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vampirchik [111]
3 years ago
7

Brainliest please helptell me if am right if not correct me ​please

Physics
1 answer:
REY [17]3 years ago
8 0

Answer:

See the answers below

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

<u>First case</u>

Vf = 6 [m/s]

Vo = 2 [m/s]

t = 2 [s]

6=2+a*2\\4=2*a\\a=2[m/s^{2} ]

<u>Second case</u>

Vf = 25 [m/s]

Vo = 5 [m/s]

a = 2 [m/s²]

25=5+2*t\\t = 10 [s]

<u>Third case</u>

Vo =4 [m/s]

a = 10 [m/s²]

t = 2 [s]

v_{f}=4+10*2\\v_{f}=24 [m/s]

<u>Fourth Case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

v_{f}=5+8*10\\v_{f}=85 [m/s]

<u>Fifth case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

8=v_{o}+4*2\\v_{0}=8-8\\v_{o}=0

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What is 0.0025 m^2 in cm^2? 1m = 100cm.
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25 cm²

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            K_{e} = ½ k x²

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b) The definition of work is

          W = ΔEm

          W = K_{ef} - K_{eo}

          W = ½ k ( x_{f}² - x₀²)

The final distance

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We calculate the work

          W = ½ 2720 (0.47² - 0.25²)

          W = 215 .4 J

We calculate the strength

          F = k x_{f}

          F = 2720 0.470

          F = 1278.4 N

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