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sammy [17]
3 years ago
13

How much work does it take to accelerate from 5 m/s to 10 m/s?

Physics
1 answer:
REY [17]3 years ago
3 0

Answer:

KE (5) = 1/2 M V^2 = 25/2 M

KE (10) = 1/2 M V^2 = 100/2 M

KE (10) - KE (5) = M/2 (100 - 25) = 75/2 / M

An object traveling at 10 m/s has 4 times the kinetic energy as an object traveling at 5 m/s

Total work would depend on the mass being accelerated

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Very large accelerations can injure the body, especially if they last for a considerable length of time. One model used to gauge
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3 years ago
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Optics of the eye: the closest object that a typical young person with normal vision can focus on clearly is closest to:______.
Natali [406]

The human eye is one of the most valuable sense organ. The closest object that a typical young person with normal vision can focus closest to is 25 cm

The lens in the eye forms the image on the retina. The image formed is real and inverted. The retina is a very delicate membrane containing numerous light sensitive cells.

This light sensitive cells gets activated when light falls on it and generates electric signals. These electric signals are sent to the brain via the optic nerves present in the eye. The brain finally processes these signals and thus we see the object clearly.

The ability of the lens of the eye to adjust its focal length accordingly is called the power of accommodation,

The minimum distance, at which the object can be seen clearly without any strain is called the least distance of distinct vision.

To see an object clearly without any strain is when we place it at about 25 cm from the eye.

Tolearn more about optics of the eye : brainly.com/question/22371574

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5 0
2 years ago
A sphere of radius 0.03m has a point charge of q= 7.6 micro C located at it’s centre. Find the electric flux through it?
GalinKa [24]

Answer:

The electric flux through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

Explanation:

Given

Radius,\ r = 0.03m\\Charge,\ q =7.6\µC

Required

Find the electric flux

Electric flux is calculated using the following formula;

Ф = q/ε

Where ε is the electric constant permitivitty

ε = 8.8542 * 10^{-12}

Substitute ε = 8.8542 * 10^{-12} and q =7.6\µC; The formula becomes

Ф = \frac{7.6\µC}{8.8542 * 10^{-12}}

Ф = \frac{7.6 * 10^{-6}}{8.8542 * 10^{-12}}

Ф = \frac{7.6}{8.8542} *\frac{10^{-6}}{10^{-12}}

Ф = \frac{7.6}{8.8542} *10^{12-6}}

Ф = 0.85834970974 *10^{12-6}}

Ф = 0.85834970974 *10^{6}}

Ф = 8.5834970974 *10^{5}}

Ф = 8.58 *10^{5} \frac{Nm^2}{C}

Hence, the electric flux through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

7 0
3 years ago
The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a
likoan [24]

Answer:

14869817.395 m

Explanation:

\theta=22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians

From Rayleigh Criterion

\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0
3 years ago
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