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nasty-shy [4]
3 years ago
5

In order to increase the amount of work completed it is necessary to

Physics
2 answers:
Dimas [21]3 years ago
5 0
Increase the input energy to the system
navik [9.2K]3 years ago
3 0
Do all you can in one big day that you have time off or work on one thing then work on the other at the same time

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Mineral salt compounds (calcium chloride, table salt)
soldier1979 [14.2K]

Answer:

Chloride and Salt

Explanation:

that make up table salt, a.k.a. sodium chloride (NaCl). ... As sodium chloride (NaCl) or calcium chloride (CaCl2) dissolve in water, ... the compound formed when a positive ion combines with a negative ion out of solution, but ... waters may have more if there is weathering or leaching from nearby mineral-rich soils and rocks.

3 0
3 years ago
4. According to Newton third law what is the action force? The swimmer pushes against the water the water pushes back on the swi
marshall27 [118]
A.The swimmer pushes the water
C. the walls force against the ball
3 0
3 years ago
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A crucial characteristic of turbines in tidal generators is that _____.
Mila [183]
<span>The blades should turn in two directions.</span>
4 0
3 years ago
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If the phase of the vibrating sources was changed so that they were vibrating completely out of phase, what effect would this ha
Over [174]

Answer:

There would be complete destructive interference.

Explanation:

This is because since the waves are completely out of phase, the phase difference is half wavelength, that is the phase angle is 180°. The vibrating sources are 180° out of phase with each other.

Since this is the case, the crest of the one source meets the trough of the other, this causes the resultant vibrational wave to cancel out, thus producing a destructive interference pattern.

Since the vibrating sources are completely out of phase, every point they meet is completely out of phase, so the resultant interference pattern would produce a complete destructive interference pattern of no wave.

4 0
3 years ago
The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

In this problem, we have

W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

Solving the formula for \Delta V, we find

\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

4 0
3 years ago
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