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3 years ago

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A fan powered by a 0.5 hp motor and delivers air at a rate of 85 m3 / min. Calculate the speed of the air moving by the fan if t

he air has a density of 1.18 kg / m3. Help me please

1 answer:

Elenna [48]3 years ago

4 0

s the following graph a bipartite graph or not, and why?

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Bruh how did i get 3 warnings and 3 brainliests removed

Brums [2.3K]

Answer:

Wait, that can happen? I'm sorry.

Explanation:

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3 years ago

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A person can jump 1.5m on the earth. How high could the person jump on a planet having the twice the mass of the earth and twice

MrMuchimi

F=mg=Gm1m2/r^2
g=Gm2/r^2
g=2Gm2/(2r)^2=2Gm2/4r^2=Gm2/2r^2
So since there is half times the gravity on this unknown planet that has twice earth's mass and twice it's radius, then the person can jump twice as high. 1.5*2= 3m high

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3 years ago

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In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

3 years ago

30 points must be a legitimate answer and correct answer or I will report

kupik [55]

Answer:

it's C

Explanation:

have a nice day............

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2 years ago

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Which most likely is a causal relationship?

const2013 [10]

Answer: C. Experimenters show a link between adult-onset diabetes and obesity
On the statement above, there is clearly a causal relationship between the adult-onset diabetes and obesity according to researches and experiments. Causal relationship shows that there is one event is the cause while the other is the result.

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3 years ago