Answer:
one quarter= 5.69 g
Explanation:
if 10 quarters = 0.1255 lbs
then 1 quarter = 0.01255 lbs (0.1255 divided by 10)
0.01255 lbs = 5.69 g
Answer:
18.2145 meters
Explanation:
Using the conservation of momentum, we have that:
m1 = m1' is the mass of the astronaut, m2=m2' is the mass of the satellite, v1 and v2 are the inicial speed of the astronaut and the satellite (v1 = v2 = 0), and v1' and v2' are the final speed of the astronaut and the satellite. Then we have that:
The negative sign of this speed just indicates the direction the astronaut goes, which is the opposite direction of the satellite.
If the astronaut takes 7.5 seconds to come into contact with the shuttle, their initial distance is:
Answer:
statement 1
Explanation:
When mass increases,KE increases and when mass decreases, KE also decreases
Answer:
The distance between gas species are assumed to be large.
Explanation:
Hello!
In this case, since the ideal gas condition is a hypothetical state of gases at which they are at low pressure and high temperature, because it is assumed the molecules are neither attracted nor repelled by each other, we can infer that the correct statement is "the distance between gas species are assumed to be large" because the low pressure and high temperature ensure the molecules are far away from each other and therefore allowing the ideal equation to be used to model the case, otherwise a rigorous equation of state such as Peng-Robinson, Redlich-Kwong and other should be used to model it as gases actually undergo interaction between their molecules.
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Answer:
Explanation:
We shall write the velocities given in vector form to make the solution easy.
The velocity of water with respect to earth that is waV(e) makes 30 degree with north or 60 degree with east so in vector form
waV(e) = 2.2 cos 60 i + 2.2 sin 60 j
waV(e) = 1.1 i + 1.9 j
Similarly , velocity of wind with respect to earth that is wiV(e) , is making 50 degree with west or - ve of x axes so we cal write it in vector form as follows
wiV(e) = - 4.5 cos 50 i - 4.5 sin 50 j
wiV(e) = - 2.89 i - 3.45 j
Now we have to calculate velocity of wind with respect to water that is
wiVwa
wiV( wa) = wiV ( e)+ eV(wa)
= wiV( e)- waV(e)
- 2.89 i - 3.45 j - 1.1 i - 1.9 j
= - 3.99 i - 5.35 j
Magnitude of this relative velocity
D² = 3.99² + 5.35²
d = 6.67 m /s