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Anna71 [15]
3 years ago
12

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a sp

eed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.
Physics
2 answers:
Alex787 [66]3 years ago
7 0

Answer:

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.

Explanation:

i want the answer i don't know

ipn [44]3 years ago
6 0

Answer:

Explanation:want answers

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Projectile motion.


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An electron (charge −e=−1.60×10−19C−e=−1.60×10−19C) is at rest at a distance 1.00×10−10 m 1.00×10−10m from the center of a nucle
shtirl [24]

Answer:

-9.45\times10^{-16} \text{ J}

Explanation:

According to Coulomb's law, the force of attraction between two point charges, q_1 and q_2, separated by a distance d is given by

F = k\dfrac{q_1q_2}{d^2}

k is a constant with a value of 9\times10^9\text{ F/m}.

When we substitute the values from the question,

F = (9\times10^9\text{ F/m})\dfrac{(-1.60\times10^{-19} \text{ C})\times(+1.31\times10^{-17}\text{ C})}{(1.00\times10^{-10}\text{ m})^2} = -1.89\times10^{-6} \text{ N}

This value is negative because it is in a direction towards the positive charge.

The work done in moving the electron from the nucleus is

W = F\times r

W = (-1.89\times 10^{-6} \text{ N})\times(5.00\times10^{-10}\text{ m}) = -9.45\times10^{-16} \text{ J}

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3 years ago
A fire hose has an inside diameter of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of 1.
inessss [21]

Answer:

The Reynolds numbers for flow in the fire hose.

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Given that,

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Using formula of the Reynolds number

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\rho =density of fluid

Put the value into the formula

n_{R}=\dfrac{2\times100\times12.44\times3.2\times10^{-2}}{1.002\times10^{-3}}

n_{R}=7.945\times10^{5}

Hence, The Reynolds numbers for flow in the fire hose.

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A sound-producing object is moving toward an observer. The sound the observer hears will have a frequency ___ that actually bein
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The answer is <span>higher than.
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