Water is a fluid, all fluids

Two waiters are trying to get through a single door of a kitchen. One pushes on one side of a door 0.567 m from the hinge with a

NISA [10]

Answer:

$275.5 N$

Explanation:

$F_{1}$ = Force on one side of the door by first waiter = 257 N

$F_{2}$ = Force on other side of the door by second waiter

$r_{1}$ = distance of first force by first waiter from hinge = 0.567 m

$r_{2}$ = distance of second force by second waiter from hinge = 0.529 m

Since the door does not move. hence the door is in equilibrium

Using equilibrium of torque by force applied by each waiter

$r_{1} F_{1} = r_{2} F_{2} \\(0.567) (257) = (0.529) F_{2}\\F_{2} = 275.5 N$

7 0

3 years ago

What is the pressure exerted by a column of fresh water 10m high<br>(p=1000kg/m3 and g=10m/s)

alexira [117]

Height (h) = 10 m
Density (ρ) = 1000 Kg/m^3
Acceleration due to gravity (g) = 10 m/s^2
We know, pressure in a fluid = hρg
Therefore, the pressure exerted by a column of fresh water
= hρg
= (10 × 1000 × 10) Pa
= 100000 Pa

<u>Answer</u><u>:</u>

Hope you could understand.

If you have any query, feel free to ask.

8 0

3 years ago

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

3 years ago

Elaborate on the classification of aluminum. A) Aluminum is a noble gas and has a full valence electron shell, is chemically non

scZoUnD [109]

The correct option is (B) <span>Aluminum is a metal and is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.

Since Aluminium is in group 13, and all the elements in group 13 are either metals or metalloids(Boron). Hence we are left with option (B) and (D). Boron is the only metalloid in group 13 and aluminium is a metal(not a metalloid); therefore, we are left with only one option which is Option (B). And Aluminium is </span>shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.<span>
</span><span>
</span>

4 0

3 years ago

Read 2 more answers

A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0 m high cliff. At the instant the ball i

scoray [572]

Answer:

The distance traveled by the woman is 34.1m

Explanation:

Given

The initial height of the cliff

yo = 45m final, positition y = 0m bottom of the cliff

y = yo + ut -1/2gt²

u = 20.0m/s initial speed

g = 9.80m/s²

0 = 45.0 + 20×t –1/2×9.8×t²

0 = 45 +20t –4.9t²

Solving quadratically or by using a calculator,

t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s

So this is the total time it takes for the ball to reach the ground from the height it was thrown.

The distance traveled by the woman is

s = vt

Given the speed of the woman v = 6.00m/s

Therefore

s = 6.00×5.69 = 34.14m

Approximately 34.1m to 3 significant figures.

3 0

3 years ago