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slava [35]
3 years ago
14

HELP PLEASE!!!!

Physics
1 answer:
olganol [36]3 years ago
7 0

The distance travelled by the ball is 43.2 m

Explanation:

The motion of the ball is a projectile motion, which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

Here, we want to find the (horizontal) distance travelled by the ball before hitting the ground. Therefore, we can just consider the horizontal motion of the ball.

Since it is a uniform motion, we can write:

v=\frac{d}{t}

where

v = 13.5 m/s is the horizontal speed

t = 3.2 s is the time of flight

Solving for d, we find

d=vt=(13.5)(3.2)=43.2 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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then it turns towards north then velocity is  

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suppose car takes t sec to change its path so average acceleration is given by

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Answer:

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Explanation:

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I \omega_i = I' \omega_f

(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f

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3\times 800 = (3+1.1)\times \omega_f

2400 = (4.1)\times \omega_f

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2) A 0.4kg ball moves in horizontal circle of radius 3 m at speed of 100m/s. What
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A cheetah can run at a maximum speed 97.8 km/h and a gazelle can run at a maximum speed of 78.2 km/h. If both animals are runnin
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(1)

Cheetah speed: v_c = 97.8 km/h=27.2 m/s

Its position at time t is given by

S_c (t)= v_c t (1)

Gazelle speed: v_g = 78.2 km/h=21.7 m/s

the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by

S_g(t)=S_0 +v_g t (2)

The cheetah reaches the gazelle when S_c=S_g. Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:

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t=\frac{S_0}{v_c-v_t}=\frac{96.8 m}{27.2 m/s-21.7 m/s}=17.6 s


(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.

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d=S_c -S_g =204 m-162.8 m=41.2 m

4 0
3 years ago
Read 2 more answers
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