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Maurinko [17]
2 years ago
11

A paintball is fired horizontally from a tower 45 m above the ground. If the paintball gun fires at 90 m/s… How long does it tak

e for the paintball to hit the ground? How far away from the tower will the paintball land?
Physics
1 answer:
ankoles [38]2 years ago
3 0

Answer:

The time taken for the paint ball to hit the ground is t =  2.143 \ s

The distance of the landing point from the tower is d = 192.86 \  m

Explanation:

From the question we are told that

   The height of the tower is h =  45 \ m

    The speed of the paintball in the horizontal direction is  v_x =  90 \  m/s

Generally from kinematic equation we have that

      h =  ut + \frac{1}{2} at^2

Here u is the initial  velocity of the paintball in the vertical direction and the value is 0 m/s , this because the ball was fired horizontally

         a is equivalent to g  = 9.8\ m/s^2

        t is the time taken for the paintball to hit the ground

   So    

       45  =  0* t + \frac{1}{2} 9.8 * t^2

=>    t =  2.143 \ s

Generally the distance of its landing position from the tower is

         d = v  *  t

=>      d = 90 *  2.143

=>    d = 192.86 \  m

     

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Answer:

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Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p
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Answer:

Acceleration, a=9.91\times 10^{15}\ m/s^2

Explanation:

It is given that,

Separation between the protons, r=3.73\ nm=3.73\times 10^{-9}\ m

Charge on protons, q=1.6\times 10^{-19}\ C

Mass of protons, m=1.67\times 10^{-27}\ kg

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

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a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}      

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3 years ago
Plz do all of it i will give brainlest and thanks to best answer<br> plz do it right
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Answer:

d

Explanation:

8 0
3 years ago
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In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th
elena55 [62]

<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird  decided to take a pit stop and slows down for 250 m. She spent 5 seconds  in the pit stop.

Here final velocity v=0 \ m/s

initial velocity u= 71 m/s  distance  

Distance covered in the slowing down phase = 250 m

v^2-u^2=2as

a= \frac {(v^2-u^2)}{2s}

a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2

v=u+at

t= \frac {(v-u)}{a}

= \frac {(0-71)}{(-10.082)}=7.042 s

t_1=7.042 s

The car is in the pit stop for 5s t_2=5 s

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2

v=u+at

t= \frac{(v-u)}{a}

t= \frac{(71-0)}{6.81}= 10.425 s

t_3=10.425 s

total time= t_1 +t_2+t_3=7.042+5+10.425=22.467 s

Distance covered by the Mercedes Benz during this time is given by s=vt=71 \times 22.467= 1595.157 m

Distance covered by the Thunderbird during this time=250+350=600 m

Difference between distance covered by the Mercedes  and Thunderbird

= 1595.157-600=995.157 m

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

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