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3 years ago

When the thermal energy of a material decreases, _____.

Physics

2 answers:

Pani-rosa [81]3 years ago

8 0

The surrounding environment cool off is the answer

Shtirlitz [24]3 years ago

4 0

D. The surrounding environment cools off

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How much force should a mother exert to lift a 5.0-kg child

andrew11 [14]

Answer:

F = 49N

Explanation:

F = mg

F = 5kg × 9.8m/s^2

F = 49kg.m/s^2 = 49N

8 0

3 years ago

A sound wave has a frequency of 265 hz and a wavelength measured at 1.3

pochemuha

Answer:

V = f λ speed of wave in terms of frequency and wavelength

t = S / V time for wave to travel a distance S

t = 91.4 m / 344.5 m/s = .265 sec time to travel 91.4 m

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1 year ago

What is a Geographic test

Genrish500 [490]

<em>It's a test on Geography!
</em>

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2 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

Assume that two of the electrons at the negative terminal have attached themselves to a nearby neutral atom. There is now a nega

devlian [24]

Answer: its negative

Explanation: becase it is

5 0

3 years ago