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Anni [7]
3 years ago
15

The magnitude of the drag force of air resistance on a certain 20kg object is proportional to its speed. If the object has a ter

minal speed 80m/s, what is the magnitude of the drag force of an object when it is falling with a speed 30m/s?
Physics
1 answer:
Anna11 [10]3 years ago
8 0
Given:
object = 20kg
terminal speed of object = 80 m/s

According to the problem, drag force is proportional to speed, so Fd = kv ; k is some constant
At terminal velocity Vt: Fg = Fdmg = kVtk = mg / Vt = (20.0)(9.8)/(80.0) = 2.45 kg/s
<span>Fd = kv = 2.45v</span>
Fd = 2.45 (30.0) = 73.5 N
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Which statement best compares the accelerations of two objects in free fall?
gregori [183]

Answer:

D) The objects have the same acceleration. With no air resistance and the same gravity, the acceleration due to gravity should be the same for both objects.

8 0
3 years ago
A 900 kg car runs into a 70 kg deer at 28 m/s. If the car transfers all of its momentum to the deer, how fast does the deer go f
Alex

Answer:

Mc = 900 Kg

Uc = 28 ms^-1

Md = 70 Kg

Ud = 0

We want Vd

Vc = Vd

This situation is elastic momentum

m _{c}u _{c}+ m _{d}u _{d} = m _{c}v _{c} + m _ dv _{d} \\  (900 \times 28) + (70 \times 0) = (900 \times v _{d} ) + (70 \times v _{d}) \\ 25200 = (900 + 70)v _{d}  \\ 25200 = 970v _{d} \\ v _{d} =  \frac{25200}{970}  \\ v _{d} = 26 \: m {s}^{ - 1}

4 0
3 years ago
If an airplane were traveling westward with a thrust force of 450 N and there was a headwind (drag) of 200 N, what would the res
Marat540 [252]

Answer:

The resulting net force on the airplane would be 250N.

Explanation:

It would be 250N because to find the amount of Newtons you would have to do 450 minus 200. 450 minus 200 equals 250. You can check by adding 200 plus 250 and it equals 450 the which is the "thrust".

6 0
2 years ago
Read 2 more answers
A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  T_R  = 11.8 \  days  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

   The semi - major axis of  Planet D is  a_D  = 60 \  AU

    The orbital  period of planet D is  T_D = 18.164 \  days

Generally from Kepler third law

          T \  \ \alpha \ \ a^{\frac{3}{2} }

Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

7 0
3 years ago
You lift a bowling ball with your hand, accelerating it upward. What are the forces on the ball and on what objects are they exe
Snowcat [4.5K]

Answer:

Explanation:

When we lift a ball with the hand the forces experienced by the ball is  

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  • Gravitational force

Resultant of these forces gives acceleration to the ball

While your hand exerts a force on the ball, the ball will also exert a force of equal magnitude but opposite in direction to the force by hand.

All these forces is exerted on the ball, hand or on the earth.

7 0
4 years ago
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