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den301095 [7]
3 years ago
5

You are a pirate working for Dread Pirate Roberts. You are in charge of a cannon that exerts a force 20,000 N on a cannon ball w

hile the ball is in the barrel of the cannon. The length of the cannon barrel is 1.84 m, and the cannon is aimed at a 50-degree angle from the ground.(1) The acceleration of gravity is 9.8 m/s2. If Dread Pirate Roberts tells you he wants the ball to leave the cannon with speed v0= 76 m/s, what mass cannon ball must you use?(2) Assuming the Dread Pirate Roberts never misses, how far from the end of the cannon is the ship that you are trying to hit (neglect dimensions of the cannon)?
Physics
1 answer:
dmitriy555 [2]3 years ago
6 0

Answer:

Explanation:

Force F=20,000 N

Length of barrel L=1.84 m

launch angle \theta =50

velocity at the time of launch v=76 m/s

using v^2-u^2=2 as

(76)^2=2\times a\times 1.84

a=1569.56 m/s^2

mass m=\frac{F}{a}=\frac{20,000}{1569.56}

m=12.74 kg

(2)Range of ball

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

R=\frac{(76)^2\sin 100}{9.8}

R=580.43 m

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A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan
PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d)  V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.  

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

                                 y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

                                 55 = 0 + 0 + 0.5*(9.81)*t^2

                                 t = sqrt ( 55 * 2 / 9.81 )

                                 t =3.349 s

- Use the second equation of motion in x direction:

                                 x(f) = x(0) + V_x,i*t

                                 150 = 0 + V_x,i*3.349

                                  V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

                                 V_y,f = V_y,i + g*t

                                 V_y,f = 0 + 9.81*3.349

                                 V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

                                 V^2 = V_y,f^2 + V_x,i^2

                                 V = sqrt (32.85^2 + 44.8^2)

                                 V = 55.55 m/s

5 0
3 years ago
Ben walks 500 meters from his house to the corner store. He then walks back toward his house but stops to talk to a neighbor whe
spin [16.1K]
Average velocity =

      (displacement) / (time for the displacement)
and
      (direction of the displacement) .

Displacement =

      (distance from the start-point to the end-point)
and
      (direction from the start-point to the end-point)   .

When Ben is 200 meters from the corner store,
he is (500 - 200) = 300 meters from his house.

His displacement is

         300 meters in the direction
                             from his house to the neighbor .


His average velocity is

         (300/910) =  0.33 meters per second, in the
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7 0
3 years ago
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LenKa [72]

Answer:

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Explanation:

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Answer:-q

Explanation:

Given

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vivado [14]
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<span>Solving the equation: </span>

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<span>This is checked by seeing if Vin is greater than Vout, which it is for a step down transformer.</span>
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