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Y_Kistochka [10]
3 years ago
11

Which expression of distance uses SI units? A. 30 miles B. 16 kilograms C. 24 feet D. 500 meters

Physics
2 answers:
Stella [2.4K]3 years ago
7 0
Hi!

SI units are physical measurements which will be in the form of kilograms, second, kelvin, metres, etc.

Since kilograms measure the weight of an object, it is out. Miles and feet are not SI units, so they are also out. This only leaves one answer left!

Hopefully, this helps! =)
lesya692 [45]3 years ago
6 0

Answer:

D) 500 meter

Explanation:

As we know that SI unit to measure the distance is meter

so here the SI unit of the measurement of the distance must be in meter units

so it is given as

D) 500 meter

while other options are

A) 30 miles : this is also measurement of distance but it is not in SI unit

B) 16 kilogram : this is measurement of mass

c) 24 feet : this is also measurement of distance but not in SI units

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Answer:

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A tow truck exerts a net horizontal force of 1050 N on a 760-kg car. What is the acceleration of the car during this time?
vovikov84 [41]
We can solve the problem by using Newton's second law of motion:
F=ma
where
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a is the acceleration of the object

In this problem, the force applied to the car is F=1050 N, while the mass of the car is m=760 kg. Therefore, we can rearrange the equation and put these numbers in, in order to find the acceleration of the car:
a= \frac{F}{m}= \frac{1050 N}{760 kg}=1.4 m/s^2

The equation also tells us that the acceleration and the force have same directions: therefore, since the force exerted on the car is horizontal, the correct answer is
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5 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:

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   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

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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

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

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     T_2 = 0.625 * 9.8

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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

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The period of the wave would be halved 
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