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3 years ago

8

A tomato of mass 0.18 kg is dropped from a tall bridge. If the tomato has a speed of 11 m/s just before it hits the ground, what

is the kinetic energy of the tomato?

1 answer:

kondor19780726 [428]3 years ago

7 0

The kinetic energy of the tomato is :

K.E = 1/2 mv^2

K.E = 1/2 x 0.18 kg x 11 m/S^2

K.E = 0.99

Hope this helps

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A battery does 2.186 J of work to transfer 0.033 C of charge from the negative to the positive terminal. What is the emf of this

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Answer:

66.24 Volts

Explanation:

W = Amount of work done in moving the charge from negative to positive terminal = 2.186 J

Q = Amount of charge being moved from negative to positive charge = 0.033 C

ΔV = EMF of the battery

Amount of work done in moving the charge from negative to positive terminal is given as

W = Q ΔV

2.186 = (0.033) ΔV

ΔV = 66.24 Volts

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A student is pedaling a stationary bicycle at the U of M Rec Center. Her alveolar PO2 = 115 mm Hg and her oxygen consumption is

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Answer:

This is an alveolar gas equation question and it is used to approximate the partial pressure of oxygen in the alveolus (PAO2):

The equation states;

PₐO₂ = P₁O₂ - (PₐCO₂/R) = [(PB - PH₂O) × F₁O₂ - (PₐCO₂/R)] ...................Eqn 1

where

PₐO₂ = Alveolar partial pressure of O2 = 115mmHg

P₁O₂ = Inspired partial pressure of O2 = 150mmHg

PB = barometric pressure,

PH₂O = Water vapor pressure (usually 747 mmHg),

F₁O₂ = fractional concentration of inspired oxygen,

and R = gas exchange ratio. (Usually around 0.8)

FₐO₂ = Fraction of alveolar O₂

(O₂) = 1L/min = 1dm³

From eqn 1. we have

PₐCO₂ = (P₁O₂ - PₐO₂)/R

= (150 - 115)x0.8

PₐCO₂ = 28mmHg

Similarly from Eqn 1, we have

F₁O₂ = (PₐO₂ + PₐCO₂/R)/(PB - PH₂O)

F₁O₂ = (115 + (28/.8))/(747 - 47)

F₁O₂ = 0.21

Now to find the Alveolar Ventilation A, we will use this equation;

O₂ = A(F₁0₂ - FₐO₂) .................Eqn 2

But FₐO₂ = PₐO₂/(PₐO₂ + PₐCO₂)

FₐO₂ = 115/ (115+28) = 0.8

A = O₂/(F₁0₂ - FₐO₂)

A = 0.001/(0.21 - 0.8)

A = 0.00169m³/min

Hence, the aveolar ventilation is 0.00169m³/min

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2 years ago

Assume that a vaulter is able to carry a vaulting pole while running as fast as Carl Lewis in his world record 100-m dash (aroun

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Answer:

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Explanation:

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2 years ago

A myopic (nearsighted) lawyer has a far point of 60.0 cm. What is the diopter value of suitable corrective contacts?

Vlada [557]

Answer:

P = -1.67 dP

Explanation:

given,

myopic lawyer has a far point of = 60.0 cm = 0.6 m

If a person is suffering from myopia then he cannot see the farthest object clearly.

Image of far object does not form on the retina so, the image appear to be blur.

using lens formula

$\dfrac{1}{v}-\dfrac{1}{u} = \dfrac{1}{f}$

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$\dfrac{1}{f} =P$

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2 years ago

If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is

viktelen [127]

To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
where
$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
where
$d$ is the distance
$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
$t=0.98$ seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
$d=V_{i}t+ \frac{1}{2} gt^2$
$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
$d=15.22$ ft

Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
$16t^2+31.21t-175.22=0$
$t=2.47$ or $t=-4.43$
Since time cannot be negative, $t=2.47$ is the time it takes the object to fall to the ground.

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
$V_{f}=V_{i}+gt$
$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

5 0

3 years ago