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fredd [130]
3 years ago
14

Which property describes the ability of one substance to dissolve in another substance?

Physics
1 answer:
oee [108]3 years ago
5 0
The answer would be solubility
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Answer:

From the given choices, the healthy snack that can provide protein after physical activity is the third option, yogurt. Yogurts are derived from fermentation of raw materials by the healthy bacteria, usually of lactobacillus origin. Other items in the choices are not able to provide protein. 

Explanation:

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Suppose the force of gravity on the electron was comparable to the electric force created by the deflection plates. how would th
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"F=Vector Sum Of The Two Forces" Is the answer.
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A 2.00-kg block of ice is moving on a frictionless horizontal surface. At t = 0 the block is moving to the right with a velocity
enyata [817]

Answer:

Explanation:

a )

We shall apply the concept of impulse .

Impulse = force x time = change in momentum

= 5 x 4 = 2 ( V - 3 )  , where V is final velocity of the object

20 = 2V - 6

V = 13 m /s

b )

Impulse applied = - 7 x 4 = - 28 kg m/s ( negative as direction of force is opposite motion )

If v be the final velocity

2 x 3 - 28 = 2 v  ( initial momentum - change in momentum = final momentum )

- 22 = 2v

v = - 11 m /s

object will move with 11 m /s in opposite direction .

4 0
3 years ago
Can you describe Saturn?
Arte-miy333 [17]
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3 years ago
Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
miv72 [106K]

Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

In this initial case

     m_A = M

     m_B = M

     a = M / (1 + 1) M g

     a = ½ g

Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

    m_B = 2M

    a = 2M / (1 + 2) M g

    a = 2/3 g

We seek tension for this case

    T’= m_A a

    T’= M 2/3 g

   

Let's look for the relationship between the tensions of the two cases

   T’/ T = 2/3 M g / (½ M g)

   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

4 0
3 years ago
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