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BlackZzzverrR [31]
2 years ago
7

Does anybody have any ideas for Wall art in a blue room.

Physics
2 answers:
Iteru [2.4K]2 years ago
8 0

Answer: Mabye like an ocean with dolphins swiming/jumping? Or even use the blue as a sky and then put green grass and do foxes or and a phoenix flying with a fox under it?

Explanation:

Just some ideas!

Sonja [21]2 years ago
5 0
Maybe like space themed with the background dark blue light blue and some white for starts
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Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work?
bagirrra123 [75]
Reduce friction because friction just makes things harder
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3 years ago
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A room has a volume of 160 m3. An air-conditioning system is to replace the air in this room every 21.3 minutes, using ducts tha
serious [3.7K]

Answer:

A. 0.204203928 m

b. 0.1443939822

Explanation:

Givens:

Volume (V) = 160 m^3

velocity 1 (v1) = 3.00 m/s

velocity 2 (v2) = 6.00 m/s

Time (t) = 21.3 minutes = 1278 seconds

Side length = ?

V = A*v*t

A = s^2

V = s^2*v*t

s= \sqrt{\frac{V}{v*t} }

A.

s = \sqrt{\frac{V}{v*t} } = \sqrt{\frac{160}{3.00*1278} } = 0.204203928 m

B.

s = \sqrt{\frac{V}{v*t} } = \sqrt{\frac{160}{6.00*1278} } = 0.1443939822 m

I hope this helped!! Have a good day.

7 0
3 years ago
For the strongest electromagnet, what wire gauge should be used?
alexandr1967 [171]

Answer:

Thick

Explanation:

Thicker wure enables higher AMP's, assuming your supply is capable, so indirectly thicker wire may help. Thicker wire takes up more space. So does more turns.

Hope this helped. Have a blessed day !

7 0
2 years ago
The series circuit shown below, what would happen if one of the light bulbs and its attached wire segment were removed?
rewona [7]

Answer:

The remaining lights would shine with the same brightness.

Explanation:

6 0
3 years ago
A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.50 × 10 4 1.50×104 rad/s to an angular s
KATRIN_1 [288]

Answer:

0.32 s

Explanation:

Initial angular speed: \omega_i = 1.50 \cdot 10^4 rad/s

Final angular speed: \omega_f = 3.35\cdot 10^4 rad/s

Angular rotation: \theta=2.02\cdot 10^4 rad

The angular acceleration of the drill can be found by using the equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

Re-arranging it, we find \alpha, the angular acceleration:

\alpha = \frac{\omega_f^2 - \omega_i^2}{2\theta}=\frac{(3.35\cdot 10^4 rad/s)^2-(1.50\cdot 10^4 rad/s)^2}{2(2.02\cdot 10^4 rad)}=22,209 rad/s^2

Now we want to know the time t the drill takes to accelerate from

\omega_i =0

to

\omega_f = 6.90\cdot 10^4 rad/s

This can be found by using the equation

\omega_f = \omega_i + \alpha t

where \alpha is the angular acceleration we found previously. Solving for t,

t=\frac{\omega_f - \omega_i}{\alpha}=\frac{22,209 rad/s^2}{6.90\cdot 10^4 rad/s}=0.32 s

5 0
3 years ago
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