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Vesna [10]
3 years ago
7

Do objects of different size but same mass have same technical velocity? why?

Physics
1 answer:
denis-greek [22]3 years ago
7 0
1. It depends on the condition you are referring to
I mean if you drop a sphere and a thin copper plate both of same mass from the Mt Everest. both will have different velocity and time period to fall on the ground because both of them experiences a drag force upwards due to their shape the more the area you have the more the drag Force if thus the lesser the speed and more the time period is
2. if you assume it to be total vaccum then both of them will fall on the same time and with same velocity no matter the mass is different or the shape is ! physics work same for them ..
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Which Lewis dot diagram shows an atom that needs 2 more electrons in its outermost shell?
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Answer: Option (c) is the correct answer.

Explanation:

Atomic number of hydrogen is 1 and one dot in its lewis structure displays only one electron. So, in order to complete its octet it needs one more electron.

Atomic number of carbon is 6 and its electronic distribution is 2, 4. Carbon needs 4 more electrons in order to complete its octet.

Atomic number of oxygen is 8 and its electronic distribution is 2, 6. In order to complete octet, oxygen needs 2 more electrons.

Atomic number of fluorine is 9 and its electronic distribution is 2, 7. In order to complete octet, fluorine needs 1 more electron.

Hence, we can conclude that out of the given options, lewis dot structure of oxygen atom needs 2 more electrons in its outermost shell.

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Starting from rest, a 0.0367 kg steel ball sinks into a vat of corn syrup. The thick syrup exerts a viscous drag force that is p
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Answer:

0.48 W

Explanation:

Given that:

the mass of the steel ball = 0.0367 kg

C = 0.270 N

g (acceleration due to gravity) = 9.8

Now;

At Terminal Velocity Weight is balance by drag force

mg =Cv

Making v the subject of the formula:we have:

v = \frac  {mg}{C}

v = \frac  {0.0367*9.8}{0.270}

v = 1.332 m/s

Thus, the Rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity is:

P=Fv

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An object with mass of 4kg is thrown with initial velocity of 20m/s from point A and follows the track of ABCD.
soldier1979 [14.2K]

<u>The distance of the length YD is 3.57m, the velocity of the object at point D is 27m/s and the time required for the object to reach the ground is 0.17s</u>

Data given;

  • Mass = 4kg
  • Initial velocity = 20m/s
  • length CD = 5m (from the image given)

a)

<h3>Determine The Length of YD</h3>

YD=CDsinθ = 5sin45=\frac{5}{\sqrt{2} } = 3.57m

b)

<h3>The velocity of the object at point D</h3>

The change in kinetic energy is given as

Δ in kinetic energy = Δ in potential energy + work done by friction

K.E - 1/2mv^2 = mgh_1 - mgh_2 + (-μmg.x)

K.E = mg(50 - 3.57) + (-mg(0.3*100) + 1/2 mv^2

\frac{1}{2}mv^2_f=mg(46.43)-mg(30)+\frac{1}{2} m(400)\\\\v^2_f=20(16.43)+400\\v^2_f=728.6\\v=\sqrt{728.6} \\v_f=26.99 = 27m/s

The velocity of the object at D with a distance of 5m.

c)

<h3>The the required for the object to reach ground</h3>

The velocity of the object in the y-axis is

v_y=vsin45=19.09

Acceleration in y-axis = 9.8

Height = 3.57m

h = ut+\frac{1}{2}at^2

3.57=19.28(t)+\frac{1}{2}(9.8)t^2\\3.57=19.28t+4.9t^2\\4.9t^2+19.28t-3.57=0\\a=4.9, b=19.28, c= -3.57\\t=\frac{-19.28+\sqrt{(19.28)^2-4(4.9)(-3.57)} }{(2*4.9)} \\t=\frac{-19.28+21.02}{9.8}

Taking the positive value

\frac{-19.28+21.02}{9.8}=0.17s

The time required for the object to reach ground is 0.17s

From the calculations above,<u> the distance YD is 3.57m, the velocity of the object at point D is 27m/s and the time required for the object to reach ground is 0.17s</u>

Learn more on projectile motion here;

brainly.com/question/1130127

<h3 />
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