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2 years ago

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The two pictures below show an area where a major geologic process has occurred. Explain which process caused the formation, giv

e evidence you see in the picture that tells you what has happened, and whether this is a slow or rapid event.

1 answer:

kozerog [31]2 years ago

8 0

Ehshbdbrbwhajiaizjcj that h the e and fun

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A taxi starts from Monument Circle and travels 5 kilometers to the east for 5 minutes. Then it travels 10 kilometers to the sout

Morgarella [4.7K]

Answer: The taxi is moving with reference to A) Monument Circle. For each leg of the trip, the taxi's A) Average speed stays the same, but it's B) Average velocity changes.

Explanation: Brainliest Please!!!!

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3 years ago

You use 42 J of energy to move a 6.0 N object in 2 seconds. How far did you move it?

bogdanovich [222]

22 that's how much I moved

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3 years ago

Sam is playing football. She kicks the ball with an average force of 75 N.

damaskus [11]

Answer:

22.5J

Explanation:

Here the force is given. Also, the displacement is given as 30cm.

First we should check if all the values are in their standard form.

Here 30cm should be converted to metre by dividing it with 100.

Which would give us 0.3m

Now we use the equation W=force x displacement =75 x 0.3=22.5J

I hope this satisfies you. If u have any further questions please let me know.

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3 years ago

An unmarked police car traveling a constant 95 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the polic

Montano1993 [528]

Answer:

Speed of the speeder will be 28 m/sec

Explanation:

In first case police car is traveling with a speed of 90 km/hr

We can change 90 km/hr in m/sec

So $90km/hr=\frac{90\times 1000m}{3600sec}=25m/sec$

Car is traveling for 1 sec with a constant speed so distance traveled in 1 sec = 25×1 = 25 m

After that car is accelerating with $a=2m/sec^2$ for 7 sec

So distance traveled by car in these 7 sec

$S=ut+\frac{1}{2}at^2=25\times 7+\frac{1}{2}\times 2\times 7^2=224m$

So total distance traveled by police car = 224 m

This distance is also same for speeder

Now let speeder is moving with constant velocity v

so $224=\left ( 1+7 \right )v$

v = 28 m/sec

4 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