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3 years ago

What is the concentration of H+ ions at a pH = 2

1 answer:

3 0

Its 1.0*10^-7M its considered a concentration because hydrogen ion is exactly equal to hydroxide ions produced by dissociation of water

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Define amoeba what are your plans

cupoosta [38]

Answer:

Amoeba (plural = amoebae) is a well known genus of unicellular organism, a protist. One of its most common species, the Amoeba Proteus, is about 0.2 to 0.3 mm large. The amoeba was first discovered by August Von Rosenhof in 1757.[1] It is a genus of protozoa that moves with false feet, called pseudopodia.

5 0

2 years ago

Read 2 more answers

A clock is designed that uses a mass on the end of a spring as a timing mechanism. If the oscillation time needed is exactly one

user100 [1]

Answer:

The value is $k = 51.34 \ N/ m$

Explanation:

From the question we are told that

The mass is $m = 1.3 \ kg$

The needed oscillation time is $T = 1 \ s$

Generally the spring constant is mathematically represented as

$k = \frac{4 \pi^2 * m }{ T^2}$

=> $k = \frac{4* 3.142^2 * 1.3 }{ 1^2}$

=> $k = 51.34 \ N/ m$

5 0

3 years ago

To find the acceleration of a glider moving down a sloping air track, you measure its velocities (V1 and V2) at two points and t

jeka94

Answer:

(a). The average acceleration is 0.08 m/s².

(b). The uncertainty in the acceleration is ±0.0135 m/s².

Explanation:

Given that,

Initial velocity $v_{1}=0.21\pm 0.05\ m/s$

Final velocity $v_{2}=0.85\pm 0.05\ m/s$

Time $t = 8.0\pm 0.1\ sec$

(a). We need to calculate the average acceleration

Using formula of acceleration

$a=\dfrac{v_{2}-v_{1}}{t}$

Put the value into the formula

$a=\dfrac{0.85-0.21}{8.0}$

$a=0.08\ m/s^2$

(b). We need to calculate the uncertainty in the velocity

Using formula of the uncertainty

$\dfrac{\Delta v}{v}=\dfrac{\Delta v}{v_{2}-v_{1}}$

Put the value into formula

$\dfrac{\Delta v}{v}=\dfrac{0.05+0.05}{0.85-0.21}$

$\dfrac{\Delta v}{v}=0.15625$

We need to calculate the uncertainty in the time

Using formula for time

$\dfrac{\Delta t}{t}=\dfrac{0.1}{8.0}$

$\dfrac{\Delta t}{t}=0.0125$

We need to calculate the uncertainty in the acceleration

Using formula for acceleration

$\dfrac{\Delta a}{a}=\dfrac{\Delta v}{v}+]\dfrac{\Delta t}{t}$

Put the value into the formula

$\dfrac{\Delta a}{a}=0.15625+0.0125$

$\dfrac{\Delta a}{a}=0.16875$

$\Delta a=0.16875\times0.08$

$\Delta a=\pm0.0135$

Hence, (a). The average acceleration is 0.08 m/s².

(b). The uncertainty in the acceleration is ±0.0135 m/s².

3 0

3 years ago

A fuzzy bunny sees a butterfly and chases it right off of a 20 m high cliff at 7m/s

bogdanovich [222]

A) 2.02 s

To find the time it takes for the bunny to reach the rocks below the cliff, we just need to analyze its vertical motion. This motion is a free fall motion, which is a uniformly accelerated motion. So we can use the suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where

s is the vertical displacement

u is the initial vertical velocity

a is the acceleration

t is the time

For the bunny here, choosing downward as positive direction,

$u_y = 0$ (initial vertical velocity is zero)

s = 20 m

$a=g=9.8 m/s^2$ (acceleration of gravity)

And solving for t, we find the time of flight:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s$

B) 14.1 m

For this part, we need to consider the horizontal motion of the bunny.

The horizontal motion of the bunny is a uniform motion with constant velocity, which is the initial velocity of the bunny:

$v_x = 7 m/s$

Therefore the distance covered after time t is given by

$d=v_x t$

And substituting the time at which the bunny hits the ground,

t = 2.02 s

We find how far the bunny went from the cliff:

$d=(7)(2.02)=14.1 m$

C) 21.0 m/s at $70.5^{\circ}$ below the horizontal

The horizontal component of the velocity is constant during the entire motion, so it will still be the same when the bunny hits the ground:

$v_x = 7 m/s$

Instead, the vertical velocity is given by

$v_y = u_y +at$

And substituting t = 2.02 s, we find the vertical velocity at the moment of impact:

$v_y = 0+(9.8)(2.02)=19.8 m/s$

So, the magnitude of the final velocity is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{7^2+(19.8)^2}=21.0 m/s$

And the angle is given by

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{19.8}{7})=70.5^{\circ}$

below the horizontal

5 0

3 years ago

A bird flies from the South Pole to the North Pole. Part of the journey is 1000 miles that takes 2 weeks. What is the bird’s vel

defon

Two weeks is
$2 weeks \times \frac{7 days}{1 week}\times \frac{24 hours}{1day}=336hours$
Then the speed is
$\frac{1000miles}{336hours}=2.97 \frac{miles}{hour}$

3 0

3 years ago