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mixer [17]
3 years ago
15

A speed boat moving at a velocity of 25 m/s runs out of gas and drifts to a stop 3 minutes later 100 meters away. What is its ra

te of deceleration?
Physics
1 answer:
mr_godi [17]3 years ago
5 0
<h2>Answer:</h2>

The rate of deceleration is -0.14m/s^{2}

<h2>Explanation:</h2>

Using one of the equations of motion;

v = u + at

where;

v = final velocity of the boat = 0m/s (since the boat decelerates to a stop)

u = initial velocity of the boat = 25m/s

a = acceleration of the boat

t = time taken for the boat to accelerate/decelerate from u to v =   3 minutes

<em>Convert the time t = 3 minutes to seconds;</em>

=> 3 minutes = 3 x 60 seconds = 180seconds.

<em>Substitute the values of v, u, t into the equation above. We have;</em>

v =  u + at

=> 0 = 25 + a(180)

=> 0 = 25 + 180a

<em>Make a the subject of the formula;</em>

=> 180a = 0 - 25

=> 180a = -25

=> a = -25/180

=> a = -0.14m/s^{2}

The negative value of a shows that the boat is decelerating.

Therefore, the rate of deceleration of the speed boat is 0.14m/s^{2}

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A catapult used by medieval armies hurls a stone of mass 32.0 kg with a velocity of 50.0 m/s at a 30.0 degree angle above the ho
NISA [10]

Answer:

The horizontal distance traveled when the stone returns to its original height = 220.81 m

Explanation:

Considering vertical motion of catapult:-

At maximum height,

Initial velocity, u =  50 sin30 = 25 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v = u + at

Substituting

    v = u + at

    0 = 25  - 9.81 x t

    t = 2.55 s

Time of flight = 2 x Time to reach maximum height = 2 x 2.55 = 3.1 s

Considering horizontal motion of catapult:-

Initial velocity, u =  50 cos30 = 43.30 m/s

Acceleration , a = 0 m/s²

Time, t = 5.10 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

   s = 43.30 x 5.10 + 0.5 x 0 x 5.10²

  s = 220.81 m

The horizontal distance traveled when the stone returns to its original height = 220.81 m

5 0
3 years ago
How is copyright infringement similar to plagiarism
Hoochie [10]

Answer:

Copyright infringement is using someone else's work without their permission amd plagiarism is claiming someone else's work as one's own

Explanation:

hope this helped you

8 0
3 years ago
Two children are riding on a merry-go-round that is rotating with a constant angular speed. Abbie is one meter from the center o
klio [65]

Answer:

The acceleration of Abbie is half of the Zak's.

Explanation:

The centripetal acceleration of an object on a circular path is given by :

a=r\omega^2

Two children are riding on a merry-go-round that is rotating with a constant angular speed. Let r_1 is distance of Abbie from the merry-go-round and r_2 is distance of Zak's from the merry-go-round. Acceleration of Abbie is :

a_1=r_1\omega^2 ...... (1)

r_1=1\ m

Acceleration of Zak's is :

a_2=r_2\omega^2 .......(2)

r_2=2\ m

Dividing equation (1) and (2) we get :

\dfrac{a_1}{a_2}=\dfrac{r_1}{r_2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{2}\\\\a_1=\dfrac{a_2}{2}

So, the acceleration of Abbie is half of the Zak's.

7 0
3 years ago
The normal boiling point of cyclohexane is 81.0 ℃. what is the vapor pressure of cyclohexane at 81.0 ℃?
Hitman42 [59]
The boiling point is defined as the temperature at which the pressure of the vapor of the liquid is equivalent to the external atmospheric pressure surrounding the liquid. Therefore, the boiling point of the liquid is dependent on the atmospheric pressure.

Based on this, the vapor pressure of cyclohexane at 81 degrees celcius will be equal to atmospheric pressure (based on barometric readings)<span />
6 0
3 years ago
Some enterprising physics students working on a catapult decide to have a water balloon fight in the school hallway. The ceiling
sergejj [24]

Answer:

\alpha =54.7º

Explanation:

From the exercise we have our initial information

y=3.4m\\v_{o}=10m/s\\g=-9.8m/s^2

When the balloon gets to the ceiling its velocity at that moment is 0 m/s. Being said that we can calculate velocity at the vertical direction

v_{y}^2=v_{oy}^2+ag(y-y_{o})

Since v_{y}=0 and y_{o}=0

0=v_{oy}^2-2(9.8m/s^2)(3.4m)

v_{oy}=\sqrt{2(9.8m/s^2)(3.4m)}=8.16m/s

Knowing that

v_{oy}=v_{o}sin\alpha

sin\alpha =\frac{v_{oy} }{v_{o} }

\alpha =sin^{-1}(\frac{v_{oy}}{v_{o}})=sin^{-1}(\frac{8.16m/s}{10m/s})=54.7º

8 0
3 years ago
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