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4 years ago

6

A 7.00 kg bowling ball is held 2.00 m above the ground. Using g = 9.80 m/s2, how much energy does the bowling ball have due to i

ts position?

2 answers:

Naily [24]4 years ago

5 0

Answer: 137.2 J

Explanation:

The gravitational potential energy of the bowling ball is given by:

$U=mgh$

where

m = 7.00 kg is the mass of the ball

g = 9.80 m/s^2 is the acceleration due to gravity

h = 2.00 m is the height above the ground

Substituting, we find

$U=(7.00 kg)(9.8 m/s^2)(2.00 m)=137.2 J$

deff fn [24]4 years ago

4 0

mgh =7*2*9.8=137.2. It should be the answer.

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Answer:

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Explanation:

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3 years ago

The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21

frutty [35]

Answer:

a) The velocity of the projectile at 2 seconds after launch is 1.9 meters per second. The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) The projectile reaches maximum height 2.192 seconds after launch.

c) The maximum height of the projectile is 26.584 meters above ground.

d) The projectile will hit the ground at 4.523 seconds after launch.

e) The velocity of the projectile right before hitting the ground in -22.871 meters per second.

Explanation:

Complete statement of problem is: <em>The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21.5 meters per second is </em> $h(t) = 3+21.5\cdot t-4.9\cdot t^{2}$ <em>after t seconds. (Round your answers to two decimal places.) </em><em>(a)</em><em> Find the velocity after 2 seconds and after 4 seconds, </em><em>(b)</em><em> When does the projectile reach its maximum height? </em><em>(c)</em><em> What is the maximum height? </em><em>(d)</em><em> When does it hit the ground? </em><em>(e)</em><em> With what velocity does it hits the ground?</em>

a) From Physics and Differential Calculus we remember that velocity is the first derivative of height. Hence, we need to differentiate the height function in time:

$v(t) = 21.5-9.8\cdot t$ (Eq. 1)

Where $v(t)$ is the velocity function, measured in meters per second.

Now we evaluate this function at given times:

t = 2 s.

$v(2) = 21.5-9.8\cdot (2)$

$v(2) = 1.9\,\frac{m}{s}$

The velocity of the projectile at 2 seconds after launch is 1.9 meters per second.

t = 4 s.

$v(4) = 21.5-9.8\cdot (4)$

$v(4) = -17.7\,\frac{m}{s}$

The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) Maximum height is reached when velocity of projectile is zero. We equalize velocity to zero and solve the expression for $t$ :

$21.5-9.81\cdot t = 0$

$t = 2.192\,s$

The projectile reaches maximum height 2.192 seconds after launch.

c) Maximum height is calculated by evaluating height function at the time found in b). That is:

$h(2.192) = 3+21.5\cdot (2.192)-4.9\cdot (2.192)^{2}$

$h (2.192) = 26.584\,m$

The maximum height of the projectile is 26.584 meters above ground.

d) In this case, we need to equalize the height function to zero and solve for $t$ . That is:

$3+21.5\cdot t-4.9\cdot t^{2} = 0$

Roots are found by means of Quadratic Formula:

$t_{1}\approx 4.523\,s$ and $t_{2}\approx -0.135\,s$

Only the first root offers a physically reasonable solution. Therefore, the projectile will hit the ground at 4.523 seconds after launch.

e) This can be found by evaluating velocity function at the time found in d):

$v(4.523) = 21.5-9.81\cdot (4.523)$

$v(4.523) = -22.871\,\frac{m}{s}$

The velocity of the projectile right before hitting the ground in -22.871 meters per second.

5 0

4 years ago

A radioactive isotope has a half-life of 5,000 years. A rock originally contained 16 grams of radioactive isotope, and now conta

mixas84 [53]

Answer:

10,000 years

Explanation:

Given that,

Half life of the radioactive isotope, $t_{\frac{1}{2}}$ = 5000 years

Initial composition, $N_0$ = 16 grams

Final composition, $N$ = 4 grams

We know that,

$N=N_0\times (\dfrac{1}{2})^\frac{t}{t_\frac{1}{2}}}\\\\4=16(0.5)^\frac{t}{5000}}\\\\\dfrac{4}{16}=(0.5)^{\frac{t}{5000}}\\\\0.25=(0.5)^\frac{t}{5000}\\\\0.5^2=(0.5)^\frac{t}{5000}\\\\2=\dfrac{t}{5000}\\\\t=2\times 5000\\\\t=10000\ yrs$

Therefore, the age of the rock is 10,000 years.

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3 years ago

#1. On a school bus traveling 22 m/s , students walks to the back of the bus at a rate of 1m/s . What is the speed of the studen

zhuklara [117]

#1 is C#2 is C aswell

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3 years ago

You are hiking in the mountains and find a shiny gold nugget. It might be the element gold, or it might be "fool's gold ," which

ra1l [238]

Answer:

The answer is: letter c, density.

Explanation:

Gold is a chemical element that is <em>very rare.</em> It has a high atomic number and is represented by the symbol "Au."

The problem above is asking about the physical property of gold that will help you determine whether a shiny gold nugget is, indeed, real gold. Remember that you are hiking and not anywhere else in the world.

If you look at choice a, appearance. This is definitely incorrect because looking at the object will not be able to help you verify whether it is true gold or "fool's gold." Considering melting point (letter b) would also be hard because you will be needing the necessary equipment to melt gold. Gold melts at 1,064 degrees Celsius. So,this sounds very impractical. (thus, melting point is incorrect) Now, if you look at letter d, physical state, it goes the same way with appearance. It will only tell you that the object is solid and that's it.

Letter c, density is the answer because if you analyze the density of gold, which is 19.32 grams per cubic meter, it is far from the density of iron pyrite (fool's gold) which is 4.8-5 grams per cubic meter. This means that gold is denser or heavier than iron pyrite. You will definitely be able to tell which such big gap in their densities.

6 0

3 years ago