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Paul [167]
2 years ago
7

Density is calculated by combining 2 units, therefore the unit we use to measure density is called a _____________________unit.

What is the unit?
Physics
1 answer:
nalin [4]2 years ago
8 0

Answer: Dependent Unit or System of Units

Explanation:

Density is calculated by dividing mass (Kg) by volume (L).

The unit of Density is Kg/L or one of their derivatives such as g/cm³.

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A car travels around a curve. What can you say about the centripetal force if the velocity is tripled?
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Answer:

F = M a = M v^2 / R

If v is increased by three the force will be increased by nine,

C) is correct

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The period of an ocean wave is 5 seconds. What is the wave's frequency?
Oksana_A [137]

Answer:

f=1/5= 0.2 Hz

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B. CO
Hunter-Best [27]

Answer:

0.6 m/s

Explanation:

2Hz = 2^-1 = 2 /s

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Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.

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The speed of a space shuttle is 10 / express this in /�
12345 [234]

Answer:

268.22m/s

Explanation:

Given;

    10mile/min to m/s

We need to convert between the two units;

  Using the dimensions;

    1 mile  = 1609.34m

     60s  = 1min

Now;

    10 x \frac{mile}{min} x \frac{1min}{60s} x \frac{1609.34m}{1mile}

  = 268.22m/s

3 0
3 years ago
A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are verti
NeX [460]

Answer:

a)  T_A = \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} ) ,  b) T_B = g [m₂ ( \frac{x}{d} -1) + m₁ ( \frac{L}{ 2d} -1) ]

c)  x = d - \frac{m_1}{m_2} \  \frac{L}{2d},  d)  m₂ = m₁  ( \frac{ L}{2d} -1)

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

      ∑ τ = 0

      T_A d - W₂ x -W₁ L/2 = 0

      T_A = \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )

b) the tension in string B

we write the expression of the translational equilibrium

       ∑ F = 0

       T_A - W₂ - W₁ - T_B = 0

       T_B = T_A -W₂ - W₁

       T_ B =   \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )  - g m₂ - g m₁

       T_B = g [m₂ ( \frac{x}{d} -1) + m₁ ( \frac{L}{ 2d} -1) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

         T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

          g m₂ (d -x) -  g m₁  (0.5 L -d) + 0 = 0

          m₂ (d-x) = m₁ (0.5 L- d)

          m₂ x = m₂ d - m₁ (0.5 L- d)

          x = d - \frac{m_1}{m_2} \  \frac{L}{2d}

 

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

           0 = d - \frac{m_1}{m_2} \  \frac{L}{2d}

         \frac{m_1}{m_2} \ (0.5L -d) = d

          \frac{m_1}{m_2} = \frac{ d}{0.5L-d}

          m₂ = m₁  \frac{0.5 L -d}{d}

          m₂ = m₁  ( \frac{ L}{2d} -1)

5 0
3 years ago
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