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arsen [322]
3 years ago
7

An object has an acceleration of 6.0 m/s/s. If the net force acting upon this object were tripled , then its new acceleration wo

uld be _____ m/s/s.
Physics
1 answer:
ch4aika [34]3 years ago
4 0

Answer:

18m/s/s

Explanation:

6 times 3= 18

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The reactivity of an element is based on it's __________.<br><br> a) protons <br> b) neutrons
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Irina finds an unlabeled box of fine needles, and wants to determine how thick they are. A standard ruler will not do the job, a
Vsevolod [243]

Answer:

d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m

Explanation:

The expression which represent the first diffraction minima by a circular aperture is given by d sin\Theta =1.22\lambda--------eqn 1

The angle through which the first minima is diffracted is given by tan\Theta =\frac{y_1}{D}---------eqn 2

As \Theta is very small so we can write sin\Theta =tan\Theta

So from eqn 1 and eqn 2 we can write

y_1=\frac{1.22\lambda D}{d}--------eqn 3

Here y_1 is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture

It is given that diameter of circular aperture is 14.7 cm so y_1=\frac{14.7}{2}=7.35 \ cm

Now putting all these value in eqn 3

d=\frac{1.22\lambda D}{y_1}

d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m

8 0
3 years ago
A cylindrical can is partially filled with water. The radius of the cylindrical can is 8 inches, and its height is 9 inches. If
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The volume of the cylindrical can is given by:

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V = volume, r = base radius, h = height

Differentiate both sides of the equation with respect to time t. The radius r doesn't change over time, so we treat it as a constant:

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-527 = π(8)²(dh/dt)

dh/dt = -2.62in/min

The height of the water is decreasing at a rate of 2.62in/min

7 0
4 years ago
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