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Andru [333]
3 years ago
12

Go to his profile and roast the mess out of him plzz 403665fl 50 points

Physics
1 answer:
drek231 [11]3 years ago
4 0

Answer:

ok

Explanation:

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If you were trying to kill a fish witha a spear, woukd you ain where you see the fish?
patriot [66]
No, you can't aim where you see the fish, because refraction will change the position of the fish and make it appear in a different position from where it actually is.

Hope this helps!
8 0
3 years ago
Read 2 more answers
A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i
Effectus [21]

Answer:

s =  1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore  

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}  

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}  

• s = \sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}

s = \sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}

s =  1.7 m

7 0
3 years ago
The average power dissipated in a 47 ω resistor is 2.0 w. what is the peak value i 0 of the ac current in the resistor?
pishuonlain [190]
The average dissipated power in a resistor in a ac circuit is:
P=I_{rms}^2 R
where R is the resistance, and I_{rms} is the root mean square current, defined as
I_{rms} =  \frac{I_0}{\sqrt{2}}
where I_0 is the peak value of the current. Substituting the second formula into the first one, we find
P=( \frac{I_0}{\sqrt{2} } )^2 R =  \frac{1}{2} I_0^2 R
and if we re-arrange this formula and use the data of the problem, we can find the value of the peak current I0:
I_0 =  \sqrt{ \frac{2 P}{R} }=  \sqrt{ \frac{2 \cdot 2.0 W}{47 \Omega} }=0.29 A
4 0
3 years ago
Gas and oil from the dead creatures, the gas, and oil are trapped under layers of rock
scoundrel [369]

Answer:

Gas is on top of oil

Explanation:

Gas is less dense ( smaller density ) than oil which is more dense. Since density is directly proportional to mass and inversely proportional to volume, gas will occupy a larger space than oil.

The larger space occupied is the space above the mixture of oil and gas, hence gas is above oil

8 0
3 years ago
A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and
iogann1982 [59]

Answer:

p_2 = 1.76 atm

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}

Putting the values

\frac{1.15*2.37}{280}  =\frac{p_2 *1.68}{304}

9.733*10^{-3} = \frac{p_2 *1.68}{304}

9.733*10^{-3}*304 = p_2*1.68

\frac{9.733*10^{-3}*304}{1.68} =p_2

p_2= 1.76 atm

7 0
4 years ago
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