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Serjik [45]
3 years ago
10

What is speed?

Physics
1 answer:
PSYCHO15rus [73]3 years ago
8 0
Speed is D. Distance per time.
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A 40kg rock falls off a cliff that is 50 meters high. How fast is the speed of the rock when it hits the ground.
olga2289 [7]

Answer:

huh?

Explanation:

6 0
3 years ago
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A(n) ______ particle consists of two protons and neutrons.
mamaluj [8]
The answer would be an alpha particle because the beta particle doesn't carry the amount of protons and neutrons.
7 0
3 years ago
A cardinal (Richmondena cardinalis) of mass 4.00×10−2 kg and a baseball of mass 0.146 kg have the same kinetic energy. What is t
sp2606 [1]

Answer:

0.5232

Explanation:

the cardinal and the baseball has the same kinetic energy

\frac{1}{2} m_{c} v_{c} ^{2}  = \frac{1}{2} m_{b} v_{b} ^{2}

m_{c}v_{c}^{2} = m_{b}v_{b}^{2}

[4.0X 10^{-2} ]v_{c} ^{2} = [0.146]v_{b} ^{2}

\frac{v_{c} }{v_{b} } = \sqrt{\frac{0.146}{4.0 X 10^{-2} } }

\frac{v_{c} }{v_{b} } = 1.910

Ratio of momentum

\frac{p_{c}}{p_{b}}  = \frac{m_{c}v_{c}}{m_{b}v_{b}} = \frac{ (4.0 X 10^{-2} )(1.910v_{b} )}{0.146v_{b} }

\frac{p_{c}}{p_{b}} = \frac{4.0 X  10 ^{-2} X 1.910 }{0.146}  = 0.5232

6 0
4 years ago
What is the ph of a buffer solution that is 0.172 m in hclo and 0.131 m in naclo? hint: the ka of hclo is 3.8 x 10-8.
nata0808 [166]

The pH of the buffer solution is 7.30 for 0.172 m in Hypochlorous acid and 0.131 m in Sodium hypochlorite.

<h3>What is a buffer solution?</h3>

A weak acid and its conjugate base, or a weak base and its conjugate acid, are mixed together to form a solution called a buffer solution, which is based on water as the solvent. They do not change in pH when diluted or when modest amounts of acid or alkali are added to them. A relatively little amount of a strong acid or strong base has little effect on the pH of buffer solutions. As a result, they are employed to maintain a steady pH.

According to the question:

Ka = 3.8×10⁻⁸

pKa = - log (Ka)

= - log(3.8×10⁻⁸)

= 7.42

pH = pKa + log {[conjugate base]/[acid]}

= 7.42+ log {0.131/0.172}

= 7.302

To know more about buffer solutions, visit:

brainly.com/question/24262133

#SPJ4

5 0
2 years ago
A stone is tied to a string and whirled at constant speed in a horizontal circle. The speed is then doubled without changing the
Tanya [424]

Answer:

Afterward the magnitude of the acceleration of the stone is four times as great

Explanation:

A stone tied to a string and whirled at a constant speed in a horizontal circle will have a centripetal acceleration given by

a = \frac{v^{2} }{r}

Where a is the centripetal acceleration

v is the speed

and r is the radius of the circle

Here, the radius of the circle is the length of the string.

Now, from the question

The speed is then doubled without changing the length of the string,

Let the new speed be v_{2}, that is

v_{2} = 2v

and without changing the length of the string means radius r is constant.

To determine the magnitude of the acceleration of the stone afterwards,

Let the new acceleration be a_{2}.

Then we can write that

a_{2} = \frac{v_{2}^{2}  }{r}

From

a = \frac{v^{2} }{r}

v = \sqrt{ar}

Recall that

v_{2} = 2v

∴ v_{2} = 2\sqrt{ar}

Now, we will put the value of v_{2} into

a_{2} = \frac{v_{2}^{2}  }{r}

Then,

a_{2} = \frac{(2\sqrt{ar}) ^{2}  }{r}

a_{2} = \frac{4ar }{r}

a_{2} = 4a

The new magnitude of the acceleration of the stone is four times the initial acceleration.

Hence,

Afterward the magnitude of the acceleration of the stone is four times as great

3 0
3 years ago
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