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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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Tammy leaves the office, drives 26 km due
fredd [130]

Answer:

72.98 km

Explanation:

Her displacement is simply the distance from her final position to her initial position.

Now, I've drawn and attached a triangle diagram to depict this her movement.

Point O is her initial starting point.

Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.

From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.

Thus;

x² = 62² + 26² - 2(62 × 26)cos 120

x² = 4520 + 806

x² = 5326

x = √5326

x = 72.98 km

4 0
3 years ago
A ray of light traveling in air strikes the surface of a liquid. if the angle of incidence is 29.7◦ and the angle of refraction
lana66690 [7]
When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
\theta_c = \arcsin ( \frac{n_r}{n_i} ) (2)
where n_r is the refractive index of the second medium and n_i is the refractive index of the first medium.

We can find the ratio n_r / n_i by using Snell's law:
n_i \sin \theta_i = n_r \sin \theta_r (1)
where
\theta_i is the angle of incidence
\theta_r is the angle of refraction

By using the data of the problem and re-arranging (1), we find
\frac{n_r}{n_i} =  \frac{\sin \theta_i}{\sin \theta_r} = \frac{\sin 16.3^{\circ}}{\sin 29.7^{\circ}} =0.566

and if we use eq.(2) we can now find the value of the critical angle:
\theta_c = \arcsin ( \frac{n_r}{n_i} ) = \arcsin (0.566) = 34.5^{\circ}
3 0
3 years ago
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Answer:

The answer is A.

Explanation:

8 0
3 years ago
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Yes, it do, for a short time.
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