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Vikki [24]
3 years ago
5

An injured monkey sits perched on a tree branch 3.0 m above the ground, while a wildlife veterinarian is kneeling down in the bu

shes 87.5 m away attempting to subdue the monkey with a tranquilizer gun. the vet knows that the moment the gun fires the monkey will be frightened and fall down from the branch. what angle up from the ground must the veterinarian aim the gun so that the tranquilizer dart will hit the falling monkey?
Physics
1 answer:
Irina18 [472]3 years ago
6 0

here in the given situation if monkey starts free fall at the same instant when veterinarian shoots towards it then we know that vertical component of motion of monkey and the dart will be same as under gravity

so here the dart will always hit the monkey because they both moves under same acceleration

so here for the angle we can use

tan\theta = \frac{H}{L}

now we have

H = 3 m

L = 87.5 m

now we will have

tan\theta = \frac{3}{87.5}

tan\theta = 0.034

\theta = 1.96 degree

so angle will be 1.96 degree above the ground

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A sailboat moves north for a distance of 10.00 km when blown by a wind from the exact south with a force of 5.00 x 10^4 n. how m
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A mixture of nitrogen and xenon gases, at a total pressure of 836 mm Hg, contains 2.80 grams of nitrogen and 24.9 grams of xenon
larisa86 [58]

Answer: Partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

Explanation:

The partial pressure of a gas is given by Raoult's law, which is:

p_A=p_T\times \chi_A

where,

p_A = partial pressure of substance A

p_T = total pressure

\chi_A = mole fraction of substance A

We are given:

m_{N_2}=2.80g

m_{Xe}=24.9g

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

And,

n_A=\frac{m_A}{M_A}

Mole fraction of nitrogen is given as:

\chi_{N_2}=\frac{\frac{m_{N_2}}{M_{N_2}}}{(\frac{m_{N_2}}{M_{N_2}}+\frac{m_{Xe}}{M_{Xe}})}

Molar mass of N_2 = 28 g/mol

Molar mass of Xe =  g/mol

Putting values in above equation, we get:

\chi_{N_2}=\frac{\frac{2.80}{28}}{\frac{2.80}{28}+\frac{24.9}{131}}

\chi_{N_2}=\frac{0.100}{0.100+0.190}=0.345

To calculate the mole fraction of xenon, we use the equation:

\chi_{Xe}+\chi_{N_2}=1\\\\\chi_{Xe}=1-0.345=0.655

p_{N_2}=836mmHg\times 0.345=288mmHg

p_{Xe}=836mmHg\times 0.655=548mmHg

Thus partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

6 0
3 years ago
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