Answer:
Distance will be 49.34 m
Explanation:
We have given wavelength 
Diameter of the antenna d = 2.7 m
Range L = 7.8 km = 7800 m
We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D
We know that distance is given by 
So distance D will be 49.34 m
Answer:
D
Explanation:
speed = frequency x wavelength
speed of light in vacuum is 3.0 x 10^8
wavelength = 5.32 x10 ^-7
3.0 x 10 ^ 8 = 5.32 x 10^-7 x frequency
frequency = 5.63909 x 10^14
round it off = 5.64 x 10^14 Hz
thus the answer is D
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Answer:
108.37°C
Explanation:
P₁ = Initial pressure = 101 kPa
V₁ = Initial volume = 530 m³
T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K
P₂ = Final pressure = 101 kPa (because it is open to atmosphere)
V₂ = Final volume = 530 m³
P₁V₁ = n₁RT₁
⇒101×530 = n₁RT₁
⇒53530 J = n₁RT₁
P₂V₂ = n₂RT₂
⇒53530 J = n₂RT₂

Dividing the first two equations we get

∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C
Answer:
Explanation:
Power (p)= 40 watt
time (t)= 3 minutes = 180 seconds
Energy (E)= ?
we know
P = E/t
40 = E / 180
E = 7200 Joules
Therefore it does 7200 joules of work.
hope it helps :)