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kykrilka [37]
3 years ago
13

When a skydiver is 1200 meters in the air they open the parachute. They are moving

Physics
1 answer:
Elis [28]3 years ago
5 0

Answer:

3m/s up

Explanation:

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A father racing his son has half the kinetic energy of the son, who has half the mass of the father.The father speeds up by 1.0
Helga [31]

Answer:

(a) v_f=2.414\ m.s^{-1}

(b) v_s=4.828\ m.s^{-1}

Explanation:

Let:

mass of father be, m_f

mass of son be, m_s=\frac{m_f}{2}

speed of son be, v_s

initial speed of father be, v_f

After speeding up, speed of father is  v_f+1

<u>We know Kinetic Energy is given as</u>

KE=\frac{1}{2} m.v^2 .....................................(1)

where:

m = mass

v = velocity

Hence, according to the initial condition the father is having kinetic energy half the kinetic energy of the son.

KE_s=2.KE_f

\frac{1}{2} m_s.v_s^2=2\times \frac{1}{2} m_f.v_f^2

(\frac{m_f}{2})\times v_s^2=2\times m_f\times v_f^2

v_s=2v_f .................................................(2)

According to the final condition:

\frac{1}{2} m_s.v_s^2= \frac{1}{2} m_f.(v_f+1)^2

(\frac{m_f}{2})\times v_s^2=m_f.(v_f+1)^2

v_s^2=2(v_f+1)^2

v_s=\sqrt{2}(v_f+1).....................................................(3)

(a)

From eq. 2 & 3

2v_f=\sqrt{2}(v_f+1)

\sqrt{2}\ v_f=(v_f+1)

v_f(\sqrt{2}-1)=1

v_f=\frac{1}{(\sqrt{2}-1)}

v_f=2.414\ m.s^{-1}

(b)

<em>putting the above value in eq. (2)</em>

v_s=4.828\ m.s^{-1}

4 0
4 years ago
Suppose a car is traveling in the negative x-direction and comes to a stop. What is the sign of that car’s acceleration? How do
Oduvanchick [21]

We have equation of motion v = u + at

Where v = final velocity, u = initial velocity, a = acceleration, and t = time

In this case car is travelling in -x direction

Velocity of car = displacement / time

Since displacement value is increasing in negative x axis it's initial velocity is negative

And it's final velocity is zero since it comes to rest, and time is also positive

So, v= u+ at => 0 = -u + at

   So, a = u/t

Which is positive and along positive X - direction

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