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Musya8 [376]
1 year ago
15

FInal physics question please help asap

Physics
1 answer:
blondinia [14]1 year ago
5 0

The cannonball will be 311.8 m far in meter when it travels horizontally

<h3>What is a Projectile ?</h3>

A projectile is any object like stone, ball or javelin that take a parabolic path when thrown into the air.

The following parameters are given from the question

  • Ф = 30°
  • u = 60 m/s
  • g = 10 m/s²
  • R = ?

Resolve the initial velocity into horizontal and vertical components. That is,

  • Horizontal component = ucosФ
  • Vertical component = usinФ

To calculate the range that is, the horizontal distance, will can first calculate the total time take by the ball to reach the ground.

Total time T = 2usinФ / g

Substitute all the parameters

T = (2 x 60sin30) / 10

T = (2 x 30) / 10

T = 60 / 10

T = 6 s

The horizontal distance R = ucosФ x T

R = 60 x cos 30 x 6

R = 311.77 m

Therefore, the cannonball will travel 311.8 m approximately horizontally.

Learn more about Projectile here: brainly.com/question/24949996

#SPJ1

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To be considered a source of water pollution, the source must include a chemical is false.

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Around 70% of the planet is covered with water and the present state of its purity is a matter of a serious thought. Looking over various pollutants, only chemical wastes are not responsible to pollute the water.

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3 years ago
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A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M a
aleksklad [387]

Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

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c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

small block

             w- T = m a

             

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

             a = g \ \frac{m - \mu M}{m+M}

             a = 9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

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