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Roman55 [17]
3 years ago
15

Kia is doing an experiment in science lab. She is given a beaker containing 100 g of liquid. The beaker has markings on the side

for measuring volume. The water comes up to the 100 mL mark. Kia puts the liquid on a hot plate by mistake. By the time she realizes the mistake, half of her liquid has evaporated.
Kia still needs 100 g of liquid for her experiment. To find out how much liquid she has to replace, she needs to re-weigh her liquid. However, the balance she used before is now broken. The teacher tells Kia that she can tell how much water is left by looking. The water now comes up to the 50 mL mark.

How much mass does Kia's remaining water have?
Physics
1 answer:
lozanna [386]3 years ago
7 0
Kia's remaining water has a mass of 50g. You can set it up as a proportion knowing that 100ml of water has a mass of 100g and thus 50ml of water would weight 50g
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Answer:

313.6 m downward

Explanation:

The distance covered by the bullet along the vertical direction can be calculated by using the equation of motion of a projectile along the y-axis.

In fact, we have:

y(t) = h +u_y t + \frac{1}{2}at^2

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h is the initial height of the projectile

u_y = 0 is the initial vertical velocity of the projectile, which is zero since the bullet is fired horizontally

t is the time

a = g = -9.8 m/s^2 is the acceleration due to gravity

We can rewrite the equation as

y(t)-h = \frac{1}{2}gt^2

where the term on the left, y(t)-h, represents the vertical displacement of the bullet. Substituting numbers and t = 8 s, we  find

y(t)-h= \frac{1}{2}(-9.8)(8)^2 = -313.6 m

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The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g
Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:  

k=\sqrt{\frac{I}{M} }

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration  k=\sqrt{\frac{I}{M} }

So, moment of rotational inertia (I) of a cylinder about it axis = \frac{MR^2}{2}

k=\sqrt{\frac{\frac{MR^2}{2}}{M} }

k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }

k=\sqrt{{\frac{R^2}{2}}

k={\frac{R}{\sqrt{2}}

k={\frac{1.20m}{\sqrt{2}}

k = 0.8455 m

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For the spherical shell of radius

(I) = \frac{2}{3}MR^2

k = \sqrt{\frac{\frac{2}{3}MR^2}{M}  }

k = \sqrt{\frac{2}{3} R^2}

k = \sqrt{\frac{2}{3} }*R

k = \sqrt{\frac{2}{3}}  *1.20

k = 0.9797 m

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For the solid sphere of  radius

(I) = \frac{2}{5}MR^2

k = \sqrt{\frac{\frac{2}{5}MR^2}{M}  }

k = \sqrt{\frac{2}{5} R^2}

k = \sqrt{\frac{2}{5} }*R

k = \sqrt{\frac{2}{5}}  *1.20

k = 0.7560

k ≅ 0.76 m

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4 years ago
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