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Zigmanuir [339]
3 years ago
7

Drag the correct labels to the images. Each label can be used more than once.

Physics
1 answer:
coldgirl [10]3 years ago
4 0

Answer:

plato answer.

Explanation:

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What is the concentration of OH– ions at a pH = 6?
makkiz [27]
<h3><u>Answer;</u></h3>

1 × 10^-8 M

<h3><u>Explanation</u>;</h3>

pH is given by the -log[H+] while

pOH is given by the -log[OH-]

But;

pH + pOH = 14

Thus; if pH is 6, then pOH = 8

pOH = 8

-log[OH-] = 8

[OH-] = 10^-8 M

The concentration of OH- ions at a pH of 6 is 1 × 10^-8 M

8 0
3 years ago
What kind of units as force measured by
Readme [11.4K]

Answer:

Newtons

Explanation:

5 0
3 years ago
Read 2 more answers
Can you help me please?
egoroff_w [7]
<h3>Answer</h3>

option B)

19N

<h3>Explanation</h3>

If the object is at equilibrium, then the net force acting upon the object should be 0 N. Thus, if all the forces are added together, horizontal and vertical forces separately, then the resultant force (the vector sum) should be 0 Newton.

As we only need to find the magnitude of x-component of force F

so find all x component/horizontal forces acting on the object.

50cos(40) - 40cos(25) + 30cos(55) + x = 0

38.30 - 36.25 + 17.21 + x + = 0

19.26 + x = 0

x = - 19.26

x ≈ 19 (magnitude only)

7 0
4 years ago
Why is accelration negative when velocity is positive.​
Sindrei [870]

Answer:

An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration).

Explanation:

8 0
3 years ago
A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of
Nastasia [14]

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = v\sqrt{\frac{m}{k} }

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

\frac{m}{2K}V² = A'²

v\sqrt{\frac{m}{2k} } = A'

Substitute v\sqrt{\frac{m}{k} } for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

8 0
3 years ago
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