According to the following formula, the answer is 2,300 g or 2.3 kg:
Volume (m)/Mass (m) Equals Density (p) (V)
Here, the density is 1.15 g/mL, allowing the formula described above to result in a mass of 2.00 L:
p=m/V
1.15 g/mL is equal to x g/2.00 L or x g/2,000 mL.
2,000 mL of x g = 1.15 g of g/mL
2.3 kg or 2,300 g for x g.
<h3>How many grams of glucose are in a 1000ml bag of glucose 5?</h3>
Its active ingredient is glucose. This medication includes 50 g of glucose per 1000 ml (equivalent to 55 g glucose monohydrate). 50 mg of glucose is present in 1 ml (equivalent to 55 mg glucose monohydrate). A transparent, nearly colourless solution of glucose in water is what is used in glucose intravenous infusion (BP) at 5% weight-to-volume.
Patients who are dehydrated or who have low blood sugar levels get glucose intravenously. Other medications may be diluted with glucose intravenous infusion before being injected into the body. Other diseases and disorders not covered above may also be treated with it.
learn more about glucose intravenous infusion refer
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<h2>The different forces acting on the ball while its in air</h2>
Amy throws a softball through the air. Applied, drag and gravitational forces are acting on the ball while it’s in the air. The softball experiences force as a result of Amy’s throw. As the ball moves, it experiences from the air it passes through.
It also experiences a downward pull because earth has the property to attract everything which is on the earth towards it. The ball is moving in the air but earth applies force on the ball to get back on the ground. Hence, in this way, gravitational force applies.
There is also a drag force which results due to friction that is present in the air. It resist to move ball in the air and there will also be applied force which is given by a person who throws by applying force.
Velocity is the answer..
hope that helps
Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>
