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MA_775_DIABLO [31]
3 years ago
11

Which is another term for the free enterprise system

Physics
1 answer:
Naily [24]3 years ago
6 0

Answer:

Another term for free enterprise system would be capitalism

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Experiments need to be repeated by the scientist and also replicated by other scientists.
BlackZzzverrR [31]

Answer:

true

Explanation:

6 0
2 years ago
Read 2 more answers
To suck lemonade of density 1040 kg/m3 up a straw to a maximum height of 4.94 cm, what minimum gauge pressure (in atmospheres) m
Lady bird [3.3K]

Answer:

The minimum gauge pressure is 0.4969 atm.

Explanation:

Given that,

Density = 1040 kg/m³

Height = 4.94 cm

We need to calculate the pressure

Using formula of pressure

P_{g}=\rho g h

Where, \rho=density

h = height

Put the value into the formula

P_{g}=1040\times9.8\times4.94

P_{g}=50348.48\ Pa

Pressure in atmospheres

1\ atm =101.3\ kPa

P_{g}=\dfrac{50348.48}{101325}

P_{g}=0.4969\ atm

Hence, The minimum gauge pressure is 0.4969 atm.

7 0
3 years ago
Read 2 more answers
If it takes you 10 seconds to increase from 0 km/h to 50 km/h, what is your acceleration?
ZanzabumX [31]
Acceleration is change in velocity over change in time. Your Δv is +13.9, since you increased speed by 50 km/h which is 13.9 m/s, and your Δt is 10s. 13.9/10 = 1.39 m/s^2, the standard units for acceleration. Make sense?
4 0
3 years ago
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The upward force exerted on an object falling through air is _____.
stira [4]

(C) Air Resistance

<u>Explanation:</u>

When an object falls through air, air resistance acts on it in upward direction. When air resistance acts, acceleration during a fall will be less than g because air resistance affects the motion of the falling objects by slowing it down. Air resistance depends on two important factors - the speed of the object and its surface area. Increasing the surface area of an object decreases its speed.

6 0
3 years ago
A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu
dalvyx [7]

Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 =  0.341 \ kg

   The mass attached to the other end of the string is  m_2 =  0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 =  m_1 *  g

substituting values

      T_1 =  0.341 * 9.8

      T_1 =  3.342 \ N

At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 =  m_2 *  g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

     \tau  = 0.2505 \ N \cdot m

5 0
3 years ago
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