The answer is: " 208 g " .
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Explanation:
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The formula/ equation for density is:
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D = m / V ; That is, "mass divided by volume" ;
Density is expressed as:
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"mass per unit volume"; in which the "mass" is expressed in units of "g" ("grams") ; and the "unit volume" is expressed in units of:
"cm³ " or "mL";
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{Note the exact equivalent: 1 cm³ = 1 mL }.
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→ The formula is: " D = m / V " ;
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in which:
"D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given);
"m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;
"V" refers to the "volume", in units of "cm³ " ;
which is: "23.4 cm³ " (given);
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We want to find the mass, "m" ; so we take the original equation/formula for the density:
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D = m / V ;
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And we rearrange; to isolate "m" (mass) on ONE side of the equation; and then we plug in our known/given values;
to solve for "m" (mass); in units of "g" (grams) ;
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Multiply each side of the equation by "V" ;
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V * { D = m / V } ; to get:
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V * D = m ; ↔ m = V * D ;
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Now, we plug in the given values for "V" (volume) and "D" (density) ; to solve for the mass, "m" ;
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m = V * D ;
m = (23.4 cm³) * (8.9 g / 1 cm³) = (23.4 * 8.9) g = 208.26 g ;
→ Round to "208 g" (3 significant figures);
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The answer is: " 208 g " .
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Answer:
The buoy will move upwards and downwards and well as left and right
Explanation:
As the wind is storing so had the capability to move the buoy in all directions as it is a light object despite being anchored down
PLEASE GIVE BRAINLIEST AND THANKS :-)
Answer:
t = 0.354 hours
Explanation:
given,
coefficient of rolling friction μr=0.002
mass of locomotive = 180,000 Kg
rolling speed = 25 m/s
The force of friction = μ mg
= (.002) x (180000) x (9.8)
= 3528 N
F = m a
now,
m a = 3528 N
180000 x a = 3528
a = 0.0196 m/s²
Then apply
v = u + at
0 = 25 - 0.0196 x t
t = 1275.51 sec
t = 1275.61/3600 hours
t = 0.354 hours
time taken by the locomotive to stop = t = 0.354 hours
The sprinter experiences muscle cramps and a stray dog on the track from A to B, followed immediately by acute nausea and a gaggle of news photographers from B to C.
If none of these selections appears on the list of choices, then perhaps we can do better if you'll permit us a glance at the picture that accompanies the question, even if you still don't let us see the choices.
