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Rzqust [24]
4 years ago
14

Discuss the validity of extrapolating the acceleration value to an angle of 90°.

Physics
2 answers:
Taya2010 [7]4 years ago
7 0
To find the acceleration when the angle of inclination is equal to 90 degrees, extrapolate the experimental values of acceleration versus angle of the incline. The acceleration at 90 degrees will equal 9.81 m/s/s, since this is the free-fall acceleration.
Ainat [17]4 years ago
3 0

Answer:

Yes, it is validated to extrapolate the acceleration value to an angle of 90 degrees.

Explanation:

Assume the friction-less inclined plane. There exists a mathematical relationship between angle of incline and the object rolling in downward direction on friction-less inclined plan.

Acceleration of object = sinθ × g  

Here, θ=90 degrees so,

Acceleration of object = g = 9.8 m/s^2

Hence, it is validated to extrapolate the acceleration value to an angle of 90 degrees .

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4 years ago
Which is the SI symbol for volume? m g L K
Bezzdna [24]

Common symbol of the volume (L)

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3 years ago
An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed o
Vesnalui [34]

Answer:

a) 75.5 degree relative to the North in north-west direction

b) 309.84 km/h

Explanation:

a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it

So the pilot should head to the West-North direction at an angle of

cos(\alpha) = 80/320 = 0.25

\alpha = cos^{-1}(0.25) = 1.32 rad = 180\frac{1.32}{\pi} = 75.5^0 relative to the North-bound.

b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is

320sin(\alpha) = 320sin(75.5^0) = 309.84 km/h

7 0
4 years ago
Read 2 more answers
A train moving at a constant speed on a surface inclined upward at 10.0° with the horizontal travels a distance of 400 meters in
Amiraneli [1.4K]
If the velocity of the train is v=s/t, where s is the distance and t is time, then v=400/5=80m/s. To get the vertical component of the velocity we need to multiply the velocity v with a sin(α): Vv=v*sin(α), where Vv is the vertical component of the velocity and α is the angle with the horizontal. So:

Vv=80*sin(10)=80*0.1736=13.888 m/s. 

So the vertical component of the velocity of the train is Vv=13.888 m/s.
7 0
4 years ago
A(n) 82.7 kg boxer has his first match in the Canal Zone with gravitational acceleration 9.782 m/s 2 and his second match at the
Annette [7]

Answer:

82.7 kg

Explanation:

the mass of the boxer remains unchanged, this is because mass is a measure of the quantity of matter in an object irrespective of its location and the gravitational force acting at its location. this means mass is independent of the gravitational acceleration hence it remains the same 82.7 kg. its unit is in kilograms (Kg).

6 0
3 years ago
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