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Sindrei [870]
3 years ago
8

What are two examples of goods and two examples of services

Physics
1 answer:
julsineya [31]3 years ago
8 0

The goods and the services make up the basis of every economy. The goods can simply be defined as merchandise or possessions. The services can be defined as the actions through which help is provided, or work is done for someone else. Example of goods are the food and furniture, with the food being crucial for the survival of the people, while the furniture is an essential part of every home and its practicality and decor. Examples of services are teaching and car repairing. The teaching is crucial for the development of the societies, as through it the people get education, while the repairing of cars is very important as lot of people have them, can not afford to buy new ones all the time, and they need for their daily movement over longer distances.

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Nataliya [291]

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A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M a
aleksklad [387]

Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

b) he two blocks must move the same distance

c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

small block

             w- T = m a

             

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

             a = g \ \frac{m - \mu M}{m+M}

             a = 9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

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Answer:

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Explanation:

We can calculate the takeoff speed easily, using the following kinematic equation.

v_{f}^{2}=v_{i}^{2} +2*a*x

where:

a = acceleration = 4[m/s^2]

x = distance = 750[m]

vi = initial velocity = 25 [m/s]

vf = final velocity

v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]

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