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Amiraneli [1.4K]
3 years ago
11

Rozważ, dlaczego telewizyjne satelity stacjonarne krążą nad równikiem. Zastanów się, czy można byłoby umieścić satelitę geostacj

onarnego tak, aby jego orbita znajdowała się cały czas nad równoleżnikiem, np. przebiegającym przez Warszawę. Wskazówka: Gdzie znajduje się środek orbity satelity?
Physics
1 answer:
Yuki888 [10]3 years ago
4 0
What language is this
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A projectile is fired from a height of 80 M above sea level, horizontally with a speed of 360 M / S, calculate: The time it take
Maslowich

Answer:

(a) The projectile takes approximately 4.420 seconds to reach the water, (b) The horizontal scope of the projectile is 1591.2 meters, (c) The remaining height to descend after 2 seconds of being launched is 63.624 meters.

Explanation:

The projectile experiments a parabolic motion, where horizontal speed remains constant and accelerates vertically due to the gravity effect. Let consider that drag can be neglected, so that kinematic equation are described below:

x = x_{o}+v_{o,x} \cdot t

y = y_{o} + v_{o,y}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}

Where:

x_{o}, y_{o} - Initial horizontal and vertical position of the projectile, measured in meters.

v_{o,x}, v_{o,y} - Initial horizontal and vertical speed of the projectile, measured in meters per second.

t - Time, measured in seconds.

g - Gravitational acceleration, measured in meters per square second.

x, y - Current horizontal and vertical position of the projectile, measured in meters.

Given that x_{o} = 0\,m, y_{o} = 80\,m, v_{o,x} = 360\,\frac{m}{s}, v_{o,y} = 0\,\frac{m}{s} and g = -9.807\,\frac{m}{s^{2}}, the kinematic equations are, respectively:

x = 360\cdot t

y = 80-4.094\cdot t^{2}

(a) If y = 0\,m, the time taken for the projectile to reach the water is:

80 - 4.094\cdot t^{2} = 0

t = \sqrt{\frac{80}{4.094} }\,s

t \approx 4.420\,s

The projectile takes approximately 4.420 seconds to reach the water.

(b) The horizontal scope is the horizontal distance done by the projectile before reaching the water. If t \approx 4.420\,s, the horizontal scope of the projectile is:

x = 360\cdot (4.420)

x = 1591.2\,m

The horizontal scope of the projectile is 1591.2 meters.

(c) If t = 2\,s, the height that remains to descend is:

y = 80-4.094\cdot (2)^{2}

y = 63.624\,m

The remaining height to descend after 2 seconds of being launched is 63.624 meters.

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An object starts moving at 10 ms and has an acceleration of 2 ms-2. Find its speed after 5 seconds
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Answer:

speed =15m/s

distance =75m

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The law of conservation of matter is that the amount of matter you begin the reaction with is the amount you will end up with.  So if you start with 500g of matter, when you burn the candle you will also have 500g of matter to end with.  SO D is your answer
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A student is trying to classify an unidentified, solid gray material as a metal or a nonmetal. Which question will best help the
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The question that will best help the student to classify if the material is metal or non metal would be answer C.
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