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3 years ago

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Kepler's____ of planetary motion states that the planets have an elliptical orbit, with the Sun at one focal point of the ellips

e. A. 1st law B. 2nd law C. 3rd law

1 answer:

pentagon [3]3 years ago

6 0

A.Kepler's first law of planetry motion.

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A mass of 2.1 kg the ball leaves her hand with the speed of 30 mass find the energy of the ball

snow_lady [41]

Ke= 1/2 x m x v^2
Ke= 1/2 x 2.1 x 30^2
Energy = 945 J

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3 years ago

Can someone please help

Maru [420]

Answer: D

Explanation: because doing a yoga desk program is physical activity, 10k steps is pysical activity, riding a bike or walking/running is also physical activity. so it should be D, all of the above.

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3 years ago

A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.

Airida [17]

Answer:

a) $v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )$ , b) $\Delta K = 9.559\times 10^{-5}\,J$

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

$(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v$

The final velocity is:

$v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )$

b) The change in the kinetic energy of the 13.5 g coin is:

$\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]$

$\Delta K = 9.559\times 10^{-5}\,J$

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3 years ago

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What is your velocity if it takes you 40 seconds to walk around a circle of 50meters 3 times?

Shtirlitz [24]

Answer:

0 m/s

Explanation:

Velocity is displacement over time. Displacement is the distance between your initial position and your final position. If you walk in a circle, you end up back where you started, so your displacement is 0. Therefore, your velocity is also 0.

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2 years ago

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The magnitude of the velocity vector of the car is ∣∣v→∣∣ = 78 ft/s. If the vector v→ forms an angle θ = 0.09 rad with the horiz

kotegsom [21]

Answer:

$\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j$

Explanation:

The x- and y- components of the velocity vector can be written as following:

$\vec{v}_x = ||\vec{v}||\cos(\theta)\^i$

$\vec{v}_y = ||\vec{v}||\sin(\theta)\^j$

Since the angle θ and the magnitude of the velocity is given, the vector representation can be written as follows:

$\vec{v} = 78\cos(0.09)\^i + 78\sin(0.09)\^j\\\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j$

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2 years ago