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3 years ago

Which unit is used to measure resistance? a. volts b. charges c. amperes d. ohms

Physics

2 answers:

Olegator [25]3 years ago

8 0

It would be D. ohms

hope this helps you

Ilia_Sergeevich [38]3 years ago

6 0

Choose last

d ohms

)))))
:)

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Matt and Anna Killian are frequent fliers on Fast-n-Go Airlines. They often fly between two cities that are a distance of 1575

marin [14]

Answer:

Speed of wind = 50mi/hr, Speed of plane in still air = 400mi/hr

Explanation:

Let the speed of the wind = Vw,

Speed of the plane in still air = Vsa,

The first trip the average speed of the plane = 1575mi/4.5hours = 350mi/hr

The coming trip the wind behind = 1575mi/3.5hrs = 450

Write the motion in equation form

First trip ( the plane flew into the wind)

Vaverage = Vsa - Vw

350 = Vsa - Vw

Second trip the wind was behind

450 = Vsa +Vw

Adding the two equation

800 = 2Vas

Vas = 800/2 = 400mi/hr

Substitute for Vas into equation 1

350mi/hr = 400mi/hr - Vw

Vw = 400-350 = 50mi/hr

6 0

3 years ago

sp2606 [1]

Answer:

Explanation:

6 0

1 year ago

The yellow balls represent neutrons and have a neutral charge. <br><br><br> true<br> false

Paul [167]

Yes the answer is true

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3 years ago

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What would be the best method to separate only iron fillings from a mixture of sand, water, iron filings, and salt

IrinaVladis [17]

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6. A garden hose attached to a nozzle is used to fill a 15-gal bucket. The inner diameter of the hose is 1.5 cm, and it reduces

Vesna [10]

Answer: 1.135 L/s; 1.35 kg/s, 22.57 m/s

Explanation:

Given

Volume of bucket $V=15\ gal\approx 56.78\ L$

time to fill it $t=50\ s$

Volume flow rate

$\dot{V}=\dfrac{56.78}{50}=1.135\ L/s\approx 1.135\times 10^{-3}\ m^3/s$

The inner diameter of the hose $D=1.5\ cm$

diameter of the nozzle exit $d=0.8\ cm$

we can volume flow rate as

$\Rightarrow \dot{V}=Av\quad \quad \text{v=average velocity through nozzle exit}\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}d^2\times v\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}(0.8\times 10^{-2})^2\times v\\\\\Rightarrow v=\dfrac{4\times 1.135\times 10^{-3}}{\pi \times 64\times 10^{-6}}=22.57\ m/s$

Mass flow rate

$\Rightarrow \dot{m}=\rho \times \dot{V}\\\Rightarrow \dot{m}=1\ kg/L\times 1.135\ L/s=1.35\ kg/s$

5 0

3 years ago