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3 years ago

A stone is thrown horizontally at a velocity of 12.38 m/s from a tree house that is

Physics

1 answer:

andrew11 [14]3 years ago

6 0

Answer:

34.8 m/s

Explanation:

Since we're only interested in vertical velocity we can ignore anything horizontal.

We have the following information (in the vertical direction)

d = 61.8

a = 9.8

vi = 0

vf = ?

The equation combining all these variables is:

$v_f^2 = v_i^2 + 2ad\\v_f^2=0+2(9.8)(61.8)\\v_f^2=1211.28\\v_f=\sqrt{1211.28}\\v_f=34.8$

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D. a telescope

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Which statement is a scientific theory?

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A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p

777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

$v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s$

Now, we can calculate the angular acceleration (w0=0rad/s)

$\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}$

$\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}$

with this value we can compute the angular velocity

$\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}$

and the tangential velocity of point B, and then the acceleration of point B:

$v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}$

hope this helps!!

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3 years ago

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Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc

Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

$W= F*d= m*g*d$

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

$W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]$

Converting from Joules to Kcals:

$\frac{10620.75}{4186} =2.537 Kcal$

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

x ---> 100%

$x=\frac{2.537*100}{20} =12.685$

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

$\frac{9000Kcals}{12.685Kcals} =709.5 times$

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

$Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\$

$P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]$

*****

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vladimir1956 [14]

Rocks- Mountains.....

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