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3 years ago

Find acceleration given the following:

Physics

1 answer:

Lorico [155]3 years ago

5 0

Answer:

Explanation:df=378

Di=0

Vi=0

T=6s

Formula

Df=Di+Vit+1/2(at)2

378=0+0+6+1/2×a×t2

378=6+1/2×a×36

378=6+18a

18a=378+6=384

a=385/18=21.33(cm/s)/s

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Why does the arrow labeled F norm point upward?

ohaa [14]

The answer is D I think

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1 year ago

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A boat has a mass of 7660 kg. Its engines generate a drive force of 4080 N due west, while the wind exerts a force of 680 N due

makvit [3.9K]

Answer:

0.29 m/s due west.

Explanation:

According to newton's second law,

Net force acting on an object = mass×acceleration

From the question,

F+F₁+F₂ = ma................ Equation 1

Where F = The force generated from the engine, F₁ = Force exerted by the wind, F₂ = Force exerted due to the water, m = mass of the boat, a = acceleration of the boat.

Given: F = 4080 N , F₁ = -680 N(east), F₂ = -1160 N(east). m = 7660 kg

substitute into equation 1

4080-680-1160 = 7660(a)

2240 = 7660a

Therefore,

a = 2440/7660

a = 0.29 m/s due west.

8 0

2 years ago

(4.3 x 10-2 )4.950 x 105)<br> O 21.285 x 103<br> O 2.13 x 104<br> O 21 x 103<br> O 2.1 x 104

Natalka [10]

Answer:

The answer is:

It's 21.285 × 103

7 0

2 years ago

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$