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mario62 [17]
3 years ago
9

QUICK TEST ENDS IN 4 MINUTES

Physics
2 answers:
vodomira [7]3 years ago
4 0

Answer: Most likely C

Explanation: We are currently using robots in outer space

ASHA 777 [7]3 years ago
4 0

Answer:

there is no possible way to prove the answer, but i would say c

Explanation:

i am only saying this because they should see what is out there before sending people.

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Which of these is an advantage of parallel circuits over series circuits?
adell [148]

Answer:

d

Explanation:

study island said so

6 0
3 years ago
It has been argued that power plants should make use of off-peak hours to generate mechanical energy and store it until it is ne
sdas [7]

Answer:

Explanation:

90 rpm = 90 / 60 rps

= 1.5 rps

= 1.5 x 2π rad /s

angular velocity of flywheel

ω = 3π rad /s

Let I be the moment of inertia of flywheel

kinetic energy = (1/2) I ω²

(1/2) I ω² = 10⁷ J

I = 2 x  10⁷ / ω²

=2 x  10⁷ / (3π)²

= 2.2538 x 10⁵ kg m²

Let radius of wheel be R

I = 1/2 M R² , M is mass of flywheel

= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .

1/2 πR⁴ x t x d = 2.2538 x 10⁵

R⁴ = 2 x 2.2538 x 10⁵ / πt d

= 4.5076 x 10⁵ / 3.14 x .1 x 7800

= 184

R= 3.683 m .

diameter = 7.366 m .

b ) centripetal accn required

= ω² R

= 9π² x 3.683

= 326.816 m /s²

3 0
3 years ago
When did the ball fly farther—when it hit the moving bat or the bat that did not move?
8_murik_8 [283]

Answer:

when it hit the moving bat

Explanation:

force equals mass times acceleration which means the moving bat will add more force to the ball.

6 0
3 years ago
A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio
Nikitich [7]

Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft  × 4  sorings

K = 100 lb/ft

Amplitude of the microscope  \frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon  \frac{\omega}{W_n})^2}]

where;

\epsilon = 0

W_n = \sqrt { \frac{k}{m}}

= \sqrt { \frac{100*32.2}{200}}

= 4.0124

replacing them into the above equation and making X the subject of the formula:

X = Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}

X = 0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}

X = 5.676*10^{-3} \ mm

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

8 0
4 years ago
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
aniked [119]

1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.

2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

a = -\omega^2 A

Where,

a = Acceleration

A = Amplitude

\omega= Angular velocity

From a reference system in which the downward acceleration is negative due to the force of gravity we will have to

a = -g

-\omega^2 A = -g

\omega = \sqrt{\frac{g}{A}}

From the definition of frequency and angular velocity we have to

\omega = 2\pi f

f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}

f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}

f = 2.5Hz

Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz

6 0
3 years ago
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