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3 years ago

Isaac newton discovered that the color white is created when light passes through a prism. t or f?

Physics

1 answer:

kap26 [50]3 years ago

7 0

Answer:

false

Explanation:

discovered colours of the rainbow

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Ganymede is one of the many moons of Jupiter. It is nearly spherical in shape. It is larger than the planet Mercury and slightly

lesya [120]

Moons revolve around a planet, and planets revolve around the sun. Ganymede is considered a moon because it revolves around the planet Jupiter, therefore, it is a moon. :) I hope this helps!

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3 years ago

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A man walks 6 hrs from his house to his office at an average speed of 2km/hr.find the difference between his house and office

laiz [17]

4 is the difference sorry if i got it wrong :( :(

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3 years ago

On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m

Nastasia [14]

Answer:

$F = 1.489*10^{-7} N$

Explanation: Weight of space probes on earth is given by: $W= m*g$

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 $\frac{m}{s^{2} }$ )

Therefore,

$m_{1} = \frac{14500}{9.81}$

$m_{1} = 1478.08 kg$

Similarly,

$m_{2} = \frac{4800}{9.81}$

$m_{2} = 489.29 kg$

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

$F = \frac{Gm_{1} m_{2}}{R^{2} }$

G= gravitational constant ( $6.67 * 10^{-11} m^{3} kg^{-1} s^{-2}$ )

$m_{1} , m_{2}$ = masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

$F = 1.489*10^{-7} N$

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.

7 0

3 years ago

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut

blsea [12.9K]

Answer:

1. $E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. $E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

$\rho (r) = \alpha (1-\frac{r}{R})$

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

$\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

$\int\, da = 2\pi r h$

where ‘h’ is the length of the imaginary Gaussian surface.

$Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

$Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3. At the boundary where r = R:

$E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}$

As can be seen from above, two E-field values are equal as predicted.

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3 years ago

Centripetal is a word used in science to mean

bogdanovich [222]

: the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation a string on the end of which a stone is whirled about exerts centripetal<span> force on the stone — compare centrifugal force.</span>

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3 years ago