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Dmitry_Shevchenko [17]
3 years ago
15

A block of aluminium of density 7900 kg/m3 has a volume of 250 cm. Find themass of the block.​

Physics
1 answer:
pashok25 [27]3 years ago
8 0

Answer:

mass of block=2.7 gm

Explanation:

concept: Density=mass/volume

given:ρ=2700 kg/m^3 and v=250 cm^3 (in cm^3 not in m^3)

=> v= convert cm^3 to m^3

there fore= 1 cm^3=1 cm*1 cm*1 cm

i.e 1 cm^3=1/100*1/00*1/100 m^3 => 1 cm^3=1/1000000 m^3

ρ=m/v

=>mass=ρ*volume

=>mass=2700*1/10^-6

=>mass=2.7*10^-3 kg =>2.7 gm

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Two dogs are pulling a disk, but the disk is not moving. The disk does not move because the forces are (select one: unbalanced,
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In a very large closed tank, the absolute pressure of the air above the water is 6.46 x 105 Pa. The water leaves the bottom of t
a_sh-v [17]

Answer:

a) 35.94 ms⁻²

b) 65.85 m

Explanation:

Take down the data:

ρ = 1000kg/m3

a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot,  at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:

Ptot = Pgas + Pwater

However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:

Ptot  = Pgas

        = 6.46 × 10⁵ Pa

The change in pressure is given by the continuity equation:

ΔP = 1/2ρv²

where v is the velocity of the water as it exits the tank.

Calculating:

6.46 × 10⁵  =1/2 ×1000×v²

solving for v, we get v = 35.94 ms⁻²

b) The Bernoulli's equation will be applicable here.

The water is coming out with the same pressure, therefore, the equation will be:

ΔP = ρgh

6.46 × 10⁵  = 1000 x 9.81 x h

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7 0
3 years ago
Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box o
Lina20 [59]

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

\Sigma F = F - F' = M\cdot a

Box with mass 2M

\Sigma F = F' - F'' = 2\cdot M \cdot a

Box with mass 3M

\Sigma F = F'' = 3\cdot M \cdot a

On the third equation, acceleration can be modelled in terms of F'':

a = \frac{F''}{3\cdot M}

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

F' = 2\cdot M \cdot a + F''

F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''

F' = \frac{5}{3}\cdot F''

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)

F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''

F = 2\cdot F''

F'' = \frac{1}{2}\cdot F

Afterwards, F' as function of the external force can be obtained by direct substitution:

F' = \frac{5}{6}\cdot F

The net forces of each block are now calculated:

Box with mass M

M\cdot a = F - \frac{5}{6}\cdot F

M\cdot a = \frac{1}{6}\cdot F

Box with mass 2M

2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F

2\cdot M \cdot a = \frac{1}{3}\cdot F

Box with mass 3M

3\cdot M \cdot a = \frac{1}{2}\cdot F

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

8 0
4 years ago
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