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FinnZ [79.3K]
3 years ago
10

What is the current in a circuit that has a resistance of 30.0 o and a power of 55.0 W?

Physics
1 answer:
salantis [7]3 years ago
8 0

Answer:

1.833a

Explanation:

power=current x resistance

p=ir

i=p/r

=55/30

=1.8333a

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What forces contribute to density
mixer [17]
No force contributes to density, the density is a physical quantity that is defined as being \rho = m/V 
 the raport between the mass of the object and its volume. However if you want to measure the density of an object you might want to determine its gravity force (weight) G=mg from which knowing the gravitational acceleratin you can find its mass m=G/g
 
where G is given in Newtons and g=9.81 is given in  (m/s^2)


5 0
3 years ago
Calculate the force of gravity between a comet with a mass of 500kg and a small asteroid with a mass of 20kg that is separated b
givi [52]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The equivalent gravitational force is ~

  • F  \approx1.48\times 10 {}^{ - 7}  \: \: N

\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}

We know that ~

\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}

where,

  • F = gravitational force

  • m_1 = mass of 1st object = 500 kg

  • m_2 = mass of 2nd object = 20kg

  • G = gravitational constant = 6.674 × {10}^ {-11}

  • r = distance between the objects = 2.12 m

Let's calculate the force ~

  • F = \dfrac{6.674   \times 10 {}^{ - 11} \times 500 \times 20}{(2.12) {}^{2} }

  • F = \dfrac{6.674  \times 10 {}^{ - 11} \times 10 {}^{4} }{4.4944}

  • F =  \dfrac{6.674}{4.4944}  \times 10 {}^{ - 7}

  • F =1.484 \times 10 {}^{ - 7}  \: \: newtons
7 0
2 years ago
If the velocity of an object is doubled, its kinetic energy is ______
swat32
Increased by a factor of 4
4 0
3 years ago
One nanometer is equal to how many centimeters?
Neko [114]
One nanometer = 1e-7
4 0
3 years ago
Read 2 more answers
Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.
olga2289 [7]

Answer:

NO_{1.499}

Explanation:

Let assume that 100 kg of the compound is tested. The quantity of kilomoles for each element are, respectively:

n_{N} = \frac{36.86\,kg}{14.006\,\frac{kg}{kmol} }

n_{N} = 2.632\,kmol

n_{O} = \frac{63.14\,kg}{15.999\,\frac{kg}{kmol} }

n_{O} = 3.946\,kmol

Ratio of kilomoles oxygen to kilomole nitrogen is:

n^{*} = \frac{3.946\,kmol}{2.632\,kmol}

n^{*}= 1.499

It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.

The empirical formula for the compound is:

NO_{1.499}

8 0
3 years ago
Read 2 more answers
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