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enyata [817]
3 years ago
9

For this problem, we assume that we are on planet-i. the radius of this planet is r =4200 km, the gravitational acceleration at

the surface is gi =9.1 m/s 2 , and the gravitational constant g = 6.67 × x10−11 n m2 /kg2 in si units. the mass of planet-i is not given. not all the quantities given here will be used. suppose a cannon ball of mass m = 5130 kg is projected vertically upward from the surface of this planet. it rises to a maximum height h =12306 km above the surface of the planet. caution: here the gravitational acceleration decreases as the cannon ball travels away from planet-i. r h determine the kinetic energy (in joules) of the cannon ball immediately after it is fired off. answer in units of j.
Physics
1 answer:
Minchanka [31]3 years ago
3 0
The expression commonly used for potential gravitational energy is just simplification. It is actually just the first term in Taylor expansion of the real expression. 
In general, the potential energy of gravitational field is defined as:
U=-G \frac{mM}{r}
Where G is universal gravitational constant, and r is the distance between the objects centers of mass. Negative sign represents the bound state.
Since we are not given the mass of the planet we have to calculate it.
F_g=G\frac{mM}{r_p^2}\\ mg=G\frac{mM}{r_p^2}\\ g=G\frac{M}{r_p^2}
This formula can be used for any planet. It gives you the gravitational acceleration on the planet's surface. We can use it to calculate the planet's mass:
g=G\frac{M}{r_p^2}\\ M=\frac{gr_p^2}{G}=2.41\cdot 10^{24}kg
Now we can calculate the potential energy of that cannonball when it reaches its maximum height.
U=-G \frac{mM}{r}\\ U=-G \frac{mM}{r_p+h}
When we plug in the numbers we get:
U=-4.99\cdot 10^{10} J
The potential energy has to be equal to the kinetic energy.
E_k=4.99\cdot 10^{10} J

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True or false The momentum of a 100kg object traveling at 3 m/s is 300 N
melomori [17]

Answer:

True

Explanation:

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Read 2 more answers
This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a
Vadim26 [7]

Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = time taken for decomposition  = 3 days

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{58}{100}\times 100=58g

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

k=\frac{2.303}{3}\log\frac{100}{58}

k=0.18days^{-1}

a) Half-life of radon-222:

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant  = 0.18days^{-1}

t = time taken for decomposition  = ?

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{15}{100}\times 100=15g

t=\frac{2.303}{0.18}\log\frac{100}{15}

t=10.54days

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

3 0
3 years ago
A value with magnitude only is a ?​
harkovskaia [24]
Scalar quantities have only a magnitude. So the answer is scalar quantities.
4 0
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