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serious [3.7K]
2 years ago
8

A barge is 10.0 m wide and 60.0 m long and has vertical sides. The bottom of the hull is 12.0 m below the water surface. What ar

e the weight of the barge and its cargo, if it is floating in freshwater?
Physics
1 answer:
SIZIF [17.4K]2 years ago
8 0

Answer:

7.06MN

Explanation:

length = 60m

Width = 10m

Height = 12 m

Let Fb = Force buoyant

Fb = pgV = mg

p(density) = rho

The density of water in this case = 1000

g = 9.8 m/s

Volume = lenght*width*height

= 60*10*12

= 7200m^3

So we have

(1000 * 9.8 * 7200) = mg

= 70560000

7.06 MN

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A man runs at a velocity of 4.5 m/s for 15.0 min. When going up an increasingly steep hill, he slows down at a constant rate of
madreJ [45]

The man ran  <u>4252.5 meters.</u>

Why?

To solve the problem, we need to divide the exercise into two movements, the first on while the was running at 4.5 m/s for 15 min, and then, while he was slowing down (going up because of the hill).

First movement: Running at 4.5 m/s for 15 min.

We need convert from minutes to seconds,

1min=60seconds\\\\15min*\frac{60seconds}{1min}=900seconds

Now, calculating the distance covered for the first movement, we have:

x_{1}=0+v_{1}*t_{1}\\\\x_{1}=4.5\frac{m}{s}*900s=4050m

So, we know that the man covered 4050m for the first movement, it will be our initial position for the second movement.

Second movement:  acceleration -0.05m/s^2 (because he's slowing down) for 90 seconds, at 4.5m/s.

x_{2}=x_{1}+v_{1}*t+\frac{1}{2}at^{2}\\\\x_{2}=4050m+4.5m\frac{m}{s}*90seconds-\frac{1}{2}*(0.05\frac{m}{s^{2}})*(90s)^{2}\\\\x_{2}=4050m+405m-(0.5*0.05\frac{m}{s^{2}}*8100s^{2})=4050m+405m-202.5m\\\\x_{2}=4252.5m

Hence, we have that he ran 4252.5 m.

Have a nice day!

4 0
2 years ago
A gold puck has a mass of 12 kg and a velocity of 5i – 4j m/s prior to a collision with a stationary blue puck whose mass is 18
Ugo [173]

Answer:

Explanation:D

6 0
2 years ago
Which of the following substances will show the smallest change in temperature when equal amounts of energy are absorbed?
SSSSS [86.1K]
It would be water because if you freeze it than you will still be able to see it and if you boil it than you will be able to see it disappear.
3 0
2 years ago
Read 2 more answers
Q 32.35: The isotope 235U decays by alpha emission with a half-life of 7.0 x 108 y. It also decays (rarely) by spontaneous fissi
Julli [10]

Answer:

rate of fission =5.89*10^3 1\Year

Explanation:

we know that

rate of fission is given asrate of fission = \frac{0.69}{(T_{1/2})_{fission}} *\frac{ Mass* Avogardo\ number}{Molar\ mass}

(T_{1/2})_{fission} = 3*10^{17} y

mass = 1.0 g

avogardo number = 6.02*10^23

molar mass of isotopes 235U =235

Putting all value to get rate of emission

rate of fission = \frac{0.69}{3*10^{17}} *\frac{1.0*6.02*10^{23}}{235}

rate of fission =5.89*10^3 1\Year

5 0
2 years ago
A cubical box, 5.00 cm on each side, is immersed in a fluid. The gauge pressure at the top surface of the box is 594 Pa and the
Zolol [24]

Answer:

The density of the fluid is 1100 kg/m³.

Explanation:

Given that,

Height = 5.00 cm

Pressure at top =594 Pa

Pressure at bottom = 1133 Pa

We need to calculate the change in pressure

Using formula of change in pressure

\Delta P=P_{b}-P_{t}

Where, P_{b} = Pressure at bottom

P_{t} = Pressure at top

put the value into the formula

\Delta P=1133-594

\Delta P=539\ Pa

Using formula of pressure for density

\Delta P = \rho g h

\rho =\dfrac{\Delta P}{gh}

Where, \rho = density

P = pressure

h = height

Put the value in to the formula

\rho =\dfrac{539}{5.00\times10^{-2}\times9.8}

\rho =1100\ kg/m^3

Hence, The density of the fluid is 1100 kg/m³.

4 0
2 years ago
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