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Rzqust [24]
3 years ago
8

Suppose a skydiver jumps out of a plane at 15,000 meters above the ground. It takes him 2.0 seconds to pull the cord to deploy t

he parachute and another 2.0 seconds for the parachute to be fully deployed. Additionally, in order to land safely on the ground, his parachute must be fully deployed 10.0 seconds before he hits the ground. With this information and assuming there is no additional air resistance or changes in atmospheric density, what is the maximum time into his fall that he can wait to pull the cord in order to make it safely to the ground?
Physics
1 answer:
Evgesh-ka [11]3 years ago
6 0

Answer:

The time he can wait to pull the cord is 41.3 s

Explanation:

The equation for the height of the skydiver at a time "t" is as follows:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

First, let´s calculate how much time will it take for the skydiver to hit the ground if he doesn´t activate the parachute.

When he reaches the ground, the height will be 0 (placing the origin of the frame of reference on the ground). Then:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 15000 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

0 m = 15000 m - 4.9 m/s² · t²

-15000 m / -4.9 m/s² = t²

t = 55.3 s

Then, if it takes 4.0 s for the parachute to be fully deployed and the parachute has to be fully deployed 10.0 s before reaching the ground, the skydiver has to pull the cord 14.0 s before reaching the ground. Then, the time he can wait before pulling the cord is (55.3 s - 14.0 s) 41.3 s.

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4 0
3 years ago
Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th
lapo4ka [179]

Answer:

T_f=5.0116^{\circ}C

Explanation:

Given:

  • mass of water, m_w=0.25\ kg
  • initial temperature of water, T_i_w=20^{\circ}C
  • initial temperature of pan, T_i_p=173^{\circ}C
  • mass of pan, m_p=0.6\ kg
  • mass of water evapourated, m_v=0.03\ kg
  • specific heat of water, c_w=4186\ J.kg^{-1}.K^{-1}
  • specific heat of aluminium pan, c_a=900\ J.kg^{-1}.K^{-1}
  • latent heat of vapourization, L=2256000\ J.kg^{-1}

<u>Using the equation of heat:</u>

<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>

m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})

0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)

T_f=5.0116^{\circ}C

5 0
3 years ago
What are two examples of common units for each of the above measurements
kap26 [50]

Density: g/mL, kg/cubic meter  

Volume: L, teaspoon  

Mass: g, MeV/sq. C

3 0
3 years ago
A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0
3 years ago
What are three ways to speed up a<br> reaction?
Ksju [112]

Explanation:

Increase the temperature in Endothermic reactions (Reactions that absorb energy, or become cold) Decrease the temperature in Exothermic reactions (Reactions that release energy, or become hot) Add a catalyst (A substance that reduces activation energy, speeding up the reaction) Increase the concentration of reactants.

source: https://socratic.org/questions/how-can-a-chemical-change-be-speeded-up

6 0
3 years ago
Read 2 more answers
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