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3 years ago

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A speedboat moving at 28 m/s approaches a no-wake buoy marker 91 m ahead. The pilot slows the boat with a constant acceleration

of 4 m/s2 by reducing the throttle. What is the velocity (in m/s) of the boat when it reaches the buoy

1 answer:

Tpy6a [65]3 years ago

6 0

Answer:

7.5 m/s

Explanation:

We can find its velocity when it reaches the buoy by applying one of Newton's equations of motion:

$v^2 = u^2 + 2as$

where v = final velocity

u = initial velocity

a = acceleration

s = distance traveled

From the question:

u = 28 m/s

a = -4 $m/s^2$

s = 91 m

Therefore:

$v^2 = 28^2 + 2 * (-4) * 91\\\\v^2 = 784 + -728 = 56\\\\v = \sqrt{56}\\ \\v = 7.5 m/s$

The velocity of the boat when it reaches the buoy is 7.5 m/s.

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a stone is thrown horizonttaly from a cliff of a hill with an initial velocity of 30m/s it hits the ground at a horizontal dista

ELEN [110]

Answer:

a) Time = 2.67 s

b) Height = 35.0 m

Explanation:

a) The time of flight can be found using the following equation:

$x_{f} = x_{0} + v_{0_{x}}t + \frac{1}{2}at^{2}$ (1)

Where:

$x_{f}$ : is the final position in the horizontal direction = 80 m

$x_{0}$ : is the initial position in the horizontal direction = 0

$v_{0_{x}}$ : is the initial velocity in the horizontal direction = 30 m/s

a: is the acceleration in the horizontal direction = 0 (the stone is only accelerated by gravity)

t: is the time =?

By entering the above values into equation (1) and solving for "t", we can find the time of flight of the stone:

$t = \frac{x_{f}}{v_{0}} = \frac{80 m}{30 m/s} = 2.67 s$

b) The height of the hill is given by:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final position in the vertical direction = 0

$y_{0}$ : is the initial position in the vertical direction =?

$v_{0_{y}}$ : is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)

g: is the acceleration due to gravity = 9.81 m/s²

Hence, the height of the hill is:

$y_{0} = \frac{1}{2}gt^{2} = \frac{1}{2}9.81 m/s^{2}*(2.67 s)^{2} = 35.0 m$

I hope it helps you!

5 0

3 years ago

A sprinter starts from rest and reaches a speed of 15 m/s in 4.25 s. Find his acceleration

kakasveta [241]

Answer:

a=3.53 m/s^2

Explanation:

Vo=0 m/s (because he is not moving at the start)

V1=15 m/s

t= 4.25 s

a = (V1-Vo) / t = 15/4.25 = 3.53 m/s^2

5 0

3 years ago

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

3 years ago

You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m

slavikrds [6]

Answer:v=3.28 m/s

Explanation:

Given

mass of rock $m=500 gm$

diameter of circle $d=2.2 m$

radius $r=\frac{2.2}{2}=1.1 m$

At highest Point

$mg+N=\frac{mv^2}{r}$

At highest Point N=0 because mass is just balanced by centripetal Force

thus $mg=\frac{mv^2}{r}$

$v=\sqrt{gr}$

$v=\sqrt{9.8\times 1.1}$

$v=\sqrt{10.78}$

$v=3.28 m/s$

6 0

3 years ago

Which statement is the correct representation of these electric field lines?

GuDViN [60]

I would say a. hope this helps

4 0

3 years ago

Read 2 more answers