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3 years ago

7

Which statement below best describes the properties of solids? A. Solids have both definite shape and definite volume. B. Solids

have definite volume but no definite shape. C. Solids have definite shape but no definite volume. D. Solids have no definite shape or volume.

1 answer:

Butoxors [25]3 years ago

7 0

The answer is A. Solids have both definite shape and denfinite volume.

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. Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A 300-cm long copp

MA_775_DIABLO [31]

Answer:

a) Yes

b) No

Explanation:

In the first case, part a, yes we can say for certainty that cylinderical symmetry holds. Why so? You may ask. This is because from the question, we are told that the length of the rod is 300 cm. And this said length is longer than the distance to the point from the center of the rod, which is 5 cm.

In the second half of the question, I beg to disagree that cylindrical symmetry holds. Again, you may ask why, this is because the length of the rod in this case, is having the same order of magnitude as the distance to the center of the rod. Thus, it is not symmetrical.

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3 years ago

What determines the state of matter for any

noname [10]

Answer:

I think the answer is B. amount of energy present but I'm not 100% sure

Explanation:

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3 years ago

Read 2 more answers

What is the relationship between the electric force and each one of the charges?

Bess [88]

Explanation: Electrostatic force is directly related to the charge of each object. So if the charge of one object is doubled, then the force will become two times greater.

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3 years ago

A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling

Setler [38]

m1=1500kg
m_2=3000kg
v_1=5m/s
v_2=7m/s

Using law of conservation of momentum

$\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3$

$\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3$

$\\ \sf\Rrightarrow 7500-21000=4500v_3$

$\\ \sf\Rrightarrow -13500=4500v_3$

$\\ \sf\Rrightarrow v_3=-3m/s$

6 0

3 years ago

A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comforta

Alex

Answer:

<em>The depth will be equal to</em> <em>6141.96 m</em>

<em></em>

Explanation:

pressure on the submarine $P_{sea}$ = 62 MPa = 62 x 10^6 Pa

we also know that $P_{sea}$ = ρgh

where

ρ is the density of sea water = 1029 kg/m^3

g is acceleration due to gravity = 9.81 m/s^2

h is the depth below the water that this pressure acts

substituting values, we have

$P_{sea}$ = 1029 x 9.81 x h = 10094.49h

The gauge pressure within the submarine $P_{g}$ = 101 kPa = 101000 Pa

this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.

Equating the pressure $P_{sea}$ , we have

62 x 10^6 = 10094.49h

depth h = <em>6141.96 m</em>

7 0

3 years ago