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MaRussiya [10]
3 years ago
11

An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its

speed, v, and obeys the equation Fdrag=(12.0N⋅s/m)v+(4.00N⋅s2/m2)v2. What is the terminal speed of this object?
Physics
1 answer:
Zinaida [17]3 years ago
3 0

Answer:

 Terminal velocity of object = 12.58 m/s

Explanation:

 We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.

Gravitational force = mg = 80 * 9.8 = 784 N

Drag force = 12.0v+4.00v^2

Equating both, we have

    784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0

  So v = 12.58 m/s or v = -15.58 m/s ( not possible)

 So terminal velocity of object = 12.58 m/s    

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Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

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525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 0
3 years ago
Key difference of the celsius scale and kevin scale
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6 0
3 years ago
Find the net force and acceleration. 10 points. Will give brainliest.
Afina-wow [57]

Answer:

a) 16 N

b) 2.13 m/s²

Explanation:

Draw a free body diagram of the tv stand.  There are four forces:

Weight force mg pulling down,

Normal force N pushing up,

Friction force Nμ pushing left,

and applied force P pulling right.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

The net force in the x direction is:

∑F = P − Nμ

∑F = P − mgμ

∑F = 25 N − (7.5 kg) (10 m/s²) (0.12)

∑F = 16 N

Net force equals mass times acceleration:

∑F = ma

16 N = (7.5 kg) a

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3 0
3 years ago
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allochka39001 [22]
That’s the color of the ball?
6 0
3 years ago
A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.
marishachu [46]

Answer:

a) the magnitude of the force is

F= Q(\frac{kqs}{r^3}) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b)  the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = \frac{kq}{r^{2} }

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = \frac{kp}{r^{3} } , where p = q × s

E(r) = \frac{kqs}{r^{3} } where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q(\frac{kqs}{r^3})

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall  F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

8 0
3 years ago
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