Phoebe's insulated foam cup is filled with 0.15 kg of coffee (mostly water) that is too hot to drink, so she adds

Does air resistance affect the motion of a falling object differently when the initial velocity of the object is greater?

DIA [1.3K]

Yes. Even greater. Air resistance or drag becomes harder the faster an object goes. This is why when cars reach their max speed they don't accelerate as fast, because they are pushing harder against the wind. If I take a tennis ball and shoot it down a bottomless pit, a 400 kph, the drag will slow the ball down till it reaches terminal velocity.

4 0

3 years ago

Read 2 more answers

A stone is dropped from from rest at the top of a mine shaft. It takes 95 seconds for the stone to fall to the bottom of the min

Marianna [84]

Distance of fall from rest,
without air resistance = (1/2) (gravity) (time)²

   = (1/2) (9.8 m/s²) (95 sec)²

   = (4.9 m/s²) (9,025 sec²)

   = 44,222.5 meters .

The depth of the mine shaft is five times the height of Mt. Everest !

6 0

3 years ago

A 0.50 kg object is attached to a spring with force constant 157 N/m so that the object is allowed to move on a horizontal frict

Genrish500 [490]

Explanation:

We have,

Mass of an object is 0.5 kg

Force constant of the spring is 157 N/m

The object is released from rest when the spring is compressed 0.19 m.

(A) The force acting on the object is given by :

F = kx

$F=157\times 0.19\\\\F=29.83\ N$

(B) The force is simply given by :

F = ma

a is acceleration at that instant

$a=\dfrac{F}{m}\\\\a=\dfrac{29.83}{0.5}\\\\a=59.66\ m/s^2$

6 0

3 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is: