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3 years ago

What to do if i accidentally added to much water conditioner to my fish tank?

1 answer:

5 0

Put the fish in another small bowl of water. Empty the tank, rinse it, and fill it with new water and the right amount of conditioner. Let it get to the right temperature for the fish, then gently put the fish back in it. And FAST ! All of this is very stressful for the fish.

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What is the difference between a low tide and a high tide

Savatey [412]

A high tide means when the water has risen and is higher up(closer to high up land). Low tide is when it’s receded

6 0

2 years ago

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If a block of wood has a density of 0.6g/cm3 and a mass of 120g what is the volume

Luda [366]

Answer:

It would be 39 :))

Explanation:

6 0

3 years ago

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.

Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 1.5 ra/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

Acceleration while slowing down is -0.0047 rad/s²

$t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s$

Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0

2 years ago

Place the following structures in the appropriate category based on their location in a cell

Studentka2010 [4]

Answer:

Cell surface.

Cilium
Plasma membrane
Microvilus.

Nucleus

Cellular DNA
Nucleolus
Chromosome.

Cytoplasm

Mitochondria
Golgi apparatus
Cytoskeleton.

3 0

2 years ago

A 5-kg projectile is fired over level ground with a velocity of 200 m/s at an angle of 25°above the horizontal. Just before it h

Nutka1998 [239]

Answer:

the change in thermal energy of the projectile is 43.8 kJ

Explanation:

Given;

mass of the object, m = 5kg

initial velocity of the projectile, v₁ = 200 m/s

final velocity of the projectile, v₂ = 150 m/s

To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.

Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²

KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)

KE = ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ

Therefore, the change in thermal energy of the projectile is 43.8 kJ

8 0

2 years ago

Read 2 more answers