7]

Answer:The answer is A,B and C

Explanation: it can be use to learn universe behavior,it is also use for the measurements of velocity of galaxy,and is can also be use to indicate,galaxy motion.

Doppler effect,measures the speed which star and galaxy are moving away from us or towards us. It shows the color of wavelength to be either red shift or blue shift.

6 0
3 years ago
Read 2 more answers
A large rocket has a mass of 2.00×10⁶ kg at takeoff, and its engines produce a thrust of 3.50×10⁷ N. Find its initial accelerati
Kazeer [188]

Answer:

17.5 m/s²

1.90476 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Force

F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{3.5\times 10^7}{2\times 10^6}\\\Rightarrow a=17.5\ m/s^2

Initial acceleration of the rocket is 17.5 m/s²

v=u+at\\\Rightarrow \frac{120}{3.6}=0+17.5t\\\Rightarrow t=\frac{\frac{120}{3.6}}{17.5}=1.90476\ s

Time taken by the rocket to reach 120 km/h is 1.90476 seconds

Change in the velocity of a rocket is given by the Tsiolkovsky rocket equation

\Delta v=v_{e}\ln \frac{m_0}{m_f}

where,

m_0 = Initial mass of rocket with fuel

m_f = Final mass of rocket without fuel

v_e = Exhaust gas velocity

Hence, the change in velocity increases as the mass decreases which changes the acceleration

4 0
3 years ago
2. A volleyball player strikes a 3 kg volleyball with a velocity of 14 m/s. What is the kinetic energy of the volleyball?
Effectus [21]

Answer:

2. 294 j

3. 37500 J

Explanation:

6 0
3 years ago
Read 2 more answers
A 1.00-kg mass is attatched to a string 1.0m long and completes a horizontal circle in 0.25s. What is the centripetal accelerati
liraira [26]

-- The string is 1 m long.  That's the radius of the circle that the mass is
traveling in.  The circumference of the circle is  (π) x (2R) = 2π meters .

-- The speed of the mass is (2π meters) / (0.25 sec) = 8π m/s .

-- Centripetal acceleration is  V²/R = (8π m/s)² / (1 m) = 64π^2 m/s²

-- Force = (mass) x (acceleration) = (1kg) x (64π^2 m/s²) =

                                                         64π^2 kg-m/s² = 64π^2 N = about <span>631.7 N .

</span>
That's it.  It takes roughly a 142-pound pull on the string to keep
1 kilogram revolving at a 1-meter radius 4 times a second !<span> 
</span>
If you eased up on the string, the kilogram could keep revolving
in the same circle, but not as fast.

You also need to be very careful with this experiment, and use a string
that can hold up to a couple hundred pounds of tension without snapping. 
If you've got that thing spinning at 4 times per second and the string breaks,
you've suddenly got a wild kilogram flying away from the circle in a straight
line, at 8π meters per second ... about 56 miles per hour !  This could definitely
be hazardous to the health of anybody who's been watching you and wondering
what you're doing.


3 0
3 years ago
The diffusion constant for oxygen diffusing through tissue is 1.0 × 10-11 m2/s. In a certain sample oxygen flows through the tis
Eduardwww [97]

Answer:

m' = 1 x 10⁻⁶ kg/s

Explanation:

Given that

Diffussion constant = 1 x 10⁻¹¹

Mass flow rate ,m = 2 x 10⁻⁶ kg/s

The diffusion is inversely proportional to the thickness of the membrane and therefore when the thickness is doubled, the mass flow rate would become half.

So new flow rate m'

m'=\dfrac{m}{2}

m'=\dfrac{2\times 10^{-6}}{2}\ kg/s

m' = 1 x 10⁻⁶ kg/s

6 0
3 years ago
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