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alexira [117]
2 years ago
6

A particle has an acceleration of 6.14 m/s2 for 0.300 s. at the end of this time the particle's velocity is 9.41 m/s. part a wha

t was the particle's initial velocity? chefggg
Physics
1 answer:
Varvara68 [4.7K]2 years ago
4 0

The initial velocity of particle is 7.568 m/s

Given:

Acceleration = a = 6.14 m/s2

Time = t = 0.300 s.

final velocity = v = 9.41 m/s.

To Find:

initial velocity = u

Solution: Velocity is the prime indicator of the position as well as the rapidity of the object.

v = u + at

u = v - at = 9.41 - 6.14*0.3 = 7.568 m/s

Hence, initial velocity of particle is 7.568 m/s

Learn more about Velocity here:

brainly.com/question/25749514

#SPJ4

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Answer:

Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy-including planets,stars,galaxies, and even light-are brought toward one another.

Explanation:

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The effect produced when two or more sound waves pass through the same point simultaneously is called?
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The answer to the question is sound
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A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth
andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

F_{D}=\frac{kq^{2}}{DC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

F_{B}=\frac{kq^{2}}{BC^{2}}

F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

F_{A}=\frac{kq^{2}}{AC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N

The direction of FA is along A to C.

The net force along +X axis

F_{x}=F_{A}Cos45-F_{D}

F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N

The net force along +Y axis

F_{y}=F_{B}-F_{A}Sin45

F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N

The resultant force is given by

F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}

F = 4.03\times10^{7}N

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

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3 years ago
List 3 cases in which potential energy becomes kinetic energy and three cases in which kinetic energy becomes potential energy.
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Answer:

when you open a can of pop

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Explanation:

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What velocity will a freefalling object have after falling 80 meters
vovangra [49]

Answer:

129.96

At a Gravitational acceleration of 32.17405 ( which is the normal rate for a freefall) you will geta velocity of 129.96 and the time of fall will be 4.039 seconds from 80 meters.

Why is the weight of a free falling body zero? It is not, an object in free fall will still have a weight, governed by the equation W = mg, where W is the object's weight, m is the object's mass, and g is acceleration due to gravity. Weight, however, has no effect on an objects free falling speed, two identically shaped objects weighing a different amount will hit the ground at the same time.

Hope this helps!! If so please mark brainliest and rate/heart to help my account if it did!!

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3 years ago
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