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3 years ago

An object with kinetic energy k explodes into two pieces, each of which moves with twice the speed of the original object.

1 answer:

6 0

<span>Assuming that the momenta of the two pieces are equal: when they have equal velocities, then the masses of the two pieces are also equal. Since there is no force from outside of the system, the center of mass moves on with the same velocity as before the equation. So the two pieces must fly at the side side of the mass center, i.e., they must always be at 90Â° to the side of the mass center. Otherwise it would not be the mass center, respectively the pieces would not have equal velocities. This is only possible, when the angle of their velocity with the initial direction is 60Â°. Because, cos (60Â°) = 1/2 = v/(2v).</span>

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In 1969, Apollo 11 landed on the Moon. It took four days for the spacecraft to travel from Earth to the Moon. In 2006, New Horiz

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Answer:

An estimate for the time it will take for a spacecraft to travel from Earth to Mars is approximately 138.8 days

Explanation:

The distance between Earth and the Moon = 684,400 km

The distance between Earth and Mars = 220.58 × 10⁶ km

The distance between Earth and Pluto = 5.2241 × 10⁹ km

The ratio of the distance between Earth and Pluto and the distance between Earth and Mars = (5.2241 × 10⁹ km)/(220.58 × 10⁶ km) ≈ 23.683

It took 2006 to 2015 (9 years) to travel from Earth to Pluto, therefore, it can take approximately (9 years)/(23.683) ≈ 0.38 of a year which is ((9 years)/(23.683)) × 365.2422 ≈ 138.8 days for a spacecraft to travel from Earth to Mars

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3 years ago

A 700 g can of beans is dropped from a shelf that is 1.5 m high. What is the gravitational potential energy of this can? Round y

OLga [1]

<span>To find the gravitational potential energy of an object, we can use this equation: GPE = mgh m is the mass of the object in kg g = 9.80 m/s^2 h is the height of the object in meters GPE = mgh GPE = (0.700 kg) (9.80 m/s^2) (1.5 m) GPE = 10.3 J The gravitational potential energy of this can is 10.3 J</span>

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Why can the atmosphere hold on to heat

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A. Air

Explanation:

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3 years ago

Need 3 examples of unbalanced forces

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1. pulling tug of war
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4 years ago

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Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago