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VLD [36.1K]
3 years ago
11

The Back Saver Sit and Reach from the Fitnessgram measure this fitness component.

Physics
1 answer:
MAVERICK [17]3 years ago
7 0
The correc answer is B.
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What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential di
yan [13]
What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×10^4v ?
3 0
2 years ago
How long will it take for a radioactive isotope with a decay constant of 0.15 (which means a half life of 4.6 days ) to decay to
jek_recluse [69]
Radioactive decay is given by:
N = No x e^(-λt)
We know that N/No has to be 0.05
λ = 0.15
0.05 = e^(-0.15t)
t = ln(0.05)/(-0.15)
t = 19.97 days
6 0
3 years ago
Argon makes up 0.93% by volume of air. Calculate its solubility (in mol/L) in water at 20°C and 1.0 atm. The Henry's law constan
vladimir2022 [97]

Answer:

S= 1.40x10⁻⁵mol/L

Explanation:

The Henry's Law is given by the next expression:

S = k_{H} \cdot p (1)

<em>where S: is the solubility or concentration of Ar in water, k_{H}: is Henry's law constant and p: is the pressure of the Ar </em>

<u>Since the argon is 0.93%, we need to multiply the equation (1) by this percent:</u>

S = 1.5 \cdot 10^{-3} \frac{mol}{L\cdot atm} \cdot 1.0atm \cdot \frac{0.93}{100} = 1.40 \cdot 10^{-5} \frac{mol}{L}

Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.

Have a nice day!          

8 0
3 years ago
HELP ASAP ILL GIVE BRAINLIST
OlgaM077 [116]

Answer:

Each of the joints represents a degree of freedom in the manipulator system and allows translation and rotary motion :) Hope this helps

4 0
2 years ago
Read 2 more answers
In the diagram, q1=+6.25 * 10^ -8 C. What is the potential difference when you go from point A to point B? Include the correct s
Nimfa-mama [501]

Answer:

Moving a unit "positive" test charge from A to B will result in a reduction in potential

V = K Q / R      potential at a point

V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q

V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8

V2 - V1 = -4.17 * 562.5 J/C

V = - 2346 Volts

7 0
2 years ago
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