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3 years ago

What is the speed of terminal velocity?

1 answer:

ivann1987 [24]3 years ago

4 0

The so-called "terminal velocity" is the fastest that something can fall
through a fluid. Even though there's a constant force pulling it through,
the friction or resistance of plowing through the surrounding substance
gets bigger as the speed grows, so there's some speed where the resistance
is equal to the pulling force, and then the falling object can't go any faster.

A few examples:
-- the terminal velocity of a sky-diver falling through air,
-- the terminal velocity of a pecan falling through honey,
-- the terminal velocity of a stone falling through water.

It's not possible to say that "the terminal velocity is ----- miles per hour".
If any of these things changes, then the terminal velocity changes too:

-- weight of the falling object
-- shape of the object
-- surface texture (smoothness) of the object
-- density of the surrounding fluid
-- viscosity of the surrounding fluid .

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Carla holds a ball 1.5 m above the ground. Daniel, leaning out of a car window, also holds a ball 1.5 m above the ground. Daniel

spayn [35]

Answer:

Carla

Explination: As Daniel's ball is dropped from the car moving at 40 mph in a horizontal direction, at the time the ball is dropped it is also moving at 40 mph in a horizontal direction due to inertia, a property of mass causing resistance to change, Daniel's ball will continue to move in a horizontal direction even after being dropped along with falling due to gravity. Daniel's ball will then fall in a projectile motion curve of sorts which will cause an overall velocity to not be straight down causing it not to fall to the ground as quickly as Carla's ball.

Sorry for the long explanation

8 0

3 years ago

Read 2 more answers

The anemometer is spinning and the pressure us high

skelet666 [1.2K]

Answer:

what help you need?????????

3 0

2 years ago

A ball is shot from the ground into the air. At a height of 8.8 m, the velocity is observed to be

Mariulka [41]

Answer:

h = 10.4 m

R = 22.48 m

v= 16,2 m/s , α = 61.7°, below the horizontal

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

Explanation:

The ball describes a parabolic path, and the equations of the movement are:

Equation of the uniform rectilinear motion (horizontal ) :

x = vx*t :

Equations of the uniformly accelerated rectilinear motion of upward (vertical ).

y = (v₀y)*t - (1/2)*g*t² Equation (2)

vfy² = v₀y² -2gy Equation (3)

vfy = v₀y -gt Equation (4)

Where:

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m

y: vertical position in meters (m)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

Known data

y= 8.8 m

v = ( (7.7)i + (5.7)j ) m/s : vx= 7.7 m/s , vy= 5.7 m/s

g = 9.8 m/s²

Calculation of the initial vertical velocity ( v₀y)

We apply Equation (3) with the known data

(vfy)² = (v₀y)² -2*g*y

(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)

(5.7)²+ 172.48 = (v₀y)²

$v_{oy} = \sqrt{(5.7)^{2}+ 172.48 }$

v₀y = 14.3 m/s

Calculation of the maximum height the ball rise (h)

In the maximum height vfy=0

We apply the Equation (3) :

(vfy)² = (v₀y)² -2*g*y

0 = (14.3)² - 2*98*h

h = (14.3)² / 19.6

h = 10.4 m

Calculation of the time it takes for the ball to the maximum height

We apply the Equation (4) :

vfy = v₀y -gt

0 = v₀y -gt

gt = v₀y

t = v₀y/g

t = 14.3/9.8

t= 1.46 s

Flight time = 2t = 2.92 s

Total horizontal distance traveled by the ball (R)

We replace data in the equation (1)

x =vx*t vx= 7.7 m/s , t =2.92 s (Flight time)

R = (7.7)* (2.92) = 22.48 m

Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground

vx = 7.7 m/s

vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s

$v= \sqrt{v_{x}^{2}+v_{y}^{2} }$

$v= \sqrt{(7.7)^{2}+ (-14.3)^{2} }$

v= 16,2 m/s

$\alpha = tan^{-1} (\frac{v_{y} }{v_{x} })$

$\alpha = tan^{-1} (\frac{-14.3 }{7.7 })$

α = -61.7°

α = 61.7°, below the horizontal

i- j components of the v

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

5 0

3 years ago

8. Mark is sitting in a chair, legs out in front of him resting on an ottoman. His knees are.....

anyanavicka [17]

His knees are flat in front of him or 180 degrees, i hope that answered your question

5 0

3 years ago

Read 2 more answers

A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. a) What is the resultant velocity of the motorb

Nostrana [21]

Explanation:

It is given that,

Velocity in East, $v_1=4\ m/s$

Velocity in North, $v_2=3\ m/s$

(a) The resultant velocity is given by :

$v=\sqrt{4^2+3^2}=5\ m/s$

(b) The width of the river is, d = 80 m

Let t is the time taken by the boat to travel shore to shore. So,

$t=\dfrac{d}{v}$

$t=\dfrac{80\ m}{5\ m/s}$

t = 16 seconds

(c) Let x is the distance covered by the boat to reach the opposite shore. So,

$x=v_2\times t$

$x=3\ m/s\times 16\ s$

x = 48 meters

Hence, this is the required solution.

4 0

3 years ago