If the water measures -5 feet at low tide and 3ft at high tide what is the tidal range

A model rocket is launched directly upward at a speed of 16 meters per second from a height of 2 meters. The function f(t)=−4.9t

Crank

Answer:

$Hmax=15.06$ meters

Explanation:The question ask for the maximum value of the function f(t) which can be find by find the maxima of the function

The maxima of the function occurs when the slope is zero. i.e.

$\frac{df}{dt} =0\\\frac{df}{dt} =\frac{d}{dt} (-4.9t^2+16t+2)\\\frac{df}{dt} =-4.9*2t+16\\-9.8t+16=0\\t=16/9.8\\t=1.63 secs$

Hence the maxima occurs at t=1.63 seconds

The maximum value of f is

$f(1.63)=-4.9(1.63^2)+16(1.63)+2\\f(1.63)=15.06\\$

hence maximum height is found to be

$Hmax=15.06$ meters

8 0

3 years ago

(a) A light-rail commuter train accelerates at a rate of 1.35 m/s2 . How long does it take to reach its top speed of 80.0 km/h,

Mashcka [7]

Answer:

(a) Time t = 16.46 sec

(b) Time t =13.466 sec

(c) Deceleration = $2.677m/sec^2$

Explanation:

(a) As the train starts from rest its initial velocity u = 0 m/sec

Acceleration $a=1.35m/sec^2$

Final speed v = 80 km/hr

$80km/hr=\frac{80\times 1000}{3600sec}=22.22m/sec$

From first equation of motion v =u+at

So $t=\frac{v-u}{a}=\frac{22.22-0}{1.35}=16.46 sec$

(b) Now initial speed u = 22.22 m/sec

As finally train comes to rest so final speed v=0 m/sec

Deceleration $a=1.65m/sec^2$

So $t=\frac{v-u}{a}=\frac{0-22.22}{-1.65}=13.466 sec$

(c) We have given that initial velocity = 80 km/hr = 22.22 m/sec

Final velocity v = 0 m/sec

Time t = 8.30 sec

So acceleration is given by

$a=\frac{v-u}{t}=\frac{0-22.22}{8.3}=-2.6771m/sec^2$

As acceleration is negative so it is a deceleration

7 0

4 years ago

Please, pretty please help me!

Andrews [41]

Answer:

umm

Explanation:

1223565 578633 =334675

8 0

3 years ago

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If the water measures -5 feet at low tide and 3ft at high tide what is the tidal range ​

If the water measures -5 feet at low tide and 3ft at high tide what is the tidal range