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xenn [34]
3 years ago
9

Marie and Calvin tried to dissolve 45 grams of KNO3 in 100 grams of water at 25oC. Only 40 grams dissolved; 5 grams settled at t

he bottom of the beaker. The solution is then said to be ________________ at 25oC.
Physics
2 answers:
Lady bird [3.3K]3 years ago
6 0
THE ANSWER IS : saturated
zloy xaker [14]3 years ago
3 0

Answer:

The answer is A.

Explanation:

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What force is needed to give a 0.25-kg arrow an acceleration of 196 m/s/s
kakasveta [241]
Force (f) = ?

Acceleration (a) = 196 m/s^2

Mass (m) = 0.25 kg

F = (m) • (a)

F = (0.25) • (196)

F = 49 N

Answer : 49 N

I hope that helps you!! Any more questions??

8 0
3 years ago
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Veins contain a network of valves to insure a one-way flow of blood but arteries do not. This is because veins must counteract t
Elena-2011 [213]
They must counteract the force of blood pressure
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3 years ago
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What is the sequence of energy transformations associated with a hydroelectric dam?
Hatshy [7]

Gravitational potential energy -> Kinetic energy -> Mechanical energy -> Electrical energy.

The water starts up (potential) and flows down (kinetic), the flowing water turns a big wheel (mechanical) which creates electricity (electrical).

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3 years ago
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A 10.0 g bullet moving at 300m/s is fired into a 1.00 kg block at rest. The bullet emerges (the bullet does not get embedded in
chubhunter [2.5K]

Answer:

v' = 1.5 m/s

Explanation:

given,

mass of the bullet, m = 10 g

initial speed of the bullet, v = 300 m/s

final speed of the bullet after collision, v' = 300/2 = 150 m/s

Mass of the block, M = 1 Kg

initial speed of the block, u = 0 m/s

velocity of the block after collision, u' = ?

using conservation of momentum

 m v + Mu = m v' + M u'

 0.01 x 300 + 0 = 0.01 x 150 + 1 x v'

v' = 0.01 x 150

v' = 1.5 m/s

Speed of the block after collision is equal to v' = 1.5 m/s

5 0
3 years ago
The electric motor of a model train accelerates the train from rest to 0.740 m/s in 30.0 ms. The total mass of the train is 560
Alex73 [517]

Answer:

= 5.1 W

Explanation:

time (t) = 30 ms = 0.03 s

mass (m) = 560 g = 0.56 kg

initial velocity (U) = 0 m/s

final velocity (V) = 0.74 m/s

power = \frac{work done}{t} = \frac{f x d}{t} = f x v = m x a x v

m x a x v = m x \frac{V-U}{t} x \frac{V + U}{2}

m x \frac{V-U}{t} x \frac{V + U}{2} = 0.56 x \frac{0.74 - 0}{0.03} x \frac{0.74+0}{2}

= 5.1 W

7 0
3 years ago
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