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Arlecino [84]
3 years ago
7

Witch of the following is not an appropriate unit for power

Physics
1 answer:
erica [24]3 years ago
6 0

There are no appropriate units for power on the list you provided

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Find the speed of a long distance runner who runs 30km in 6 hours
slava [35]

30mi/6hrs is a speed of 5 mph, which converts to a pace of 12 min/mi.

7 0
3 years ago
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Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

Hence, The line charge density is 1.59\times10^{-4}\ C/m

4 0
3 years ago
What force is necessary to accelerate a 2500 kg care from rest to 20 m/s over 10 seconds?
EleoNora [17]
Force = mass x acceleration
force = 2500kg x (20m/s / 10m/s)
force = 2500kg x 2m/s^2
force = 5000kg m/s^2 = 5kN

i hope this is right (^^)
4 0
2 years ago
Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and
dexar [7]

Kepler’s
third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496
× 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work
with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel
them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The
closest answer is 1.99
× 10^30

(it may vary
a little with rounding – the difference is less than 1%)


8 0
3 years ago
Read 2 more answers
A Carnot engine's operating temperatures are 240 ∘C and 20 ∘C. The engine's power output is 910 W . Part A Calculate the rate of
scoray [572]

Answer:1200

Explanation:

Given data

Upper Temprature\left ( T_H\right )=240^{\circ}\approx 513

Lower Temprature \left ( T_L\right )=20^{\circ}\approx 293

Engine power ouput\left ( W\right )=910 W

Efficiency of carnot cycle is given by

\eta =1-\frac{T_L}{T_H}

\eta =\frac{W_s}{Q_s}

1-\frac{293}{513}=\frac{910}{Q_s}

Q_s=2121.954 W

Q_r=1211.954 W

rounding off to two significant figures

Q_r=1200 W

5 0
3 years ago
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