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2 years ago

Determine the half-life of uranium-238 based on the graph.

2 answers:

7 0

From the curve on the graph we can see, when y=50 x=4.5
the unit of x-axis is in "billions of years", so the answer is 4.5 billions of years

Wittaler [7]2 years ago

3 0

Answer:

4.5 billion years

Explanation:

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A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,

n200080 [17]

Answer:

a. $t = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$ b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

$t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0} +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁ = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0

3 years ago

Which electrode in Thomson’s vacuum tube was positively charged

Taya2010 [7]

Anode

Explanation:

The anode in the gas discharge tube used by Thomson in his experiment was the positively charged electrode.

Using the gas discharge tube, Thomson made the remarkable discovery of cathode rays.

The rays moves from the negatively charged cathode to the positively charged anode. This indicated that the rays carry positive charges.

Some parts of the tube are:

Cathode - negatively charged electrode
Power source
Gas at low pressure
Outlet to vacuum pump

Learn more:

cathode brainly.com/question/12747250

#learnwithBrainly

8 0

2 years ago

What term best describes the geologic event taking place in the above illustration?

grigory [225]

u lying you made me get it wrong, for ya'll out there who want the real answer is sea floor spreading

8 0

2 years ago

Read 2 more answers

Question 10 of 34

labwork [276]

Julia walks from the park, which is six blocks east of her house, to the store, which is three blocks east of her house. Julia walks for 5 minutes. This walk's average speed will be 1.2 blocks per minute. Option B is correct.

<h3>What is displacement?</h3>

Displacement is defined as the shortest distance between the two points. Distance is the horizontal length covered by the body. While displacement is the shortest distance between the two points.

Displacement is a vector quantity .its unit is m.

The average velocity on this walk will be;

$\rm v_{avg}= \frac{d}{t} \\\\ \rm v_{avg}= \frac{6 \ block+ 3 \ block }{5 \ minute } \\\\ v_{avg}=1.4 \ block /min$

Hence option B is correct.

To learn more about displacement refer to the link; brainly.com/question/10919017

#SPJ1

3 0

1 year ago

If the student hit a drum in his bedroom how would the sound wave behave differently than if he hit it in a swimming pool? Pleas

zlopas [31]

The sound wave would behave differently in a swimming pool than in his bedroom because sound waves travel faster in more dense mediums; such as water. The wave will travel faster in water, and slower in air.

4 0

3 years ago