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tatiyna
3 years ago
10

A sky diver, with parachute unopened, falls 625 m in 15.0 s. Then she opens her parachute and falls another 362 m in 139 s. What

is her average velocity (both magnitude and direction) for the entire fall?
Magnitude m/s
So for the free fall v = (a)(t) and because a = g in this instance I did v = (9.8)(15.0) = 147 m/s
So for the free fall velocity = 147
The second part is confusing, The parachute is open so a != 9.8 because of the wind drag on the chute, does a = 1/2gt^2? Cause if it does I think I can figure it out.
Physics
1 answer:
Jobisdone [24]3 years ago
4 0

Answer:

v_{avg} = 6.41 m/s

Explanation:

Average velocity is defined as the ratio of total displacement of the motion and total time taken in that motion

here we know that initially the sky diver drops without opening parachute by total displacement 625 m

then she open her parachute and drop another 362 m

so first it took time t = 15 s to drop without open parachute

then it took t = 139 s to drop next displacement

so here total displacement is given as

d = 625 m + 362 m

total time is given as

t = 15 s + 139 s

so average velocity is given as

v_{avg} = \frac{625 + 362}{15 + 139}

v_{avg} = 6.41 m/s

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Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

In this initial case

     m_A = M

     m_B = M

     a = M / (1 + 1) M g

     a = ½ g

Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

    m_B = 2M

    a = 2M / (1 + 2) M g

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We seek tension for this case

    T’= m_A a

    T’= M 2/3 g

   

Let's look for the relationship between the tensions of the two cases

   T’/ T = 2/3 M g / (½ M g)

   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

4 0
3 years ago
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