Answer:
The scientist will be looking for the velocity of the wave in air which is equivalent to 10^7m/s
Explanation:
If an object in space is giving off a frequency of 10^13Hz and wavelength of 10^-6m then the scientist will be looking for the velocity of the object in air.
The relationship between the frequency (f) of a wave, the wavelength (¶) and the velocity of the wave in air(v) is expressed as;
v = f¶
Given f = 10^13Hz and ¶ = 10^-6m,
v = 10¹³ × 10^-6
v = 10^7 m/s
The value of the velocity of the object in space that the scientist will be looking for is 10^7m/s
The final atmospheric pressure is 
Explanation:
Assuming that the temperature of the air does not change, we can use Boyle's law, which states that for a gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume. In formula,

where
p is the gas pressure
V is the volume
The equation can also be rewritten as

where in our problem we have:
is the initial pressure (the atmospheric pressure at sea level)
is the initial volume
is the final pressure
is the final volume
Solving the equation for p2, we find the final pressure:

Learn more about ideal gases:
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Answer:
The impulse on the object is 60Ns.
Explanation:
Impulse is defined as the product of the force applied on an object and the time at which it acts. It is also the change in the momentum of a body.
F = m a
F = m(
)
⇒ Ft = m(
-
)
where: F is the dorce on the object, t is the time at which it acts, m is the mass of the object,
is its initialvelocity and
is the final velocity of the object.
Therefore,
impulse = Ft = m(
-
)
From the question, m = 3kg,
= 0m/s and
= 20m/s.
So that,
Impulse = 3 (20 - 0)
= 3(20)
= 60Ns
The impulse on the object is 60Ns.
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i think so ,but i am not sure</span>