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Bond [772]
3 years ago
11

Why are objects that fall near earth’s surface rarely in free fall?

Physics
1 answer:
Sloan [31]3 years ago
3 0

Answer:

Because of the presence of air resistance

Explanation:

When an object is in free fall, ideally there is only one force acting on it:

- The force of gravity, W = mg, that pushes the object downward (m= mass of the object, g = acceleration of gravity)

However, this is true only in absence of air (so, in a vacuum). When air is present, it exerts a frictional force on the object (called air resistance) with upward direction (opposite to the motion of free fall) and whose magnitude is proportional to the speed of the object.

Therefore, it turns out that as the object falls, its speed increases, and therefore the air resistance acting against it increases too; as a result, the at some point the air resistance becomes equal (in magnitude) to the force of gravity: when this happens, the net acceleration of the object becomes zero, and so the speed of the object does not increase anymore. This speed reached by the object is called terminal velocity.

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Which two characteristics are important for soil or land that is used for farming ?
Tanzania [10]

Answer:

deep, fertile soils. not too dry and not too wet. warm climate. land suitable for machinery.

I hope it's helpful!

8 0
2 years ago
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If 180 grams of potassium iodide is dissolved in 100 cm3 of water at 30oC, a(n) _______________ solution is formed. A) saturated
kenny6666 [7]

Answer: D)supersaturated

Explanation: Solubility is defined as the amount of solute in grams which can dissolve in 100 g of the liquid to form a saturated solution at that particular temperature.

At 30^0C , the solubility of KI is 153g/100 ml.

Thus if 180 grams is dissolved, it contains more amount of solute than it can hold at that that temperature, and thus is supersaturated solution.

A saturated solution is a solution containing the maximum concentration of a solute dissolved in the solvent. ​The additional solute does not dissolve in a saturated solution.

An unsaturated solution is solution in which the solute concentration is lower than its equilibrium solubility.

A supersaturated solution is one that has more solute than it can hold at a certain temperature.

6 0
3 years ago
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Student swings a small rubber stopper attached to a string over her head in a horizontal, circular path. The string is 1.50 mete
harkovskaia [24]

Answer:

v = 18.84 m/s

Explanation:

Given that,

The length of the string, r = 1.5 m (it will act as radius)

The rubber stopper makes 120 complete circles every minute.

Since, 1 minute = 60 seconds

It means, its frequency is 2 circles every second.

Let we need to find the average speed of the rubber stopper. It can be calculated as follows :

v=\dfrac{d}{T}

d is distance, d=2\pi r and 1/T = f (frequency)

v=2\pi rf\\\\=2\pi \times 1.5\times 2\\\\=18.84\ m/s

So, the average speed of the rubber stopper is 18.84 m/s.  

4 0
3 years ago
Newton’s 3rd law of motion states that “every action has an equal and opposite reaction.” A golf ball was hit with a force of 20
DedPeter [7]

Answer:

<u>Example of Newton's III law</u>

  • In the, golf the ball was hit by a club with certain force. As the club hits the ball it's the action. When the ball flies away its the reaction.
  • When a person swings a golf club at the ball, when it hits the ball, it causes the ball to roll up the face of the club and into the air towards the target.
8 0
3 years ago
In a second experiment, you decide to connect a string which has length L from a pivot to the side of block A (which has width d
Salsk061 [2.6K]

Answer:

The answer is in the explanation

Explanation:

A)

i) The blocks will come to rest when all their initial kinetic energy is dissipated by the friction force acting on them. Since block A has higher initial kinetic energy, on account of having larger mass, therefore one can argue that block A will go farther befoe coming to rest.

ii) The force on friction acting on the blocks is proportional to their mass, since mass of block B is less than block A, the force of friction acting on block B is also less. Hence, one might argue that block B will go farther along the table before coming to rest.

B) The equation of motion for block A is

m_{A}\frac{\mathrm{d} v}{\mathrm{d} t} = -m_{A}g\nu_{s}\Rightarrow \frac{\mathrm{d} v}{\mathrm{d} t} = -\nu_{s}g \quad (1)

Here, \nu_{s} is the coefficient of friction between the block and the surface of the table. Equation (1) can be easily integrated to get

v(t) = C-\nu_{s}gt \quad (2)

Here, C is the constant of integration, which can be determined by using the initial condition

v(t=0) = v_{0}\Rightarrow C = v_{0} \quad (3)

Hence

v(t) = v_{0} - \nu_{s}gt \quad (4)

Block A will stop when its velocity will become zero,i.e

0 = v_{0}-\nu_{s}gT\Rightarrow T = \frac{v_{0}}{\nu_{s}g} \quad (5)

Going back to equation (4), we can write it as

\frac{\mathrm{d} x}{\mathrm{d} t} = v_{0}-\nu_{s}gt\Rightarrow x(t) = v_{0}t-\nu_{s}g\frac{t^{2}}{2}+D \quad (6)

Here, x(t) is the distance travelled by the block and D is again a constant of integration which can be determined by imposing the initial condition

x(t=0) = 0\Rightarrow D = 0 \quad (7)

The distance travelled by block A before stopping is

x(t=T) = v_{0}T-\nu_{s}g\frac{T^{2}}{2} = v_{0}\frac{v_{0}}{\nu_{s}g}-\nu_{s}g\frac{v_{0}^{2}}{2\nu_{s}^{2}g^{2}} = \frac{v_{0}^{2}}{2\nu_{s}g} \quad (8)

C) We can see that the expression for the distance travelled for block A is independent of its mass, therefore if we do the calculation for block B we will get the same result. Hence the reasoning for Student A and Student B are both correct, the effect of having larger initial energy due to larger mass is cancelled out by the effect of larger frictional force due to larger mass.

D)

i) The block A is moving in a circle of radius L+\frac{d}{2} , centered at the pivot, this is the distance of pivot from the center of mass of the block (assuming the block has uniform mass density). Because of circular motion there must be a centripetal force acting on the block in the radial direction, that must be provided by the tension in the string. Hence

T = \frac{m_{A}v^{2}}{L+\frac{d}{2}} \quad (9)

The speed of the block decreases with time due to friction, hence the speed of the block is maximum at the beginning of the motion, therfore the maximum tension is

T_{max} = \frac{m_{A}v_{0}^{2}}{L+\frac{d}{2}} \quad (10)

ii) The forces acting on the block are

a) Tension: Acting in the radially inwards direction, hence it is always perpendicular to the velocity of the block, therefore it does not change the speed of the block.

b) Friction: Acting tangentially, in the direction opposite to the velocity of the block at any given time, therefore it decreases the speed of the block.

The speed decreases linearly with time in the same manner as derived in part (C), using the expression for tension in part (D)(i) we can see that the tension in the string also decreases with time (in a quadratic manner to be specific).

8 0
3 years ago
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