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disa [49]
2 years ago
15

A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 1

20 m/s. He misses his target and the cannonball splashes into the briny deep. How far did the cannonball travel? (please show work)
Physics
1 answer:
Bumek [7]2 years ago
4 0

Answer:

202.8m

Explanation:

Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.

First calculate the total time travelled by using the second equation of motion

h = Ut + 1/2gt^2

Let assume that u = 0

And h = 3.5

Substitute all the parameters into the formula

3.5 = 1/2 × 9.8 × t^2

3.5 = 4.9t^2

t^2 = 3.5/4.9

t^2 = 0.7

t = 0.845s

To know how far the cannonball travel, let's use the equation

S = UT + 1/2at^2

But acceleration a = 0

T = 2t

T = 1.69s

S = 120 × 1.69

S = 202.834 m

Therefore, the distance travelled by the cannon ball is approximately 202.8m.

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Josie walks 15 meters north down the freshman hallway. She then walks 25 meters east. Lastly, she walks 15 meters south. What is
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A spring with a constant k=400n/m shoots a 1. 00kg ball up a frictionless incline after being compressed 0. 150m. what is the ma
Zielflug [23.3K]

The maximum height reached by the ball is 0.46m.

To find the answer, we have to know about the potential energy of a spring mass system.

<h3>How to find the maximum height reached by the ball?</h3>
  • It is given that,

                       k=400N/m\\m=1kg\\x=0.150m\\h=?\\

  • We have to find the maximum height reached by the ball.
  • Thus, we have the expression for potential energy of a spring mass system and that of gravitational field as,

                             U=\frac{1}{2}kx^2 \\U=mgh

  • Combining both, we get,

                    h=\frac{kx^2}{2mg} =\frac{400*(0.15)^2}{2*1*9.8} =0.46m

Thus, we can conclude that, the maximum height reached by the ball is 0.46m.

Learn more about the potential energy here:

brainly.com/question/26962934

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Two resistors of resistances R1 and R2, with R2&gt;R1, are connected to a voltage source with voltage V0. When the resistors are
ehidna [41]

Answer

The Value of  r  = 0.127

Explanation:

The mathematical representation of the two resistors connected in series is

                               R_T = R_1 +R_2

 And from Ohm law

                           I_s =\frac{ V}{R_T}

                            I_s  = \frac{V_0}{R_1 +R_2} ---(1)

The mathematical representation of the two resistors connected in parallel  is

                    R_T = \frac{1}{R_1} +\frac{1}{R_2}

                          = \frac{R_1 R_2}{R_1 +R_2}

From the question I_p =10I_s

          =>                 I_p =10I_s = \frac{V_0 }{\frac{R_1R_2}{R_1 +R_2} }  = \frac{V_0 (R_1 +R_2)}{R_1 R_2}---(2)

     Dividing equation 2 with equation 1

       =>                 \frac{10I_s}{I_s} =\frac{\frac{V_0 (R_1 +R_2)}{R_1 R_2}}{\frac{V_0}{R_1 +R_2}}

                                  10 = \frac{(R_1+R_2)^2}{R_1 R_2}----(3)

We are told that    r = \frac{R_1}{R_2} \ \ \ \ \  = > R_1 = rR_2

From equation 3  

                            10 = \frac{(1-r)^2}{r}

=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  1+r^2 + 2r = 10r

=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r^2 -8r+1 = 0

Using the quadratic formula

                             r =\frac{-b\pm \sqrt{(b^2 - 4ac)} }{2a}

        a = 1  b = -8 c =1  

                              =  \frac{8 \pm\sqrt{((-8)^2- (4*1*1))} }{2*1}

                               r= \frac{8+ \sqrt{60} }{2}  \ or \  r = \frac{8 - \sqrt{60} }{2}

                              r = \ 7.87\ or \  r \  = \ 0.127

Now  r =  0.127 because it is the least value among the obtained values

                               

                                   

                             

4 0
3 years ago
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