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levacccp [35]
3 years ago
10

What is the maximum distance we can shoot a dart, provided our toy dart gun gives a maximum initial velocity of 5.84 m/s?

Physics
1 answer:
poizon [28]3 years ago
4 0
1) The maximum distance can be reached when the dart is shot with an angle of \theta=45^{\circ} above the horizontal (see demonstration of this fact at the end).

2) The motion of the dart is an uniform motion on the x-axis (horizontal direction) with constant velocity v_i = 5.84~m/s and it is an uniformly accelerated motion on the y-axis (vertical direction), with the gravitational acceleration g=9.81~m/s^2 acting downwards. So we can write the laws of motion on both directions:
S_x(t) =( v_i \cos{\theta}) t
S_y(t) = (v_i \sin{\theta})t - \frac{1}{2} gt^2
where the negative sign means that g points downwards.

3) First of all we can find the time at which the dart reaches the ground. This can be found by requiring S_y(t)=0:
(v_i \sin{\theta})t - \frac{1}{2} gt^2=0
From this we find two solutions: t=0~s, corresponding to the beginning of the motion (so we are not interested in this one), and 
t= \frac{2 v_i \sin{\theta}}{g} =0.84~s

4) Now that we now when the dart reaches the ground, we can use this information to find the distance covered on the x-axis, by using t=0.84~s inside the equation of S_x(t) written at point 2:
S_x(0.84~s)= 5.84~m/s \cdot \cos{45^{\circ}} \cdot 0.84~s }=3.47~m

--------------------------
DEMONSTRATION OF POINT 1:

Calling v the initial velocity of the dart, and using a coordinate system where the x-axis coincides with the horizontal direction and the y-axis with the vertical direction, we can write the law of motion on both directions:
S_x(t) = v cos{\theta} t
S_y(t) = v sin{\theta} t + \frac{1}{2} g t^2
where t is the time, \theta is the initial angle of the dart, and g= -9.81 m/s^2 is the gravitational acceleration.
The maximum horizontal distance  can be found by requiring that S_y=0. This condition occurs twice: when the motion starts (t=0) and when the dart falls to the ground (let’s call t_s the time at which this happens). Therefore we can find t_s by requiring S_y(t_s)=0, i.e.:
v sin{\theta} t_s + \frac{1}{2} g t_s^2 =0
which has two solutions: t_s=0 (beginning of the motion) and t_s = - \frac{2 v sin{\theta}}{g}.So we can find the maximum horizontal distance covered by the dart by substituting this t_s into the law of motion for S_x(t):
S_x(t_s) =  v cos\theta t_s = -\frac{2 v^2 cos\theta sin\theta}{g} 
Since
sin2\theta = 2 cos\theta sin\theta,
we can write
S_x(t_s) = - \frac{v^2 sin2\theta}{g}
Since g is negative, the maximum of this function occurs for sin2\theta=1, and this happens when \theta=45^{\circ}.
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umka2103 [35]

Answer:

a) F=2.048\times 10^{-7}\ N

b) a=0.1138\ m.s^{-2}

Explanation:

Given:

  • mass of raindrops, m=1.8\times 10^{-6}\ kg
  • charge on the raindrops, q=+21\times 10^{-12}\ C
  • horizontal distance between the raindrops, r=0.0044\ m

A)

<u>From the Coulomb's Law the force between the charges is given as:</u>

F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}

we have:

\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}

<em>Now force:</em>

F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}

F=2.048\times 10^{-7}\ N

B)

<u>Now the acceleration on the raindrops due to the electrostatic force:</u>

a=\frac{F}{m}

a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}

a=0.1138\ m.s^{-2}

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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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Answer:

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