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Ber [7]
3 years ago
9

Of all the Group 13 metals, I am the one with the least mass.

Physics
2 answers:
Orlov [11]3 years ago
8 0
Boron has the least mass
posledela3 years ago
6 0

Answer:

Boron has the least mass

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Please select the word from the list that best fits the definition site of sea-floor spreading
serious [3.7K]
<h3><u>Answer;</u></h3>

Mid-ocean ridges

<h3><u>Explanation</u>;</h3>
  • Sea-floor spreading is the process by which molten material adds new oceanic crust to the ocean floor.
  • Mid-ocean ridge is an undersea mountain chain where new ocean floor is produced at a divergent plate boundary.
  • Mid ocean ridge occurs when convection currents rise in the mantle beneath the oceanic crust and create magma where two tectonic plates meet at a divergent boundary.

8 0
3 years ago
Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in
beks73 [17]

Answer:

a. the force between them quadruples

Explanation:

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, the charges on both objects are doubled, so

q_1' = 2q_1\\q_2' = 2q_2

While the distance does not change, so the new force will be

F'= k \frac{(2q_1)(2q_2)}{r^2}=4 (k\frac{q_1 q_2}{r^2})=4 F

so, the force will quadruple.

4 0
3 years ago
How are light years used to measure distances in the universe
FinnZ [79.3K]
A light year is a unit of distance. It is a distance that light can travel in a years time which is six trillion miles. It is used to measure the distances in space. To take one example, the distance to the next nearest big galaxy, the Andromeda Galaxy, from earth is 21,000,000,000,000,000,000 km. 
Do you understand it? <span />
4 0
3 years ago
A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de
EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/πr^{2}

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / πr^{2}ex

E = \frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6}    } = 17.368 * 10^{9} Pa = 17.4 GPa

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = \frac{1}{E} (бy - v (бx + бz)) = -\frac{v}{E}бx

= \frac{vFx}{Epir^{2} } = \frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9}  } = -0.63 *10^{-6}

Finally

ey = Δr / r

Δr = ey * r = 10 * -0.63* 10^{-6} mm = -6.3 * 10^{-6} mm

Δd = 2Δr = -12.6 * 10^{-6} mm

Explanation:

5 0
3 years ago
An object has an acceleration of 6.0 m/s/s. if the net force acting upon this object were doubled, then its new acceleration wou
Flauer [41]
F has direct relation with a
then doubling F cause acc. to get double i:e 6×2=12
5 0
4 years ago
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