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3 years ago

10

A high jumper jumps 2.09m. If the jumper has a mass of 72kg, what is his gravitational potential energy at the highest point of

the jump?

1 answer:

Furkat [3]3 years ago

4 0

Si Ko alam hehehehe sorrry men

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What is the molarity of 5.60 mol of sodium carbonate in 1500 ml of solution?

FrozenT [24]

Answer:

3.74 M

Explanation:

We know that molarity is moles divided by liters. The first thing to do here is convert your 1500 mL of solution to L. There's 1,000 mL in 1 L, so you need to divide 1500 by 1000:

1500 ÷ 1000 = 1.50

Now you can plug your values into the equation for molarity:

5.60 mol ÷ 1.50 L = 3.74 M

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3 years ago

PO3−4 what's the formula name?

masha68 [24]

Answer:

Phosphate ion

Explanation:

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3 years ago

Atoms full valence electron shells are chemically stable. Most noble gases have eight valence electrons. So, one bottle for chem

kap26 [50]

Gain or lose.
The exchange of electrons in chemical bonding seeks to fulfill the octet rule. There are some exceptions, such as with hydrogen and helium, whose valence shells have a capacity of two electrons.

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3 years ago

Read 2 more answers

What are prevailing winds?

Tems11 [23]

Answer:

D) winds that blow in the same direction at a consistent speed

Explanation:

i took the quiz got it right so i know the answer please trust me i know this is right i promise with all my heart

7 0

3 years ago

The half life for the decay of carbon-14 is 5.73 x 10 years. Suppose the activity due to the radioactive decay of the carbon-14

Elena-2011 [213]

Answer:

Age of the atifact is $4.2\times 10^{2}$ years

Explanation:

For first -order radioactive decay- $A_{t}=A_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$
$A_{t}$ represents activity of radioactive nuclide after t time, $A_{0}$ represents initial activity of radioactive nuclide and $t_{\frac{1}{2}}$ represents half-life
Here, $A_{t}=19Bq$ , $A_{0}=20Bq$ and $t_{\frac{1}{2}}=5.73\times 10^{3}years$

Plug-in all the given values in the above equation-

$19=20\times (\frac{1}{2})^{\frac{t}{5.73\times 10^{3}}}$

or, $t=4.2\times 10^{2}$

So, age of the atifact is $4.2\times 10^{2}$ years

6 0

3 years ago