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Studentka2010 [4]

3 years ago

9

A pendulum has 366 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o

f its swing?

1 answer:

Ilya [14]3 years ago

5 0

366J. Energy is conserved and is transformed from completely potential energy at the top to completely kinetic at the bottom.

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Cold air holds less water than warm air.<br> True<br> False

Vsevolod [243]

Answer:

The answer is for your question is :

Explanation:

True

4 0

2 years ago

A mass of 0.54 kg attached to a vertical spring stretches the spring 36 cm from its original equilibrium position. The accelerat

UNO [17]

<h2>Spring constant is 14.72 N/m</h2>

Explanation:

We have for a spring

Force = Spring constant x Elongation

F = kx

Here force is weight of mass

F = W = mg = 0.54 x 9.81 = 5.3 N

Elongation, x = 36 cm = 0.36 m

Substituting

F = kx

5.3 = k x 0.36

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Spring constant is 14.72 N/m

6 0

3 years ago

ABCD next 3 letters??????

jasenka [17]

Answer:EFG

Explanation:AYEEE

8 0

3 years ago

Read 2 more answers

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

The forces that hold different atoms or ions together are

denis23 [38]

Answer:

Chemical bonds

Explanation:

The chemical bonds hold the different type atoms or ions together.

The type of chemical bonds :

1. Ionic bond ;

Ions or atoms changes the electron and form ionic bond.Those atoms gains the electron gets negative charge and those atoms donate the electron gets positive charge.

2.Covalent bond :

In this type of bond atoms share the electrons and form covalent bonds.

3. Metallic bond :

This type of bond are present in the metals.In this atoms are used free electrons to form bonds.

Therefore answer is --

Chemical bonds

7 0

3 years ago