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vovikov84 [41]
3 years ago
14

What could you do to increase the electric force between two charged particles by a factor of 16?

Physics
1 answer:
oee [108]3 years ago
5 0
Hopefully this will help you. 

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Calculate (a) the torque, (b) the energy, and (c) the average power required to accelerate Earth in 4.0 days from rest to its pr
natima [27]
<h2>Answer:</h2>

Torque = <em>2.05 x 10²⁸ Nm</em>

Energy = <em>3.54 x 10³³ J</em>

Average power = <em>1.02 x 10²⁸ W</em>

<h2>Explanation:</h2>

(a) Torque (τ) is the rotational effect of a given force.  

It is given by

τ = I x α          -------------(i)

Where;

I = rotational inertia of the object

α = angular acceleration of the object.

In this case, the object is the Earth. Therefore,

I = 9.71 x 10³⁷ kg m²

α = ω / t

Where;

ω = angular velocity of earth = 2π rad / day

<em>Since </em>

<em>1 day = 24 hours and 1 hour = 3600seconds</em>

<em>1 day = 24 x 3600 seconds = 86400seconds</em>

<em>=> ω = 2π rad / 86400seconds</em>

<em>=> ω = 7.29 × 10⁻⁵ rad/s</em>

<em />

t = 4 days = 4 x 24 x 3600 seconds = 345600 seconds

=> α = ω / t

=> α = 7.29 × 10⁻⁵ / 345600

=> α = (7.29 × 10⁻⁵) / (3.456 x 10⁵)

=> α = (7.29 × 10⁻⁵⁻⁵) / (3.456)

=> α = (7.29 × 10⁻¹⁰) / (3.456)

=> α = 2.11 × 10⁻¹⁰ rad/s²

Now substitute the values of I and α into equation (i)

τ = 9.71 x 10³⁷ x 2.11 × 10⁻¹⁰

τ = 9.71 x 10²⁷ x 2.11

τ = 20.5 x 10²⁷ Nm

τ = 2.05 x 10²⁸ Nm

(ii) The energy (rotational energy) E is given by;

E = \frac{1}{2} x I x ω

E = \frac{1}{2} x 9.71 x 10³⁷ x 7.29 × 10⁻⁵

E = 35.4 x 10³² J

E = 3.54 x 10³³ J

(iii) The average power P, is given by;

P = E / t

P = 3.54 x 10³³ / 345600

P = 1.02 x 10²⁸ W

5 0
3 years ago
On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The
Cloud [144]

Answer:

Explanation:

a is the acceleration

μ is the coefficient of friction

Acceleration of the object is given by

a = g (\sin  \theta -\mu \cos \theta)\\\\=10( \sin 30 - 0.4 \cos 30)\\\\=10(0.5-0.3464)\\\\=1.54m/s^2

Velocity at the bottom

v^2=u^2+2as\\\\u=0\\\\v^2=2as\\\\=2*1.54*8\\\\=24.576\\\\v=4.96m/s

after travelling 4m , its velocity becomes 0

a=\frac{v^2-u^2}{2s}

a=\frac{0-u^2}{2s}

a=\frac{-(-4.96)^2}{2*4} \\\\=-3.075m/s^2

Coefficient of kinetic friction

μ = F/N

=\frac{ma}{mg} \\\\=\frac{3.075}{10} \\\\=0.31

Therefore, the Coefficient of kinetic friction is 0.31

8 0
2 years ago
Read 2 more answers
Which of these is a TRUE statement? A) Sound travels through gases easier than liquids. B) Sound travels through solids easier t
lakkis [162]
It’s B. Sound travels faster through solids than liquids. Have you ever put your head on a desk, and tap the desk? That’s an example of it going faster through solids
6 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP!!!! A huge thanks to anyone who can help me with 14 problems. I'll do anything to return the favor. All true an
snow_lady [41]
Hello, I see you are in a jam. Lemme help.

1.) True
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3.) True
4.) True
5.) True

LOL these are all true ;)
4 0
3 years ago
A normal blood pressure reading is less than 120/80 where both numbers are gauge pressures measured in millimeters of mercury (m
KIM [24]

The pressure value is given by the equation,

P= \rho gh

Where,

\rho represents the density of the liquid

g= gravity

h= Heigth

A) For the measurement of the guage pressure we have the data data,

\rho_{mercury} = 13.6*10^3kg/m^3

h=0.0930m

g=9.8m/s^2

Replacing we get,

P_g=\rho_{mercury}gh

[tex]P_g = (13.6*10^3)(9.8)(0.0930)P_g = 12395Pa[/tex]

In order to find the Absolute pressure, we perform a sum between the atmospheric pressure and that of the Gauge,

B) The atmospheric pressure at sea level is 101325Pa, assuming ideal conditions, we will take this pressure for our calculation, so

P_T = P_a+P_g\\P_T = 101325Pa+12395Pa\\P_T = 113720Pa\\P_T = 113.7kPa

6 0
3 years ago
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