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Dennis_Churaev [7]

3 years ago

12

When Dr. Hewitt immerses an object in water the second time and catches the water that is displaced by the object, how does the

weight lost by the object compare to the weight of the water displaced?

1 answer:

Dimas [21]3 years ago

5 0

Answer:

Explanation:

- The volume of water displaced by immersing the object is equal the amount of water spilled and caught by Dr. Hewitt.

- The amount of water is proportional to the volume of object of fraction of object immersed in water will lead to the same fraction of water displaced and caught by Dr. Hewitt.

- When the object is immersed the force of Buoyancy acts against the weight and reducing the scale weight.

- The amount of Buoyancy Force is proportional to the fraction of Volume of object immersed in water; hence, the same amount is spilled/lost.

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How many total atoms are in 2CuSO4 A.2 B.6 C.8 D.12

makvit [3.9K]

That depends on how much of it you have, but it's a hugely enormous gigantic number in even the smallest sample.

You can see from the formula that there are 12 atoms in every molecule of the stuff !

5 0

3 years ago

HELP ASAP!!!!! WILL GIVE BRAINLIEST FOR TOP ANSWER!!!! What occurs in lymphocytes in the thymus as they develop into T-cells?

tester [92]

Answer:

They learn to produce other blood cells

Explanation:

I just look it up on so I hope its right

8 0

3 years ago

Read 2 more answers

A 60.0-kg person rides in an elevator while standing on a scale. the scale reads 400 n. the acceleration of the elevator is clos

Keith_Richards [23]

Reading of the scale will be given as Normal force on the man

So here we can say that

weight of the mas will be

$W = mg = 60*10 = 600 N$

normal force on the man will be given as

$F_n = 400 N$

now net force is given by Newton's II law

$F_{net} = ma$

$F_g - F_n = ma$

$600 - 400 = 60*a$

$a = \frac{200}{60} = \frac{10}{3}$

a = 3.33 m/s^2

so elevator is accelerating downwards with an acceleration a = 3.33 m/s^2

4 0

4 years ago

Read 2 more answers

You will begin with a relatively standard calculation.Consider a concave spherical mirror with a radius of curvature equal to 60

Strike441 [17]

a) 30.0 cm

For any mirror, the radius of curvature is twice the focal length:

r = 2f

where

r is the radius of curvature

f is the focal length

For the mirror in this problem, we have

r = 60.0 cm is the radius of curvature

Therefore, solving the equation above for f, we find its focal length:

$f=\frac{r}{2}=\frac{60.0 cm}{2}=30.0 cm$

b) 90 cm

The mirror equation is:

$\frac{1}{s'}=\frac{1}{f}-\frac{1}{s}$

where

s' is the distance of the image from the mirror

f is the focal length

s is the distance of the object from the mirror

For the situation in the problem, we have

f = +30.0 cm is the focal length (positive for a concave mirror)

s = 45.0 cm is the object distance from the mirror

Solving the formula for s', we find

$\frac{1}{s'}=\frac{1}{30.0 cm}-\frac{1}{45.0 cm}=0.011 cm^{-1}$

$s'=\frac{1}{0.011 cm^{-1}}=90 cm$

c) -2

The magnification of the mirror is given by

$M=-\frac{s'}{s}$

where in this problem we have

s' = 90 cm is the image distance

s = 45.0 cm is the object distance

Solving the equation, we find:

$M=-\frac{90 cm}{45 cm}=-2$

So, the magnification is -2.

d) -12.0 cm

The magnification can also be rewritten as

$M=\frac{y'}{y}$

where

y' is the height of the image

y is the heigth of the object

In this problem, we know

y = 6.0 cm is the height of the object

M = -2 is the magnification

Solving the equation for y', we find

$y'=My=(-2)(6.0 cm)=-12.0 cm$

and the negative sign means that the image is inverted.

Part e and f are exactly identical as part b) and c).

6 0

4 years ago

The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much

aleksklad [387]

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

$\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}$

Upon substituting our given values, we will get:

$\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}$

$\frac{(T_1)^2}{(T_2)^2}=0.0073878361246365$

Taking square root of both sides, we will get:

$\frac{T_1}{T_2}=\sqrt{0.0073878361246365}$

$\frac{T_1}{T_2}=0.0859525225030452495$

$\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}$

$\frac{T_2}{T_1}=11.634329870476699\approx 11.634$

$T_2=11.634\cdot T_1$

This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.

So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

8 0

3 years ago