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azamat
3 years ago
12

Everybody can you post the lyrics to your favorite song? I'll give brainliest if I know the song or if I like the lyrics, I'll b

e doing a few of these so.... yeah have fun!>
Physics
2 answers:
Setler [38]3 years ago
5 0

Answer:

I cant post da lyrics lol but this is the song

Explanation:

chill bill by rob stone feat. J. Davis and spooks

andrezito [222]3 years ago
4 0

Answer:

Show and tell

I'm on display for all you (bad word) to see

Show and tell

Harsh words if you don't get a pic with me

Buy and sell (buy and sell me, baby)

Like I'm a product to society

Art don't sell

Unless you (bad word) every authority

Explanation:

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A ball is dropped from a 10 story building (30.0 m). How long does it take to hit the ground?
tensa zangetsu [6.8K]
Answer: 1.428 seconds.
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3 years ago
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As shown in Fig. 4, an ideal gas of monatomic molecules expands from its initial state A to a state B through an isobaric proces
cupoosta [38]
The picture is very blurry
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3 years ago
An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t
mojhsa [17]

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a)  the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F_L = C_L\frac{1}{2}ρV²A = W

we substitute

0.45 × \frac{1}{2} × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F_D = T = C_D\frac{1}{2}ρV²A

we substitute

= 0.065 × \frac{1}{2} × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

8 0
3 years ago
A student pushes a laptop cart down the hallway by applying 20 Newtons force. The student pushes it 10 meters and
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Answer:

None.

Explanation:

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        Since the laptop cart is at rest while he stops at the water fountain, no  

       net work done is on the laptop cart.

4 0
3 years ago
WILL GIVE BRAINLIEST
krek1111 [17]

Answer:

I think d

Explanation:

6 0
3 years ago
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