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mihalych1998 [28]
3 years ago
8

A metaphysical poet is a writer whose

Physics
1 answer:
EastWind [94]3 years ago
8 0
Not really sure but...

<span>A metaphysical poet is a writer whose </span>focus is on universal human experiences. 
hope this helps!...
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If 14:15=42:x find the value of x​
Yuliya22 [10]

Answer:

x=45

Explanation:

by taking 42 and dividing it by 14 you get 1/3, because 14 is 1/3 of 42 you can then see that the ratio is multiplied by 3. si then you can just multiply 15 by 3 to get 45

3 0
3 years ago
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The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla
Rama09 [41]
Please elaborate more on your question so I can help you
7 0
3 years ago
0.125 C of charge flow out of a
zimovet [89]

Answer:

Resistance = 252.53 Ohms

Explanation:

Given the following data;

Charge = 0.125 C

Voltage = 5 V

Time = 6.3 seconds

To find the resistance;

First of all, we would determine the current flowing through the battery;

Quantity of charge, Q = current * time

0.125 = current * 6.3

Current = 0.125/6.3

Current = 0.0198 A

Next, we find the resistance;

Resistance = voltage/current

Resistance = 5/0.0198

Resistance = 252.53 Ohms

3 0
3 years ago
The drag force that resists the motion of a car traveling at 80 km h^- 1 is 300 N.
kobusy [5.1K]

The power require to keep the car traveling is 6,666 W.

The power of the engine at the given efficiency is 3,999.6 W.

<h3>What is Instantaneous power?</h3>

This the product of force and velocity of the given object.

The power require to keep the car traveling is calculated as follows;

P = Fv

P = 300\ N \ \times  \ \frac{80 \ kmh^{-1}}{3.6 \ km h^{-1}/m/s} \\\\&#10;P = 300 \ N \times 22.22 \ m/s\\\\&#10;P = 6,666 \ W

The power of the engine at the given efficiency is calculated as follows;

E = \frac{P_{out}}{P _{in}} \times 100\%\\\\&#10;60\% = \frac{P_{out}}{6,666} \times 100\%\\\\&#10;0.6 = \frac{P_{out}}{6,666} \\\\&#10;P_{out} = 3,999.6 \ W

Learn more about efficiency here: brainly.com/question/15418098

8 0
2 years ago
14. The design for a rotating spacecraft below consists of two rings. The outer ring with a radius of 30 m holds the living quar
Zolol [24]

Answer:

T= 11.0003s

Explanation:

From the question we are told that

The outer ring with a radius of 30 m

inner Gravity Approximately 9.80 m/s'

Outer Gravity Approximately 5.35 m/s.

Generally  the equation for centripetal force is given mathematically as

Centripetal acceleration enables Rotation therefore?

     \omega ^2 r =Angular\ acc

Considering the outer ring,

 \omega ^2 r = 9.8

  \omega ^2= \frac{9.8}{30}

 \omega = \sqrt{\frac{9.8}{30}}

 \omega= 0.571 rad/s

Therefore solving for  Period T

Generally the equation for solving Period T is mathematically given as

 T= \frac{2\pi}{\omega}

 T= \frac{2\pi}{0.571 rad/s}

 T= 11.0003s

5 0
3 years ago
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