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Juliette [100K]
2 years ago
5

An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50 newtons. Calcuate the acceleration

of the object
Physics
1 answer:
rjkz [21]2 years ago
6 0

Answer:

this answer you question

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Science Seminar Question: Why did Vehicle 2 fall off the cliff in Claire's test of the collision scene but Vehicle 2 did not fal
asambeis [7]

Complete Question:

Check the file attached to get the complete question

Answer:

In the film Ice word Revenge, vehicle 2 did not fall of the cliff because, Weight_{vehicle 1} < Weight_{vehicle 2} but in Claire's test, vehicle 2 off the cliff because Weight_{vehicle 1} \geq Weight_{vehicle 2}

Explanation:

In Claire's test, the weight of vehicle 1 is either equal to or greater than the weight of vehicle 2, so it was sufficient to push it down the cliff.  In the film Ice word revenge, the weight of vehicle 1 is less than the weight of vehicle 2, it is not sufficient to make it fall off the cliff ( Note: Looking exactly the same in the movie, as Claire claimed, does not mean they have the same mass). Therefore if Claire wants a collision that will not make the vehicle 2 fall off the cliff, he should collide it with a vehicle of lesser mass/weight.

8 0
2 years ago
Please help on this one?
dezoksy [38]

the answer is heat engine

3 0
3 years ago
How are problems best solved by scientists?
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One scientist proposes an idea and other scientists repeat his or her experiments so they can Accept the Idea.
3 0
2 years ago
Read 2 more answers
Hello I need some help.Two guys enetr in a room,a guy come from outside(there is cold)and another guy come from a room(there is
slavikrds [6]
I would feel warm because putting cold and hot together is gonna be warm because with the cold it is cooling down the hot to make warm
5 0
3 years ago
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
zvonat [6]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

3 0
2 years ago
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