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arsen [322]
3 years ago
13

A dart leaves a toy dart gun with initial velocity of 7.76 m/s, regardless of the angle it is fired. What is the maximum horizon

tal distance that the dart can travel? Assume that the dart gun is fired at ground level.
A. 3.88m

B. More Information is needed

C. 6.14m

D. 12.28m
Physics
1 answer:
geniusboy [140]3 years ago
5 0

Answer:

C. 6.14 m

Explanation:

The maximum horizontal distance travelled by a projectile is given by:

d=\frac{v^2}{g}sin (2\theta)

where

v is the initial velocity

g=9.8 m/s^2 is the acceleration of gravity

\theta is the angle of launch

From the equation, we see that the maximum range is achieved when the projectile is fired at \theta=45^{\circ}.

The dart in the problem has an initial velocity of

v = 7.76 m/s

Substituting into the formula, we find the maximum horizontal distance:

d=\frac{7.76^2}{9.8}sin (2\cdot 45^{\circ}) = 6.14 m

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[C] Velocity.

<h3>Explanation:</h3>

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A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
bagirrra123 [75]

Answer:

a) 520m

b) 10.30 s

c) 100,95 m/s

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a) According the given information, the rocket suddenly stops when it reach the height of 520m, because the engines fail, and then it begins the free fall.

This means the maximum height this rocket reached before falling  was 520 m.

b) As we are dealing with constant acceleration (due gravity) g=9.8 \frac{m}{s^{2}} we can use the following formula:

y=y_{o}+V_{o} t-\frac{gt^{2}}{2}   (1)

Where:

y_{o}=520 m  is the initial height of the rocket (at the exact moment in which it stops due engines fail)

y=0  is the final height of the rocket (when it finally hits the launch pad)

V_{o}=0 is the initial velocity of the rocket (at the exact moment in which it stops the velocity is zero and then it begins to fall)

g=9.8m/s^{2}  is the acceleration due gravity

t is the time it takes to the rocket to hit the launch pad

Clearing t:

0=520 m+0-\frac{9.8m/s^{2} t^{2}}{2}   (2)

t^{2}=\frac{-520 m}{-4.9 m/s^{2}}   (3)

t=\sqrt{106.12 s^{2}   (4)

t=10.30 s   (5)  This is the time

c) Now we need to find the final velocity V_{f} for this rocket, and the following equation will be perfect to find it:

V_{f}=V_{o}-gt  (6)

V_{f}=0-(9.8 m/s^{2})(10.30 s)  (7)

V_{f}=-100.95 m/s  (8) This is the final velocity of the rocket. Note the negative sign indicates its direction is downwards (to the launch pad)

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A fast Humvee drove from Desert A to Desert B. for the first 12 hours and traveled at an average speed of 185 km/h. For the next
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Answer:

The value is v_t = 172 \  km/h

Explanation:

From the question we are told that

    The average speed for the first 12 hours is u =  185 km/h

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Generally the total time taken is mathematically represented as

     t_t =  12 + 13

=> t_t =  25 \  h

The distance covered in the first movement is

          D =  u * 12

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The distance covered in the first movement is

          d=  v * 13

         d =  160 * 13

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The total distance traveled is  

       D_t  = D + d

       D_t  = 2220 +2080

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The average  of the whole journey is

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    v_t = 172 \  km/h

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Answer:

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