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3 years ago

Select the correct answer.

Physics

2 answers:

Dahasolnce [82]3 years ago

5 0

Answer:

a is the correct choice

Explanation:

Nina [5.8K]3 years ago

4 0

Answer:

The answer is A

Explanation:

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Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1306p(1/2), where p is the pressure (in psi). If th

shepuryov [24]

Answer:

$a=38.5 ft/sec^{2}$

Explanation:

Note that acceleration is the rate change of velocity i.e

$acceleration=\frac{change in velocity}{change in time}\\a=\frac{dv}{dt} \\$ .

Since the velocity is giving as a variable dependent on the pressure, we have to differentiate implicitly both side with respect to time,i.e

$\frac{dv}{dt}=1306*(1/2)p^{-1/2}\frac{dp}{dt} \\$

if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have

$a=653*0.1667*0.354\\$

$a=\frac{dv}{dt}=653(36)^{-1/2}*0.354\\ a=38.5 ft/sec^{2}$

3 0

3 years ago

What is energy converted to if it is not used to do work?

Romashka [77]

Answer:

the human body isn't very efficient at converting food into useful work. The human body is less than 5% efficient most of the time. The rest of the energy is converted to heat, which may or may not be useful, depending on how cool or warm a person wants to be.

Explanation:

3 0

2 years ago

Which will result in positive buoyancy and cause the object to float?

Wewaii [24]

the answer should be:

When the buoyant force is equal to the force of gravity

4 0

3 years ago

Read 2 more answers

SHOW WORK

Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

$T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\$

For Point b:

$E=\frac{1}{2} m a^2 w^2\\\\$

$=\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J$

For Point C:

$V_{max}= a w$

$= (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}$

For point D:

$X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm$

5 0

3 years ago

A train moving at a constant speed on a surface inclined upward at 10.0° with the horizontal travels a distance of 400 meters in

Amiraneli [1.4K]

If the velocity of the train is v=s/t, where s is the distance and t is time, then v=400/5=80m/s. To get the vertical component of the velocity we need to multiply the velocity v with a sin(α): Vv=v*sin(α), where Vv is the vertical component of the velocity and α is the angle with the horizontal. So:

Vv=80*sin(10)=80*0.1736=13.888 m/s.

So the vertical component of the velocity of the train is Vv=13.888 m/s.

7 0

3 years ago