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Semenov [28]
3 years ago
15

A 13,000 kg helicopter accelerates upward a 0.5 m/s^2 while lifting a 2000 pound car. to the nearest newton, what is the lift fo

rce exerted by the air on the rotors? use 4.45 n = 1 lb.
Physics
1 answer:
Vinil7 [7]3 years ago
8 0

Lift force exerted by the air on the rotors=143244 N

Explanation:

we use Newtons second law

F- (M+m)g=(M+m)a

F= lift force

m= mass of helicopter= 13000 Kg

M= mass of car= 2000 lb=907.2 kg

a= acceleration= 0.5 m/s²

g= acceleration due to gravity

F- (M+m)g=(M+m)a

F=(M+m)(a+g)

F=(13000+907.2)(0.5+9.8)

F=143244 N

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Answer:

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Light enters a substance from air at 30 degrees to thenormal.
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Answer:

Explanation:

The angle of incidence and refraction are both measured from the normal

angle of incidence = 30°

angle of refraction = 23°

refractive index(n) = sini / sinr

n = sin30°/sin23°

n = 1.27965

refractive index (n) = 1/sinC

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5 0
4 years ago
Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an
ddd [48]

Answer:

Al's mass is 102.92  kg  

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

F_{net} = 0

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

F_{net} = \frac{dp_{horizontal}}{dt} = 0

p_{horizontal_i }= p_{horizontal_f}

where the suffix i and f means initial and final respectively.

The initial momentum will be:

p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}

But, as they are at rest, initially

p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0

p_{horizontal_i} = 0

So, this means:

p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0

We know that the have an combined mass of 195 kg:

m_{total} = m_{Al} + m_{Jo} = 195 \ kg.

so:

m_{Jo} = 195 \ kg - m_{Al}.

m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0

m_{Al} \  v_{Al_f} - m_{Al} \  v_{Jo_f}= - 195 \ kg \  v_{Jo_f}

m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}

m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } {  v_{Al_f} - v_{Jo_f} }

m_{Al} = \frac{195 \ kg  \ v_{Jo_f} } {    v_{Jo_f} - v_{Al_f} }

Now, we can use the values:

v_{Al_f}= 10.2 \frac{m}{s}

v_{Jo_f}= - 11.4 \frac{m}{s}

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m_{Al} = \frac{195 \ kg  ( - 11.4 \frac{m}{s} ) } {   (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }

m_{Al} = 102.92 \ kg

and this is the Al's mass.

5 0
3 years ago
If hot air rises then why is it cold in space please explain in detail thank you
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The required spring constant:

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