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Tju [1.3M]
2 years ago
6

Two spheres have a gravitational force between

Physics
1 answer:
luda_lava [24]2 years ago
8 0

Answer:

<em>The new force is 2.8125 N</em>

Explanation:

<u>Newton’s Law of Universal Gravitation </u>

Objects attract each other with a force that is proportional to their masses and inversely proportional to the square of the distance.  

\displaystyle F=G{\frac {m_{1}m_{2}}{r^{2}}}

Where:

m1 = mass of object 1

m2 = mass of object 2

r     = distance between the objects' center of masses

G   = gravitational constant: 6.67\cdot 10^{-11}~Nw*m^2/Kg^2

Suppose two spheres have a gravitational force between them of F = 45 N. Now increase the distance to r'=4r. The new force F' is:

\displaystyle F'=G{\frac {m_{1}m_{2}}{(4r)^{2}}}

\displaystyle F'=G{\frac {m_{1}m_{2}}{16r^{2}}}

\displaystyle F'=\frac{1}{16}\ G{\frac {m_{1}m_{2}}{r^{2}}}

Substituting the original value of the force:

\displaystyle F'=\frac{1}{16}\ 45 N

F' = 2.8125 N

The new force is 2.8125 N

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klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

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μn = 1200 cm^2 / V s

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The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

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3 years ago
How can I shorten this and for it to sill make sense?
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Answer:

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Explanation:

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Explain how the basic unit are combined to give the derived units of force, velocity, pressure and work
LuckyWell [14K]

Velocity:

Velocity is change in displacement with respect to time:

\frac{\Delta x}{\Delta t}

Analysing the units, meters (displacement) and seconds (time) are basic units:

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Therefore the unit of velocity is m/s

Force:

Newton's second law of motion:

F = ma

Kilogram (mass) is a basic unit, and accelerations unit can be found using the equation:

a=\frac{\Delta v}{\Delta t}

Analysing the units:

\frac{\frac{m}{s}}{s}=\frac{m}{s^2}

Therefore, the unit of force is:

kg\frac{m}{s^2}

Pressure:

Pressure is given by the equation:

P=\frac{F}{S} where S is area of effect, F is force

Area for a basic rectangle (geometric shape is arbitrary for dimensional analysis) is found by multiplying two lengths:

[l^2]=m^2, the unit of area

Dividing the aforementioned unit of force by the unit of area:

\frac{kg\frac{m}{s^2}}{m^2}=\frac{kg}{ms^2}, the unit of pressure

Work:

Work is given by the equation:

W=\vec{F}\cdot \vec{x}, (dot product may be assumed as normal multiplication for the purposes of unit analysis)

Knowing displacement's (x) unit is m:

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2 years ago
Keisha finds instructions for a demonstration on gas laws. 1. Place a small marshmallow in a large plastic syringe. 2. Cap the s
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The correct answer is option C. <span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger. 
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Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity \omega_o = 0

angular acceleration \alpha = 4.44 rad/s²

Using the formula:

\omega = \omega_o+ \alpha t

Making t the subject of the formula:

t= \dfrac{\omega- \omega_o}{ \alpha }

where;

\omega = 1.53 \ rad/s^2

∴

t= \dfrac{1.53-0}{4.44 }

t = 0.345 s

b)

Using the formula:

\omega ^2 = \omega _o^2 + 2 \alpha \theta

here;

\theta = angular displacement

∴

\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }

\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }

\theta =0.264 \ rad

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

x = \dfrac{0.264 \times 1}{2 \pi}

x = 0.042 revolutions

c)

Here; force = 270 N

radius = 1.20 m

The torque = F * r

\tau = 270 \times 1.20 \\ \\  \tau = 324 \ Nm

However;

From the moment of inertia;

Torque( \tau) = I \alpha \\ \\  Since( I \alpha) = 324 \ Nm. \\ \\  Then; \\ \\  \alpha= \dfrac{324}{I}

given that;

I = 84.4 kg.m²

\alpha= \dfrac{324}{84.4} \\ \\  \alpha=3.84 \ rad/s^2

For re-tardation; \alpha=-3.84 \ rad/s^2

Using the equation

t= \dfrac{\omega- \omega_o}{ \alpha }

t= \dfrac{0-1.53}{ -3.84 }

t= \dfrac{1.53}{ 3.84 }

t = 0.398s

The required time it takes= 0.398s

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