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4 years ago

In order to determine the power, you must know..

Physics

2 answers:

amm18124 years ago

8 0

Answer:

the right answer is...…..…..…..…..…..…

A. current and voltage

Explanation:

Rudik [331]4 years ago

6 0

Choices 'a' and 'd' will both do the job.

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At t=0 a ball is projected vertically upward from a roof of a building of height 68.1 meters with a velocity of 42.4 m/s. The ba

Finger [1]

Answer:

7.45 s.

Explanation:

Given:

h = 68.1 m

vi = 0 m/s

vf = 42.4 m/s

g = 9.81 m/s^2

Using,

h = vi*t +1/2*(a*t^2)

68.1 = 1/2 * (9.81*t^2)

t = sqrt((68.1*2)/9.81)

= 3.726 s.

Total time of flight = 2*t

= 2 * 3.726

= 7.45 s.

4 0

3 years ago

Read 2 more answers

A .100 kg bullet is flying at 200 m/s. How much energy is the bullet storing due to its motion?

morpeh [17]

KE = 2000 J

Explanation:

KE = (1/2)mv^2

= (1/2)(0.100 kg)(200 m/s)^2

= 2000 J

4 0

3 years ago

What force is required to accelerate a 385 kg couch at 0.2 m/s^2 ?

Juliette [100K]

Answer:

It takes 77 N

Explanation:

Using Newton's second law of motion, F=ma (Force equals mass times acceleration. Since the mass of the couch is 385 kg and the target acceleration is 0.2 m/s, you simply multiply mass times acceleration (ma) to get the total force, or 77 N.

6 0

3 years ago

A force of 60 N is applied to a skier to pull him along a horizontal surface so that his speed remains constant. If the coeffici

matrenka [14]

Answer:

$1200\ \text{N}$

Explanation:

F = Force on the skier = 60 N

$\mu$ = Coefficient of friction = 0.05

w = Weight of skier

Force is given by

$F=\mu w$

$\Rightarrow w=\dfrac{F}{\mu}$

$\Rightarrow w=\dfrac{60}{0.05}=\dfrac{6000}{5}$

$\Rightarrow w=1200\ \text{N}$

Weight of the skier on which the force is being applied is $1200\ \text{N}$ .

7 0

3 years ago

Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt

Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

$= \frac{4}{3}\pi(6400+ 50)^3 - \frac{4}{3}\pi (6400)
^3\\\\= \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3$

Hence the volume of atmosphere is $2.6\times 10^{19} m^3$

(b)

Write the ideal gas equation as foll ows:

$PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles$

$no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules
$

Hence the required molecules is $13.3\times10^{43} molecules
$

(c)

Write the ideal gas equation as follows:

$PV =nRT
\\\\n=\frac{1.0 atm \times 0.5L
}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles$

$no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}
{Imole}= 1.2\times 10^{23} molecules$

Hence the required molecules in Caesar breath is $1.2\times 10^{23} molecules$

(d)

Volume fraction in Caesar last breath is as follows:

$Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}$

(e)

Since the volume capacity of the human body is 500 mL.

$Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath$

5 0

3 years ago