166x5.75= 954.5 meters in 5.75 seconds.
If you could please give me a already given speed I could estimate it. since there is no speed shown you wouldn't be able to estimate the speed of the moving train.
Answer:
Explanation:
Initial momentum is 1.5e6(3) = 4.5e6 kg•m/s
An impulse results in a change of momentum
The tug applied impulse is 12000(10) = 120000 N•s or 0.12e6 kg•m/s
The remaining momentum is 4.5e6 - 0.12e6 = 4.38e6 kg•m/s
The barge velocity is now 4.38e6 / 1.5e6 = 2.92 m/s
The tug applies 0.012e6 N•s of impulse each second.
The initial barge momentum will be zero in
t = 4.5e6 / 0.012e6 = 375 s or 6 minutes and 15 seconds
To stop the barge in one minute(60 s), the tug would have to apply
4.5e6 / 60 = 75000 N•s /s or 75 000 N
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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The statement which best describes how light waves travel in an uniform medium is in straight lines. The correct answer will be A.