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3 years ago

Please help. All the information is in the image.

Physics

1 answer:

kondaur [170]3 years ago

5 0

The tiger's position at time t is given by

x (horizontal) = (4.5 m/s) t

y (vertical) = 7.5 m - 1/2 gt ²

Solve y = 0 for t to find the time it takes for the tiger to reach the ground :

0 = 7.5 m - 1/2 (9.8 m/s²) t ²

===> t = √(2 (7.5 m) / (9.8 m/s²)) ≈ 1.2 s

Evaluate x at this time :

x = (4.5 m/s) (1.2 s) ≈ 5.6 m

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Read 2 more answers

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

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3 years ago