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Sliva [168]
3 years ago
11

A barge floating in fresh water (rho = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A = 550 m2 and heigh

t H = 2.0 m. When empty the bottom of the barge is located H0 = 0.45 m below the surface of the water. When fully loaded with coal the bottom of the barge is located H1 = 1.4 m below the surface.a) Write an equation for the buoyant force on the empty barge in terms of the known data.b) Determine the mass of the barge in kilograms.c) Find the mass of the coal in terms of the given data.d) Find the mass of the coal in kilograms.e) How far would the barge be submerged (in meters) if mc,2 = 250000 kg of coal had been placed in the empty barge?
Physics
1 answer:
lapo4ka [179]3 years ago
8 0

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At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi
Amanda [17]

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

3 0
3 years ago
A car speeds up from 12.0 m/s to 16.0 m/s in 8.0 s. the acceleration of the car is?
maria [59]

Vi=12m/s Vf=16m/s t=8s a=? a=Vf-Vi/t=16-12/8=4/8=1/2 a=0.5m/s^2

6 0
3 years ago
Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative
SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

Explanation:

Given there are three blocks of masses M_{a}, M_{b} and M_{c} (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *a    ................(equation 1)

Also it is given that M_{a} does not move with respect to M_{c}, which gives tension T  is exerted on pulley  by M_{a} only, Hence tension T is

T = M_{a} *a    ..........(equation 2)

There is also also tension exerted by M_{b}. There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = M_{b} \sqrt{a^{2} +g^{2} }   ................(equation 3)

From equation 2 and 3, we get

M_{a} *a  = M_{b} \sqrt{a^{2} +g^{2} }  

Squaring both sides we get

M_{a} ^{2} *a^{2} = M_{b} ^{2} * (a^{2}+g^{2})

M_{a} ^{2} *a^{2} = (M_{b} ^{2} * a^{2})+ (M_{b} ^{2} *g^{2})

(M_{a} ^{2}  -  M_{b} ^{2}) * a^{2} = M_{b} ^{2} *g^{2}

a^{2} = M_{b} ^{2} *g^{2}/(M_{a} ^{2}  -  M_{b} ^{2})

Taking square root on both sides, we get acceleration a

a = M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}})

Hence substituting the value of a in equation 1, we get

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

3 0
3 years ago
a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance tra
LenaWriter [7]

Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

a=\dfrac{\Delta v}{\Delta t}=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.47\,\mathrm s}=11.2\,\dfrac{\mathrm m}{\mathrm s^2}

Now, we have that

{v_f}^2-{v_0}^2=2a\Delta x

so we end up with a distance traveled of

\left(46.1\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(11.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x

\implies\Delta x=79.6\,\mathrm m

6 0
3 years ago
2 a A pile of 60 sheets of paper is 6 mm high. Calculate the average thickness of a sheet of the paper.
fiasKO [112]

The average thickness of a sheet of the paper is 0.1 mm.

The number of ice blocks that can be stored in the freezer is 80 blocks of ice.

<h3>Average thickness of a sheet of the paper</h3>

The average thickness of a sheet of the paper is calculated as follows;

average thickness = 6 mm/60 sheets = 0.1 mm /sheet

Thus, the average thickness of a sheet of the paper is 0.1 mm.

<h3>Volume of each block of ice</h3>

Volume = 10 cm x 10 cm x 4 cm

Volume = 400 cm³

<h3>Volume of the freezer</h3>

Volume = 40 cm x 40 cm x 20 cm = 32,000 cm³

<h3>Number of ice blocks that can be stored</h3>

n = 32,000 cm³/400 cm³

n = 80 blocks of ice

Thus, the number of ice blocks that can be stored in the freezer is 80 blocks of ice.

Learn more about average thickness here: brainly.com/question/24268651

#SPJ1

6 0
1 year ago
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