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3 years ago

If the distance between two masses is tripled, the gravitational force between changes by a factor of

Physics

2 answers:

maw [93]3 years ago

6 0

A. 1/9

Explanation:

The gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1 and m2 are the two masses

r is the distance between the two masses

From the formula, we see that the magnitude of the force is inversely proportional to the square of the distance: therefore, if the distance is tripled (increased by a factor 3), the magnitude of the force changes by a factor

$\frac{1}{r^2}=\frac{1}{3^2}=\frac{1}{9}$

icang [17]3 years ago

3 0

A. 1/9 is the answer.

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A mass m = 3.9 kg hangs from a massless string wrapped around a uniform cylinder with mass Mp = 10.53 and radius R= 0.96 m. The

luda_lava [24]

Answer:

the rotational inertia of the cylinder = 4.85 kgm²

the mass moved 7.942 m/s

Explanation:

Formula for calculating Inertia can be expressed as:

$I =\frac{1}{2}mR^2$

For calculating the rotational inertia of the cylinder ; we have;

$I = \frac{1}{2}m_pR^2$

$I = \frac{1}{2}*10.53*(0.96)^2$

$I=5.265*(0.96)^2$

$I=4.852224$

I ≅ 4.85 kgm²

mg - T ma and RT = I ∝

T = $\frac{Ia}{R^2}$

$a = \frac{g}{1+\frac{I}{mR^2}}$

$a = \frac{9.8}{1+\frac{4.85}{3.9*(0.96)^2}}$

a = 4.1713 m/s²

Using the equation of motion

$v^2 = u^2+2as \\ \\ v^2 = 2as \\ \\ v = \sqrt{2*a*s} \\ \\ v= \sqrt{2*4.1713*7.56} \\ \\ v = 7.942 \ m/s$

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2 years ago

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Electric circuits and electric current

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an electric current is a flow of electric charge in electric circuits this is carried by moving electrons in a wire and an electric circuit is an electrical network of electrical components and model of interconnection consisting electrical elements

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2 years ago

Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem

statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

$x=x_o+v_1t$ (1)

xo = 10 km

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car 2:

$x'=v_2t$ (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

$x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}$

$t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s$

The position in which both cars coincides is:

$x=(55km/h)(0.33h)=18.33km$

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Answer:

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2 years ago

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