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slamgirl [31]
2 years ago
15

While 75-kg Anthony roller skates at 3 m/s, 60-kg Sue jumps into his arms with a velocity of 5 m/s. How fast does the pair go to

gether?
Velocity: m/s

Direction:
Physics
1 answer:
AlladinOne [14]2 years ago
5 0

inelastic collision

75.3+60.5=(75+60)v'

225+300=135v'

v'=3.8 m/s

direction = positive = right

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A car start from rest and move 6km due east and 8km due south. Calculate the distance covered
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for this you use the pythagoreom theorem

6^2 + 8^2

36 + 64 = 100

the square root of 100 is 10

10 is the answer


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4 years ago
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Which strategy should you use if your research question is too broad for the scope of your project? (1 O narrow the focus of res
Nesterboy [21]

Answer:

"Narrow the focus of research question"

Explanation:

O Narrow the focus of research question

    This is good! You can still use your question, but focus in on something so you have a proper research project.

O Add another research question

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3 0
2 years ago
A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is
pochemuha

Answer:

2856.96 J

0

0

\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f

6.78822 m/s

Explanation:

v_i = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J

At height y = 0 the potential energy is 0 as

P=mgy\\\Rightarrow P=mg0=0

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f

Half of maximum height

\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}

h_i=\frac{v_i^2}{2g}

v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s

The velocity of the athlete at half the maximum height is 6.78822 m/s

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Answer:

Hope it helps

Explanation:

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