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OleMash [197]
3 years ago
6

A helicopter travels from point C to point A to perform a medical supply drop. The helicopter then needs to land at point B. How

many degrees should the helicopter turn at point A to travel towards point B? A right-angled triangle A B C, with an exterior angle. The right angle is formed at angle B and angle C is marked as 32 degrees. Angle B and angle C is a pair of opposite inferior angles to the exterior angle. The helicopter should turn
Physics
1 answer:
Ainat [17]3 years ago
8 0

Answer:

32 degrees

Explanation:

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A snowball is rolling down a hill at 4.5 m/s and accumulating snow as it goes. Its diameter begins at 0.50 m and ends at the bot
Reil [10]
To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.

The formula for centripetal acceleration is:
Centripetal Acceleration = v squared divided by r

where:
v = velocity, m/s
r= radium, m

assuming the velocity does not change:

at the top of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 0.25 m
                                      = 81 m/s^2

at the bottom of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 1.25 m
                                      = 16.2 m/s^2

to find the change in centripetal acceleration, take the difference of the two.
change in centripetal acceleration = centripetal acceleration at the top of the hill - centripetal acceleration at the bottom of the hill

= 81 m/s^2 - 16.2 m/s^2
= 64.8 m/s^2 or 65 m/s^2
6 0
3 years ago
2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the
frozen [14]

Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

 

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum =  2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = \frac{1}{2} mv²

v² = 2gh₁  

v = √(2gh₁)

from the second equation; s = ut +  \frac{1}{2} at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )    

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b)

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

 

c)

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{3}mR²) ω²

v = √( \frac{6}{5}gh₁ )

so we substitute √( \frac{6}{5}gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( \frac{6}{5}gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

8 0
3 years ago
Explain why asleel meedle does not csink on water​
Juli2301 [7.4K]

Answer:

upthrust

Explanation:

i think it does not sink in water because of the force pulling it upwards

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2 years ago
Which group of measurements is the most precise? A) 2 g, 3 g, 4 g B) 2 g, 2.5 g, 3 g C) 2.0 g, 3.0 g, 4.0 g, 5.0 g D) 2.0 g, 3.0
siniylev [52]
The answer to your question is A
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2 years ago
Which state of matter is the least compressible
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Incompressible. Compressibility is determine by the amount of space between particles in each state.  
7 0
3 years ago
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