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Strike441 [17]
3 years ago
12

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics
2 answers:
dalvyx [7]3 years ago
4 0

Answer:

The answer is A

Explanation:

I just C in and it was wrong, The answer is A.

Nonamiya [84]3 years ago
3 0

Based on the graph, the soccer ball is not moving from C) from 4.5 to 5.0 seconds

\texttt{ }

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

\texttt{ }

This problem is about Average Speed

<em>We will use this following formula:</em>

\texttt{Average Speed} = \texttt{Distance} \div \texttt{Time Taken}

From 4.5 to 5.0 seconds , there is no change in distance of the ball.

t₁ = 4.5 → s₁ = 65 m

t₂ = 5.0 → s₂ = 65 m

\texttt{Average Speed} = (65 - 65) \div (5.0 - 4.5)

\texttt{Average Speed} = 0 \div 0.5

\texttt{Average Speed} = 0 \texttt{ m/s}

Since the average speed is zero, the ball doesn't move from 4.5 to 5.0 seconds.

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

\texttt{ }

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Projectile , Motion , Horizontal , Vertical , Release , Point , Ball , Wall

#LearnWithBrainly

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Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
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The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

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