Balance the following equation Fe203+HCL fecl3+H20

andrew11 [14]

Answer:

Gametes are the cells used during sexual reproduction to produce a new individual organism.

Explanation:

The male gamete or sperm, is a smaller, mobile cell that meets up with the much larger and less mobile female gamete, egg or ova.

6 0

2 years ago

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A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and

alexira [117]

Answer:

A) $\omega_f=17.503\ rad.s^{-1}$

B) $t=55.6822\ s$

C) $\theta=1312\ rad$

Explanation:

Given:

mass of flywheel, $m=40\ kg$

diameter of flywheel, $d=0.72\ m$

rotational speed of flywheel, $N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}$

duration for which the power is off, $t_0=35\ s$

no. of revolutions made during the power is off, $\theta=180\times 2\pi=360\pi\ rad$

<u>Using equation of motion:</u>

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2$

$\alpha=-0.8463\ rad.s^{-2}$

Negative sign denotes deceleration.

A)

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$\omega_f=15\pi-0.8463\times 35$

$\omega_f=17.503\ rad.s^{-1}$ is the angular velocity of the flywheel when the power comes back.

B)

Here:

$\omega_f=0\ rad.s^{-1}$

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$0=15\pi-0.8463\times t$

$t=55.6822\ s$ is the time after which the flywheel stops.

C)

Using the equation of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2$

$\theta=1312\ rad$ revolutions are made before stopping.

3 0

3 years ago

A karate master wants to break a board by hitting the board swiftly with his hand. The master's hand has a mass of 0.30 kg, and

slava [35]

Answer:

f=-1380N

Explanation:

A karate master wants to break a board by hitting the board swiftly with his hand. The master's hand has a mass of 0.30 kg, and as it strikes the board, his hand has a velocity of 23.3 m/s. The master contacts the board for 0.0050 seconds

.the concluding part to the question should be

What is the impact force (impulse) on the board?

solution

from the Newton's second law of motion which states that

the rate of change in momentum is directly proportional to the force applied

f=m(v-u)/t

f=0.3(0-23.3)/0.005

f=-1380N

f=force impact

m=mass of the karates master's hand

t=time for the impact

v=0m/s final velocity

u=initial velocity

6 0

3 years ago

The area of land where water is drained downhill into a body of water is known as a drainage basin, or _______.

GenaCL600 [577]

Water sheds i hope this helps give me a brainiest answer

6 0

3 years ago

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A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceler

Pepsi [2]

Answer:

$331665750000\ m/s^2$

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = $1.989\times 10^{30}\ kg$

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2$

The gravitational acceleration at the surface of such a star is $331665750000\ m/s^2$

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s$

The velocity of the object would be 3257806.62409 m/s

8 0

3 years ago