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3 years ago

How does air resistance affect how fast a feather falls?

Physics

1 answer:

Korvikt [17]3 years ago

6 0

Answer: Ain't its because how light the feather is ? It's not as heavy.

Explanation: With air resistance, acceleration throughout a fall gets less than gravity (g) because air resistance affects the movement of the falling object by slowing it down. ... Usually, resistance is not very high at low speed or for small or sharp objects (Google Source if needed to prove yours answer)

You might be interested in

Explain how you can use the factors of production to produce a fruit juice in a production company

frozen [14]

Answer:

The factors of production include Land, Labour, Capital and Enterpreneurship

Explanation:

The fruit could be apple, orange , pineapple etc which are usually grown on land . They are tended to by people to ensure there is maximum yield. These people provide the required labour needed.

The cost of planting and payment of workers usually comes from the capital which is often used in running the business by the owner which makes certain decisions to ensure the fruit company is in place. All these factors work hand in hand to ensure production of fruit in a production company is possible.

7 0

3 years ago

The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff

stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet m = 45 kg

coefficient of static friction μ = 0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

$\sum f_y =0$

The upward force and the downward force is :

$N_A+N_B = mg$

where;

$\mathbf { N_A \ and \ N_B}$ are the normal contact force at center point A and B respectively .

$N_A+N_B = 45*9.8$

$N_A+N_B = 441$ ------- equation (1)

Considering the forces on the horizontal axis:

$\sum f_x = 0$

$F_A +F_B = P$

where ;

$\mathbf{ F_A \ and \ F_B }$ are the static friction at center point A and B respectively.

which can be written also as:

$\mu_s N_A + \mu_s N_B = P$

$\mu_s( N_A + N_B) = P$

replacing our value from equation (1)

P = 0.30 ( 441)

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at $N_A$ and it is shifted to $N_B$ :