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Lyrx [107]
2 years ago
8

1. I drop a penny from the top of the tower at the front of Fort Collins High School and it takes 1.85 seconds to hit the ground

. Calculate the velocity in m/s after 1.10 seconds of freefall and calculate the velocity at impact in mi/hr.
Physics
1 answer:
ladessa [460]2 years ago
3 0

The acceleration of gravity on Earth is  9.8 m/s² .
The speed of a falling object keeps increasing smoothly,
in such a way that the speed is always 9.8 m/s faster than
it was one second earlier.

If you 'drop' the penny, then it starts out with zero speed. 
If you also start the clock at the same instant, then

         After  1.10 sec,  Speed = (1.10 x 9.8) = 10.78 meters/sec


         After  1.85 sec,  Speed = (1.85 x 9.8) = 18.13 meters/sec

But you want this second one given in a different unit of speed.
OK then:

     =  (18.13 meter/sec) x (3,600 sec/hr) x (1 mile/1609.344 meter)

     =    (18.13 x 3,600 / 1609.344)  (mile/hr)  =  40.56 mph  (rounded)

We did notice that in an apparent effort to make the question
sound more erudite and sophisticated, you decided to phrase
it in terms of 'velocity'.  We can answer it in those terms, if we
ASSUME that there is no wind, and the penny therefore doesn't
acquire any horizontal component of motion on its way down.

With that assumption in force, we are able to state unequivocally
and without fear of contradiction that each 'speed' described above ...
with the word 'downward' appended to it ... does become a 'velocity'.

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2 years ago
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When boiling water, a hot plate takes an average of 8 minutes and 55 seconds to boil 100 milliliters of water. Assume the temper
alexandr1967 [171]

Answer:

90.9 seconds

Explanation:

m = Mass of liquid = Volume×Density

c = Specific heat

\Delta T = Change in temperature

t = Time taken

Room temperature = 75 °F

Converting to Celsius

(75-32)\times \frac{5}{9}=23.889\ ^{\circ}C

Heat required to raise the temperature of water

Q=mc\Delta T\\\Rightarrow Q=100\times 10^{-6}\times 1000\times 4186\times (100-23.889)\\\Rightarrow Q=31860.0646\ J

Power

P=\frac{Q}{t}\\\Rightarrow P=\frac{31860.0646}{8\times 60+55}\\\Rightarrow P=59.55152\ W

Efficiency of the plate

\frac{59.5512}{283}\times 100=21.04282\%

Heat required to raise the temperature of water

Q=mc\Delta T\\\Rightarrow Q=100\times 10^{-6}\times 784\times 2150\times (56-23.889)\\\Rightarrow Q=5412.63016\ J

P=\frac{Q}{t}\\\Rightarrow t=\frac{Q}{P}\\\Rightarrow t=\frac{5412.63016}{0.2104282\times 283}\\\Rightarrow t=90.9\ s

Time taken to heat the aceton is 90.9 seconds

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2 years ago
Any ferrous metal object within or near the mri magnet has the potential of becoming a projectile. this is commonly referred to
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2 years ago
In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of you
olchik [2.2K]

PART A)

horizontal distance that will be moved = 14 m

Height of the fence = 5.0 m

height from which it is thrown = 1.60 m

angle of projection = 54 degree

So here we can say that stone will travel vertically up by distance

\Delta y = 5 - 1.6 = 3.40 m

now we will have displacement in horizontal direction

\Delta x = 14 m

now we know that

v_x = vcos54

v_y = vsin54

now we will have

\Delta x = v_x t

14 = (vcos54)t

also for y direction

\Delta y = v_y t + \frac{1}{2}at^2

3.40 = (vsin54)t - \frac{1}{2}(9.8) t^2

now from the two equations we will have

3.40 = (vsin54)(\frac{14}{vcos54}) - 4.9 t^2

3.40 = 14 tan54 - 4.9 t^2

3.40 = 19.3 - 4.9 t^2

t = 1.8 s

now from above equations

14 = vcos54 (1.8)

v = 13.2 m/s

So the minimum speed will be 13.2 m/s

Part B)

Total time of the motion after which it will land on the ground will be "t"

so its vertical displacement will be

\Delta y = -1.60 m

now we will have

-1.60 = v_y t + \frac{1}{2}at^2

-1.60 = (13.2sin54)t - \frac{1}{2}(9.8)t^2

4.9 t^2 - 10.7t - 1.60 = 0

t = 2.3 s

Now the time after which it will reach the fence will be t1 = 1.8 s

so total time after which it will fall on other side of fence

t_2 = t - t_1

t_2 = 2.3 - 1.8 = 0.5 s

now the displacement on the other side is given as

\Delta x = (vcos54) t_2

\Delta x = (13.2 cos54)(0.5)

\Delta x = 3.88 m

4 0
2 years ago
A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text
antoniya [11.8K]

Answer:

The energy of the system is 15 J.

Explanation:

Given that,

Energy E = 2.5 J

Amplitude = 10 cm

We need to calculate the spring constant

Using formula of mechanical energy of the system

E=\dfrac{1}{2}kA^2

Put the value into the formula

2.5=\dfrac{1}{2}k\times(10\times10^{-2})^2

k=\dfrac{2.5\times2}{(10\times10^{-2})^2}

k=500\ N/m

If the block is replaced by a block with twice the mass of the original block

Amplitude = 6 cm

We need to calculate the energy

Using formula of mechanical energy

E=\dfrac{1}{2}kA^2

Put the value into the formula

E=\dfrac{1}{2}\times500\times(6\times10^{-2})

E=15\ J

Hence, The energy of the system is 15 J.

8 0
2 years ago
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