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3 years ago

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A 100-kg astronaut floating in space holds a 10-kg object. Then the object is tossed at 50 m/s. How fast does the astronaut go a

fter tossing the object?

1 answer:

zlopas [31]3 years ago

4 0

Answer:

velocity is 5 m/s

Explanation:

mass of astronaut (M)=100kg

mass of object (m)=10kg

Initially they are rest , so initial velocity (u)= 0 m/s

Final velocity of object ; $v_1=10m/s$

let final velocity of astronaut = $v_2$ (Unknown)

Now , By conservation of linear momentum

Initial momentum = Final momentum

$(m + M)\times u = (m\times v_1) + (M\times v_2)$

Now insert values

$0 = (10 kg \times 50 m/s ) + (100 kg \: v_2)$

Further simplifying

$v_2=-\frac{500 kg-m/s}{100kg}$

So,

$v_2= -5\:m/s$

So, velocity is 5 m/s (the negative sign is due to direction)

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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

Which goes into which category?

Pavel [41]

Evidence of Chemical Change:

A car rusting

Leaves changing in October

The glow of a light bulb

A white cloudy substance occurs after mixing two substances

Burning toasts

A candle being lit

Evidence of Physical Change:

Water boiling

Yellow and blue paints mixed together

Wax melting

Crushing a rock with a hammer

Hope this helps!

Have a great day!

6 0

3 years ago