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Alona [7]
3 years ago
6

13. A steel block has a volume of 0.08 m³ and a density of 7,840 kg/m³. What is the force of gravity acting on the block (the we

ight) in water?
A. 6,700.56 N
B. 7,600.18 N
C. 6,150.64 N
D. 5,362.56 N

Student Answer: C
Answer: Incorrect

the correct answer is D. 5,362.56....cause i took a test
Physics
2 answers:
Travka [436]3 years ago
8 0
D.5326.52n is the correct answer on Penn Foster
Contact [7]3 years ago
5 0
A. 6,7000.56n is the correct choice according to siri lol
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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
An object completes one and half revolution of a circle of radius R calculate the displacement and distance
Papessa [141]

Answer:

Distance is path length covered by particle. When particle moves along half circle, it covers half the circumference therefore distance covered is (2×pi×r)/2 = pi× r. ... Hence displacement is equal to diameter or 2 times the radius of circle.

8 0
3 years ago
A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ
Elena L [17]

Answer:

The ocean is 6485.6m deep when measured from the vessel

Explanation:

v=1474m/s

t=8.88s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

velocity=\frac{distance}{time}\\ v=\frac{2d}{t} \\vt=2d\\d=\frac{vt}{2}

d=\frac{1474*8.8}{2}

d= 6485.6m

7 0
3 years ago
Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:
Natasha2012 [34]

The work done by \vec F along the given path <em>C</em> from <em>A</em> to <em>B</em> is given by the line integral,

\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r

I assume the path itself is a line segment, which can be parameterized by

\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath

with 0 ≤ <em>t</em> ≤ 1. Then the work performed by <em>F</em> along <em>C</em> is

\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}

7 0
3 years ago
10 points
enot [183]

Answer:

B

Explanation:

I just took the test

8 0
3 years ago
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