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3 years ago

an object placed in a graduated cylinder raises the volume from 12.2mL to 14.5 mL. Find the volume of the object

Physics

2 answers:

stich3 [128]3 years ago

7 0

Answer:

2.3 milliliter the volume of the object.

Explanation:

Initial volume in the graduated cylinder ,V= 12.2 mL

Final volume of the graduated cylinder after placing an object ,V'= 14.5

Volume occupied by an object = v

Volume changed by the placing an object in the graduate cylinder will be equal to the volume of an object.So,:

v = V' - V = 14.5 mL - 12.2 mL = 2.3 mL

2.3 milliliter the volume of the object.

Irina18 [472]3 years ago

4 0

Honestly, look it up on google. It will give you the formula instantly. Like what I did, and here's your answer V(Volume) = Pie(3.14) x R(Radius)Squared(x2) x H(Height)

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Given :

Object A is 71 degrees and object B is 75 degrees .

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How will thermal energy flow.

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So, in the given question energy will flow from object B from object A.

Hence, this is the required solution.

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3 years ago

A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the obje

kobusy [5.1K]

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

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3 years ago

When light is reflected by a mirror, the angle of incidence is always A. equal to the angle of reflection. B. less than the angl

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The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how

liraira [26]

Answer:

$x = 0.198456 \ m$

$h = 1.3061 \ m$

$v = 5.06 \ m/s$

$d = 4.0273 \ m$

Explanation:

Considering the first question

From the question we are told that

The spring constant is $k = 32.50 N/m$

The potential energy is $PE = 0.640 \ J$

Generally the potential energy stored in spring is mathematically represented as $PE = \frac{1}{2} * k * x^2$

=> $0.640= \frac{1}{2} * 32.50 * x^2$

=> $x = \sqrt{0.03938}$

=> $x = 0.198456 \ m$

Considering the second question

From the question we are told that

The mass of the dart is m = 0.050 kg

Generally from the law of energy conservation

$PE = mgh$

=> $0.640 = 0.050 * 9.8 * h$

=> $h = 1.3061 \ m$

Considering the third question

The height at which the dart was fired horizontally is $H = 3.90\ m$

Generally from the law of energy conservation

$PE = KE$

Here KE is kinetic energy of the dart which is mathematical represented as

$KE = \frac{1}{2} * mv^2$

=> $0.640 = \frac{1}{2} * 0.050 * v^2$

=> $v^2 = 25.6$

=> $v = 5.06 \ m/s$

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

$t = \frac{ 2 * H }{g}$

=> $t = \frac{ 2 * 3.90 }{9.8 }$

=> $t = 0.7959 \ s$

Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as

$d = v * t$

=> $d = 5.06 * 0.7959$

=> $d = 4.0273 \ m$

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