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3 years ago

10

I need help woth this question: List energy sources that are fossil fuels.

1 answer:

sertanlavr [38]3 years ago

3 0

Fossil fuels are formed when prehistoric plants died and were gradually buried under layers of rock. Some examples are:

Crude Oil
Coal
Natural Gas

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6. A light ray strikes a reflective plane surface at an angle of 560 with the surface.

Zolol [24]

Answer:

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Explanation:

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3 years ago

An 80-cm uniform 10-kg bar is resting on two scales, one at either end. A smaller 4-kg mass (m) is placed at a distance of d = 2

Varvara68 [4.7K]

Answer

given,

length of bar = 80 cm

mass of the bar = 10 kg

smaller mass = 4 kg

distance = 20 cm

$s_1 + s_2 = 10 + 4$

$s_1 + s_2 = 14\ kg$

taking moment about B

$s_1 \times 0.8 - 10 \times 0.4 - 4 \times 0.6 = 0$

$s_1 \times 0.8 = 6.4$

$s_1 = 8\ N$

$s_2 = 14 - s_1$

$s_2 = 14 - 8$

$s_2 = 6 N$

difference between two scale = 8 - 6

= 2 N

7 0

4 years ago

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.

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4 years ago

A 2.0-kilogram cantaloupe rolling north at 4.0 meters per second collides head on with a 1.0-kilogram orange rolling south at 8.

Murrr4er [49]

Answer:

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8 0

3 years ago

Read 2 more answers

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Aleks [24]

Answer:

cần cung cấp 70 độ vì nước sôi ở 100°C

Explanation:

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3 years ago