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Mumz [18]
4 years ago
10

Which describes how the same force affects a small mass and a large mass

Physics
1 answer:
goldenfox [79]4 years ago
5 0
You shall use the equation for force given by the second Law of Newton, this is F = m*a, where F is the net force that acts over the object, m is the mass of the object and a is the acceleration that the object will acquire. From that equation you can find a = F/m, which means that a is direct proportional to F and invsersely related to m. So, small masses accelerate faster than large masses, and <span>the answer is the option B. the small mass accelerates faster.</span>
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Answer:

4

Explanation:

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1. Ignore friction and determine an expression for the distance d the boxes travel before coming to rest. Choose the floor as th
Nadusha1986 [10]

Newton's second law allows us to find the results for the displacement and work on box 1 are:

       1) The displacement is   x = \frac{v_o^2 m_2}{2(m_1+m_2)} g

      2) The work is  W= \frac{1}{2} m_1v_o^2

1) Newton's second law says that force is directly proportional to the mass and acceleration of bodies.

         F = ma

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the bodies

The reference system is a coordinate system with respect to which the decompositions of the forces are carried out and all the measurements are made, in this case we take a system where the x axis is horizontal, the y axis is vertical and the zero of the system is located at soil

In the attached we can see a free body diagram of the system where the external forces are indicated, let's apply Newton's second law to body 1

body 1

x-axis

          -T = m₁ a

body 2

 y-axis

           T - W₂ = m₂ a

           W₂ = m₂ g

let's solve the system

        - m₂ g = (m₁ + m₂) a

           a = - \frac{m_2}{m_1+m_2} \ g- m2 / m1 + m2 g

Kinematics studies the movement of bodies, let's use the expression

            v² = v_o^2 - 2 a x

When the body stops the velocity is zero

            x = \frac{v_o^2}{2a}

We substitute

            x = \frac{v_o^2}{2} \frac{m_2}{m_1+m_2} \ g

Since the two boxes are connected by a rope, they both travel the same distance

           

2) They ask for Tension work on box 1

Work is defined by the scalar product of force and displacement

        W = F. d

Where the bold letters indicate vectors, W is the work that is a scalar, F the force and d the displacement

We can write this expression by developing the dot product

       W  = F d cos θ

Where θ is the angle between force and displacement.

In this case the force is the tension of the rope, from the attached graph we see that the force is directed to the right and the displacement is to the left, therefore the angle is 180º and the cos  180 is equal to -1

We look for the tension from Newton's second law for box 1

          T = - m₁ a

Let's substitute

          T =- m_1 \ ( - \frac{m_2}{m_1+m_2} ) \ g  

          T = \frac{m_1m_2}{m_1+m_2} \ g

         

We calculate the work

         W = - \frac{m_1m_2}{m_1+m_2 } \ g ( \frac{v_o^2 (m1+m_2)}{2 m_2 g} )

         W = - ½ m₁ v₀²

In conclusion using Newton's second law we can find the results for the displacement and work on box 1 are:

1)  The displacement is  x= \frac{v_o^2 m_2}{2(m_1+m_2)} g

2) The work is  W = - ½ m₁ v₀²

Learn more here: brainly.com/question/17290735

3 0
2 years ago
About how much time elapses between high tides on the same day?
Pani-rosa [81]
There is about 12 hours
6 0
4 years ago
1.A Radio station broadcasts modern song on medium wave 350 Hz every day at ten o’clock in the morning. The velocity of radio wa
love history [14]

Answer:

ans \:  = \boxed{{4.8 \times 10}^{ - 4}  Hz}

Explanation:

given \to \\  f_{r} = 350 \:  \\ v_{r} =  {3 \times 10}^{8}  \\ but \to \\ v = f \gamma   \to \:  \gamma  =  \frac{v}{f}  : hence \to \\  \gamma _{r} =  \frac{v_{r}}{f_{r}}   =  \frac{3 \times 10^{8} }{350}   =  \boxed{857,142.85714 \: m}\\ therefore \to \\ given \to \\  f_{w} = water \: frequency = \:  \boxed{  ?}\:  \\ v_{w} =  14 50 \\ but \to \\ v = f \gamma   \to \:  \gamma  =  \frac{v}{f}  : hence \to \\  \gamma _{w} =  \frac{v_{w}}{f_{w}}   =  \frac{1}{100}  \times \gamma _{r}  =  \frac{1}{100}  \times 857,142.85714  \\\gamma _{w}  =  \boxed{8,571.4285714 \: m} : hence \to \:  \\ f_{w} =  \frac{v_{w}}{ \gamma _{w}}  =  \frac{1450}{8,571.4285714}  =  \boxed{0.1691666667} \\ if \: the \: number \: of \: times = \boxed{ x} \\ f_{r} (x)=f_{w} \\ (x) =  \frac{f_{w}}{f_{r}}  =  \frac{0.1691666667}{350}  = 0.0004833333 \\ hence \to \\ the  \: frequency  \: of \:  the \:  radio  \: wave  \: is \to \:   \boxed{{4.8 \times 10}^{ - 4}  }\:  \\ that  \: of  \: the \:  wave  \: created  \: in  \: the  \: water.

♨Rage♨

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3 years ago
Why cant soy sauce freeze
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The salt content in soy
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