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3 years ago

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Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu

lar to each other (take A initially moving to the right and B initially moving up). Use momentum conservation (make a complete Momentum Chart) to find the velocity (magnitude and direction with-respect-to the velocity asteroid A had before the collision) of the asteroids after the collision.

1 answer:

11111nata11111 [884]3 years ago

4 0

Answer:

velocity = 62.89 m/s in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $10^5$ kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $10^5$

So, $V_x = \frac{10^5}{3000}$ = $\frac{100}{3}$ m/s

and $V_y=\frac{160}{3}$ m/s

Therefore, velocity is = $\sqrt{V_x^2 + V_y^2}$

= $\sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2}$

= 62.89 m/s

And direction is

tan θ = $\frac{V_y}{V_x}$ = 1.6

therefore, $\theta = \tan^{-1}1.6$

= $58 ^{\circ}$ from x-axis

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Answer:

The velocity of the blades is 88.185 m/s.

Explanation:

Given;

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$V = \omega r\\\\V = (\frac{1 \ rev}{5.7 \ s} \times \frac{2 \pi \ rad}{ 1 \ rev} )(80 \ m)\\\\V = 88.185 \ m/s$

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3 years ago

Nadya [2.5K]

1) The mass of the continent is $3.3\cdot 10^{21}kg$

2) The kinetic energy of the continent is 1683 J

3) The speed of the jogger must be 6.57 m/s

Explanation:

1)

The continent can be represented as a slab of size

$d=5850 km = 5.85\cdot 10^6 m$

and depth

$t = 35 km = 3.5\cdot 10^4 m$

So its volume is

$V=d^2 t = (5.85\cdot 10^6)^2(3.5\cdot 10^4)=1.20\cdot 10^{18} m^3$

We also know that the density of the continent is

$\rho = 2750 kg/m^3$

Therefore, we can calculate its mass as:

$m=\rho V=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg$

2)

The kinetic energy of the continent is given by

$K=\frac{1}{2}mv^2$

where

m is its mass

v is its speed

We have already calculate its mass, while the speed is

v = 3.2 cm/year

We have to convert into SI units first, as follows:

$v=3.2 \frac{cm}{year} \cdot \frac{1}{100 cm/m} \cdot \frac{1}{(365 d/y)(24h/d)(60min/h)(60 s/min)}=1.01\cdot 10^{-9} m/s$

The mass is

$m=3.3\cdot 10^{21} kg$

So, the kinetic energy of the continent is

$K=\frac{1}{2}(3.3\cdot 10^{21})(1.01\cdot 10^{-9})^2=1683 J$

3)

Here we have a jogger having the same kinetic energy of the continent, so

$K=1683 J$

And the kinetic energy of the jogger can be expressed as

$K=\frac{1}{2}mv^2$

where

m = 78 kg is the mass of the jogger

v is his speed

We can therefore re-arrange the equation to find the speed of the man, and we get:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1683)}{78}}=6.57 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

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3 years ago

Four point charges of magnitudes +3q, -q, +2q, and -4q are arranged in the corners of a square of side length L. The charge -q c

mafiozo [28]

Answer:

d) 0 V

Explanation:

It can be showed that the potential due to a point charge q, to a distance d from the charge, can be expressed as follows:

$V = \frac{k*q}{r}$

where k = $\frac{1}{4*\pi*\epsilon0} = 9e9 N*m2/C2$

As the potential is an scalar, and is linear with the charge, we can apply the superposition principle, which means that we can find the potential due to one of the charges, as if the other were not present.

By symmetry, all four charges are at the same distance from the center, so we can write the total potential, as follows:

$V = \frac{k}{d} ( q1 + q2 + q3 + q4) (1)$

where d, is the semi-diagonal of the square, that we can find applying Pythagorean theorem, as follows:

$d = \sqrt{\frac{L^{2}}{4} + \frac{L^{2}}{4} } = L*\frac{\sqrt{2}}{2}$

Replacing by the values in (1) we have:

$V = \frac{9e9N*m2/C2}{\frac{L}{2}*\sqrt{2} }* ( +3q -q + 2q + -4q) = 0 V$

which is equal to the option d).

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4 years ago

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Answer:

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Answer:

4.825 N

Explanation:

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mass of moon, Mm = 7.34 x 10^22 kg

Mass of satellite, Ms = 1250 kg

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A.

force on space craft by the earth

$F_{1}=\frac{GM_{e}M_{s}}{a^{2}}$

By substituting the values

$F_{1}=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 1250}{\left (3.84\times 10^_{8} \right )^{2}}$

F1 = 3.39 N

force on space craft by the moon

$F_{1}=\frac{GM_{m}M_{s}}{a^{2}}$

By substituting the values

$F_{1}=\frac{6.67\times 10^{-11}\times 7.34\times 10^{22}\times 1250}{\left (3.84\times 10^_{8} \right )^{2}}$

F2 = 0.042 N

These two forces act at 60° with each other

$F = \sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}Cos60}$

$F = \sqrt{3.39^{2}+0.042^{2}+2\times 3.39\times 0.042\times Cos60}$

F = 4.825 N

5 0

3 years ago