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Ugo [173]
3 years ago
15

The most common salt in ocean water is calcium surface. True or False

Physics
2 answers:
dmitriy555 [2]3 years ago
4 0
False: The most common salt found in ocean water would be sodium chloride. Seawater or ocean water contains many types of salt and the most abundant type of dissolved ions are sodium and chloride ions which when crystallized becomes NaCl. 
sergiy2304 [10]3 years ago
3 0
False, sodium and chloride are the most common ions on the ocean and they are salty causing the water to be salty
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Block A can slide relative to block B which, in turn, can slide on a perfectly smooth horizontal plane. If the initial velocity
Anna11 [10]

Answer:

the final velocity of the two blocks is v = \frac{mv_o}{m+M}

the distance that A slides relative to B is S = \frac{v_o^2M}{2 \mu g (M+m)}  

Explanation:

From the diagram below;

acceleration of A relative to B is : a = - ( \mu g  + \frac{ \mu mg}{M})

where

v = u + at

0 = v_o + ( - \mu g - \frac{\mu m g }{M})t

Making t the subject of the formula; we have:

t = \frac{v_o M}{(\mu g )(M+m)}

v^2 = u^2 +2 as\\\\0^2 = v_o^2 - 2 (\mu g ) (\frac{M+m}{M})S\\\\

S = \frac{v_o^2M}{2 \mu g (M+m)}  which implies the distance that A slides relative to B.

The final velocities of the two blocks can be determined as follows:

v = u + at

v = v_o - \mu g \frac{v_oM}{\mu g (M+m)}\\\\v = \frac{\mu g mv_o}{m+M}\\\\

v = \frac{mv_o}{m+M}

Thus, the final velocity of the two blocks is v = \frac{mv_o}{m+M}

4 0
2 years ago
A converging lens brings rays of light together at a focal point. the bending of light rays is the result of
disa [49]
The bending of light rays is the result of refraction of light passing through the lens. A converging lens is curved on both sides such that the rays coming out of it come together at a point (converge). The point at which the right rays meet after refraction is called the focal point which is a real in the convex lens. 
3 0
3 years ago
Read 2 more answers
In a basketball game, a player shoots a jump shot. Which force actually causes the player to jump? the player pushing down on th
Gemiola [76]

Answer:  the player pushing down on the floor

Explanation:

4 0
3 years ago
Read 2 more answers
While jumping on a trampoline you calculate that at the highest peak of your jump you have 900 joules of gravitational potential
BabaBlast [244]

Jumping on a trampoline is a classic example of conservation of energy, from potential into kinetic. It also shows Hooke's laws and the spring constant. Furthermore, it verifies and illustrates each of Newton's three laws of motion.

<u>Explanation</u>

When we jump on a trampoline, our body has kinetic energy that changes over time. Our kinetic energy is greatest, just before we hit the trampoline on the way down and when you leave the trampoline surface on the way up. Our kinetic energy is 0 when you reach the height of your jump and begin to descend and when are on the trampoline, about to propel upwards.

Potential energy changes along with kinetic energy. At any time, your total energy is equal to your potential energy plus your kinetic energy. As we go up, the kinetic energy converts into potential energy.

Hooke's law is another form of potential energy. Just as the trampoline is about to propel us up, your kinetic energy is 0 but your potential energy is maximized, even though we are at a minimum height. This is because our potential energy is related to the spring constant and Hooke's Law.

8 0
3 years ago
A transformer has two sets of coils, the primary with N1 = 160 turns and the secondary with N2 = 1400 turns. The input rms volta
vovikov84 [41]

To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.

So things our data are given by

N_1 = 160

N_2 = 1400

\Delta V_{1rms} = 62V

PART A) Since it is a system in equilibrium the relationship between the two transformers would be given by

\frac{N_1}{N_2} = \frac{\Delta V_{1rms}}{\Delta V_{2rms}}

So the voltage for transformer 2 would be given by,

\Delta V_{2rms} = \frac{N_2}{N_1} \Delta V_{1rms}

PART B) To express the number value we proceed to replace with the previously given values, that is to say

\Delta V_{2rms} = \frac{N_1}{N_2} \Delta V_{1rms}

\Delta V_{2rms} = \frac{1400}{160} 62V

\Delta V_{2rms} = 1446.66V

7 0
3 years ago
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