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2 years ago

If two balls have the same volume,

1 answer:

8 0

Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.

1. Ball A will have the greater density
2. Ball C and Ball D have the same density.
3. Ball Q will have the greater density.
4. Ball X and Y will have the same density

The density of an object is given as its mass per unit volume of the object.

Mathematically;.

Density = Mass/Volume.

For Case 1:

Va = Vb and Ma = 2Mb
D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
Therefore, the density of ball A,
D(a) = 2D(b).
Therefore, ball A has the greater density.

For Case 2:

Vc = 3Vd,

Vd = (1/3)Vc

Md = (1/3)Mc

D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd

D(c) = D(d).

Therefore, ball C and D have the same density

For Case 3:

Vp = 2Vq and Mp = Mq
D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
Therefore, the density of ball P is half the density of ball Q
Therefore, ball Q has the greater density.

For case 4:

Mx = (1/2)My
Vx = Vy

Therefore, Ball X and Ball Y have the same density.

Read more:

brainly.com/question/18110802

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A boy is going down a slide. As he reaches the bottom, friction causes him to slow down and stop. Which one of Newton's laws is

lara31 [8.8K]

I think it’s the second law

8 0

3 years ago

Two concrete spans of a 380 m long bridge are

Mazyrski [523]

Answer:

4.163 m

Explanation:

Since the length of the bridge is

L = 380 m

And the bridge consists of 2 spans, the initial length of each span is

$L_i = \frac{L}{2}=\frac{380}{2}=190 m$

Due to the increase in temperature, the length of each span increases according to:

$L_f = L_i(1+ \alpha \Delta T)$

where

$L_i = 190 m$ is the initial length of one span

$\alpha =1.2\cdot 10^{-5} ^{\circ}C^{-1}$ is the temperature coefficient of thermal expansion

$\Delta T=20^{\circ}C$ is the increase in temperature

Substituting,

$L_f=(190)(1+(1.2\cdot 10^{-5})(20))=190.0456 m$

By using Pythagorean's theorem, we can find by how much the height of each span rises due to this thermal expansion (in fact, the new length corresponds to the hypothenuse of a right triangle, in which the base is the original length of the spand, and the rise in heigth is the other side); so we find:

$h=\sqrt{L_f^2-L_i^2}=\sqrt{(190.0456)^2-(190)^2}=4.163 m$