During an earthquake, where is the greatest motion felt on the surface?

PLS ANSWER FAST TIMED TEST WILL GIVE BRAINLY!!!!!!!!!

lutik1710 [3]

The answer would be true

7 0

3 years ago

Read 2 more answers

What happens to balloon filled with air when it goes very high attitude from surface of earth why

vlada-n [284]

Answer:

The balloon will continue to expand and eventually burst.

Explanation:

Simply, the reason for this is because the density of the atmosphere decreases gradually as you increase in altitude closer to space. This means that the air on the outside of the balloon can't provide enough pressure over the surface of the balloon in order to counteract the gas on the inside of the balloon from expanding.

3 0

3 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.

4 0

3 years ago

Urgent need help 100 points

Sati [7]

Answer: The velocity with which the sand throw is 24.2 m/s.

Explanation:

Explanation:

acceleration due to gravity, a = 3.9 m/s2

height, h = 75 m

final velocity, v = 0

Let the initial velocity at the time of throw is u.

Use third equation of motion

The velocity with which the sand throw is 24.2 m/s.

7 0

2 years ago

A snail travels 300 cm in 4 minutes.calculate speed of snail in m/s

LekaFEV [45]

Answer:

$\frac{0.0125m}{s}$

Explanation:

In order to solve this question we need to know that $speed = \frac{distance}{time}$ . Then we need to convert 4 minutes into seconds and cm into m. We can do that by multiplying the number of minutes by 60 (because there is 60 seconds in one minute) and dividing the number of cm by 100 (because there is 100 cm in one m). So.......

4min = 4 x 60s = 240s

300cm = 300/100 m = 3m

Now we know that distance = 300m, and that the time = 4min = 240s ⇒

⇒ $speed=\frac{distance}{time} = \frac{3m}{240s} = \frac{0.0125m}{s}$

5 0

3 years ago