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barxatty [35]
3 years ago
15

A flare is launched from a life raft with an initial velocity of 192 ft/sec. How many seconds will it take for the flare to retu

rn to the sea if h = 1 9 2 t − 1 6 t 2 represents the height of the flare in feet above the sea t sec after it is launched?
Physics
1 answer:
DIA [1.3K]3 years ago
8 0

We use the formula,

h= ut- 16 t^2

Here, h is the  variable  represents the height of the flare  in feet when it returns to the sea so, h = 0 and u is the initial velocity of the flare, in feet per second and its value of 192 ft/sec.

Substituting these values in above equation, we get

0 = 192 t - 16 t^2  \\\\ 16 t( 12 - t ) =0 \\\\ t = 12 s.

Here, t= 0 neglect because it is  the time when the flare is launched.

Thus, flare return to the sea in 12 s.

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Yes, yea, yep, you don't make much sense.

7 0
3 years ago
A quarterback claims that he can throw the football a horizontal distance of 167 m. Furthermore, he claims that he can do this b
pychu [463]

Answer:u=42.29 m/s

Explanation:

Given

Horizontal distance=167 m

launch angle=33.1^{\circ}

Let u be the initial speed of ball

Range=\frac{u^2\sin 2\theta }{g}

167=\frac{u^2\sin (66.2)}{9.8}

u^2=1788.71

u=\sqrt{1788.71}

u=42.29 m/s

7 0
2 years ago
. a boat can travel in still water. (a) if the boat points directly across a stream whose current is what is the velocity (magni
mixer [17]

a) The velocity of the boat relative to the shore is 3.40 m/s and b) The position of the boat relative to its point of origin after 3s is 10.20m.

Here it is given that the speed of the boat (x) = 2.20m/s

The speed of the stream current (y) = 1.20m/s

a) We have to find the velocity of the boat relative to the shore.

The speed of the boat = x + y

                                    = 2.20 + 1.20

                                    = 3.40m/s

b) Now we have to find the position of the boat after 3s

The formula for speed:

Speed = Distance/ Time

distance = speed × time

speed = 3.40m/s

Time = 3s

distance = 3.40 × 3

              = 10.20 m

Therefore we get a) speed as 3.40m/s and b) distance as 10.20m.

To know more about the boat and stream refer to the link given below:

brainly.com/question/382952

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8 0
1 year ago
A 1022kg Caprice car stopped at an intersection is rear-ended by a 1620kg ranger truck moving with a speed of 14.5m/s. If the ca
Alika [10]

Answer:

Explanation:

mass of car, m = 1022 kg

mass of truck, M = 1620 kg

initial velocity of truck, U = 14.5 m/s

initial velocity of car, u = 0 m/s

Let the final velocity of car is v and the final velocity of truck is V.

Collision is elastic, so the coefficient of restitution, e = 1

Use conservation of momentum

initial momentum of car + initial momentum of truck = final momentum of car + final momentum of truck

m x u + M x U = m x v + M x V

0 + 1620 x 14.5 = 1022 v + 1620 V

23490 = 1022 v + 1620 V ..... (1)

Use the formula of coefficient of restitution

e = \frac{V_{1}-V_{2}}{u_{2}-u_{1}}

1 (14.5 - 0) = v - V

14.5 = v - V

V = v - 14.5 .... (2)

Put in equation (1)

23490 = 1022 v + 1620 (v - 14.5)

23490 = 1022 v + 1620 v - 23490

46980 = 2642 v

v = 17.8 m/s

Put in equation (2)

V = 17.8 - 14.5

V = 3.3 m/s

Thus, the speed of car is 17.8 m/s and the velocity of truck is 3.3 m/s after collision.

8 0
2 years ago
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azamat
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