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svlad2 [7]
2 years ago
10

Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object

moves with constant acceleration. Which part of a velocity vs. time graph can be used to calculate the displacement of the object? A. the area of the rectangle under the line B. the area of the rectangle above the line C. the area of the rectangle plus the area of the triangle under the line D. the area of the rectangle plus the area of the triangle above the line

Physics
2 answers:
Tatiana [17]2 years ago
6 0

Answer:

C. the area of the rectangle plus the area of the triangle under the line

Explanation:

Based on the information provided, the velocity vs. time graph is a line with a positive slope and a y-intercept of (0, 3).  The displacement is the area under this line.  This area can be divided into a triangle and a rectangle.  So of the options available, C is the correct one.

vesna_86 [32]2 years ago
6 0

Answer:

C

Explanation:

There is no explanation for the answer.

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Answer:

Angular acceleration = 0.167 rad/s^2

Explanation:

Given

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Final Angular velocity (w2) = 0 rad/s

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Angular acceleration is equal to the change in angular velocity to the time taken for making thing change

Hence, Angular acceleration

\frac{w_2 -w_1}{t} \\\frac{5-0}{30}\\0.167rad/s^2

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A 65-cm segment of conducting wire carries a current of 0.35 A. The wire is placed in a uniform magnetic field that has a magnit
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A book rests on the shelf of a bookcase. The reaction force to the force of gravity acting on the book is 1. The force of the sh
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Answer:

1. The force of the shelf holding the book up.

Explanation:

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1 - The weight of the book towards downwards

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Since the book is at rest, these two forces are equal to each other and according to Newton's Third Law the reaction force to the force of gravity is equal but opposite to the weight of the book. This reaction force is the one that holds the book up on the shelf.

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A box of mass 5.0 kg is accelerated from rest by a force across a floor at a rate of 2.0 m/s2 for 7.0 s. find the net work done
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A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and
victus00 [196]

Answer:

a)  t=195.948N.m

b)  \phi=13.6 \textdegree

Explanation:

From the question we are told that:

Density \rho=1.225kg/m^2

Velocity of wind v=14m/s

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient \mu=2.05

a)

Generally the equation for Force is mathematically given by

F=\frac{1}{2}\muA\rhov^2

F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2

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Therefore Torque

t=F*r*sin\theta

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b)

Generally the equation for torque due to weight is mathematically given by

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Where

d=sin \phi

Therefore

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195.948=833sin \phi

\phi=sin^{-1}\frac{195.948}{833}

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