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2 years ago

A fish tank is a cube of size L × L × L, where L = 1 m, filled with water. Find

Physics

1 answer:

Murljashka [212]2 years ago

3 0

At the bottom of the tank :

P = ρgH

P = (1000 kg/m³)(10 m/s²)(1 m)

P = 10000 N/m²

F = P • A

F = (10000 N/m²)(1 m²)

F = 10000 N

At the side of the tank :

Pav = ½ρgH

Pav = ½(1000 kg/m³)(10 m/s²)(1 m)

Pav = 5000 N/m²

F = P • A

F = (5000 N/m²)(1 m²)

F = 5000 N

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Suppose scientists discover two new moons.The average surface temperature of one of the moons is –180°C, but the temperature can

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3 years ago

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20 quantities and classified them in vector and scaler quantities

ss7ja [257]

Answer:

A scalar quantity is defined as the physical quantity that has only magnitude, for example, mass and electric charge. On the other hand, a vector quantity is defined as the physical quantity that has both magnitude as well as direction like force and weight.

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2 years ago

The speed at which a light aircraft can take off is 120 km/h. (A) What is the minimum constant acceleration required for the pla

kondor19780726 [428]

Answer:

A) a = 2.31[m/s^2]; B) t = 14.4 [s]

Explanation:

We can solve this problem using the kinematic equations, but firts we must identify the data:

Vf= final velocity = take off velocity = 120[km/h]

Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

$v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\ a=2.31[m/s^2]\\$

To find the time we must use another kinematic equation.

$v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]$

7 0

3 years ago

A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand

enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

$v=gt$

And the distance traveled downwards is:

$\displaystyle y=\frac{gt^2}{2}$

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Replacing into the first equation:

$\displaystyle v=g\sqrt{\frac{2y}{g}}$

Rationalizing:

$\displaystyle v=\sqrt{2gy}$

Let's call v1 the final speed of the package dropped from a height H. Thus:

$\displaystyle v_1=\sqrt{2gH}$

Let v2 be the final speed of the package dropped from a height 4H. Thus:

$\displaystyle v_2=\sqrt{2g(4H)}$

Taking out the square root of 4:

$\displaystyle v_2=2\sqrt{2gH}$

Dividing v2/v1 we can compare the final speeds:

$\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}$

Simplifying:

$\displaystyle v_2/v_1=2$

The final speed of the second package is twice as much as the final speed of the first package.

5 0

3 years ago

Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their

geniusboy [140]

Answer:

a. 11 m/s at 76° with respect to the original direction of the lighter car.

Explanation:

In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

$m_{1} v_{x1}+m_{2}v_{x2}=(m_{1}+m_{2})v_{fx}\\m_{1} v_{y1}+m_{2}v_{y2} =(m_{1}+m_{2})v_{fy}$

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

$1*13 + 4*0 = (1+4) v_{fx}\\v_{fx}=\frac{13}{5} =2.6\\1*0 + 4*13 = (1+4) v_{fy}\\v_{fy}=\frac{13*4}{5} =10.4\\$

The magnitude of the final velocity of the wreck can be found as:

$v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72$

The final velocity has an intensity of roughly 11 m/s

As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

$\theta = cos^{-1}(\frac{v_{fx} }{v_{f}} )\\\theta = cos^{-1}(\frac{2.6}{10.7} )\\\theta = 76^{o}$

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.

5 0

3 years ago