What evidence of a chemical change is observed when rust forms on iron

A group of rescue workers is rescuing a dog from a condemned building. Due to demolition, the building has no stairs leading to

zaharov [31]

Answer:

Minimize the elevation or jump distance

Explanation:

The only action that minimizes, the impact force is to reduce as much as possible the height of the jump, the dog, ie the height from the jump point of the building. Since at the time of the jump its speed will increase every second at the rate of 9 [m/ s], that is this low effect of the gravitational acceleration of 9 [m/s^2]

$v_{f}=\sqrt{2*g*h}$

where:

vf = final velocity [m/s]

g = gravity [m/s^2]

h = elevation [m]

As we can see while there is higher height, at a higher speed will impact the ground.

6 0

4 years ago

What force (in N) must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2400 kg car (a large car

makvit [3.9K]

Answer:

Fm= 91.88 N

Explanation:

Pascal principle

The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.

The pressure is definited like this:

P=F/A

Where:

P: Pressure in pascals (Pa)

F: Force acting in the area (N)

A : Area where the force acts (m²)

Pascal principle

Pm=Ps

Fm/ Am= Fs/ As Formula (1)

Where :

Pm : Pressure on the master piston

Ps : Pressure on the slave piston

Fm : Force on the master piston (N)

Fs: Force on the slave piston ((N)

Am: master piston area (m²)

As: slave piston area (m²)

Area Formula (A)

A= π*R²

R : piston radius

Calculation of the weight of the car (W)

W= m*g= 2400 kg*9.8m/s²= 23520 N

W = Fs

Data

Fs = 23520 N

Dm = 1.5 cm

Ds = 24 cm

Rm = 0.75 cm

Rs = 12 cm

Am = π*Rm² = π*(0.75)²

As = π*Rs² = π*(12)²

Force exerted on the master cylinder

We replace data in the formula (1)

$\frac{F_{m} }{A_{m} } = \frac{F_{s} }{A_{s} }$

$F_{m} = \frac{F_{s}*A_{m} }{A_{s}}$

$F_{m} = \frac{(23520 N)*(\pi *(0.75)^{2})(cm^{2})}{(\pi *(12)^{2})(cm^{2})}$

$F_{m} = (23520 N)*\frac{(0.75)^{2} }{(12)^{2} }$

Fm= 91.88 N

8 0

3 years ago

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

(c) $F_{se} = 3.521\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

(c) The force exerted by the Sun on the Earth is given by eqn (1):

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

3 years ago

When one person was talking in a small room, the sound intensity level was 60 dB everywhere within the room. Then, there were 14

dolphi86 [110]

Answer:

E= 71dB

Explanation:

See attached file for step by step calculation

3 0

3 years ago

A newton meter is a measuee of work also known as the what

Sladkaya [172]

The "newton-meter" that is a unit of work or energy
is identical to the Joule.

3 0

4 years ago