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labwork [276]
2 years ago
14

According to Newton's Universal Law of Gravitation, when the distance between two interacting objects doubles, the gravitational

force is
Physics
1 answer:
maksim [4K]2 years ago
5 0

Answer:

<em>If the distance doubles, the gravitational force is divided by 4</em>

Explanation:

<u>Newton’s Universal Law of 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

If the distance between the interacting objects doubles to 2r, the new force F' is:

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

Operating:

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

Substituting the original value of F:

\displaystyle F'=\frac{1}{4}F

If the distance doubles, the gravitational force is divided by 4

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Answer:

735 J/kg/C

Explanation:

Q = mcT

943 = (0.447)( c )(2.87)

1.28289c = 943

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2 years ago
Suppose that a certain battery produces a voltage of 1.55V without a load connected (open circuit) and a current of 500mA when s
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Explanation:

Let the internal resistance be r .

Since in open circuit the volt is 1.55 V , this will be the source voltage .

Source voltage = 1.55

If external resistance be R .

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So sum of internal resistance and external resistance will be 3.1 ohm.

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3 years ago
Mrs. Howel, the principal, can expel a student for misconduct. In terms of social pressure, Mrs. Howel is capable of eliciting
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On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
IceJOKER [234]

Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
7 0
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Artyom0805 [142]

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Explanation:

Solar cells work by the photoelectric effect, which consists of the emission of electrons (electric current) when light (electromagnetic waves) falls on a metal surface under certain conditions.  

In this sense, the portion of the electromagnetic spectrum this cells use is Ultraviolet light (UV) from the Sun, whose wavelength is approximately between 100 nm and 380 nm.

It is important to note, this is a type of electromagnetic radiation that is not visible to the human eye.

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