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ankoles [38]
3 years ago
14

In February 1955, a paratrooper fell 365 m from an airplane without being able to open his chute but happened to land in snow, s

uffering only minor injuries. Assume that his speed at impact was 56 m/s (terminal speed), that his mass (including gear) was 85 kg, and that the magnitude of the force on him from the snow was at the survivable limit of 1.2 x 10^5 N.What is the minimum depth of snow that would have stopped him safely?What is the magnitude of the impulse on him from the snow?
Physics
1 answer:
DaniilM [7]3 years ago
7 0

Answer:

a. i=4760 kg*m/s

b. D_U= 1.11 m

Explanation:

a)

F= 120,000N

Kinetic energy @ impact = 120,000*depth

K_E= (1/2)*85kg*(56m/s)^2

K_E=133280 J

D_U= \frac{133280J}{120000N} = 1.11 m

b)

The momentum is equal to the impulse on him from the snow so:

p=m*v

p=85kg*56m/s

i=4760 kg*m/s

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Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh

here we know that for spherical shell and pure rolling conditions

v = R \omega

I = \frac{2}{3}mR^2

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh

\frac{5}{6}mv^2 = mgh

h = \frac{5v^2}{6g}

h = \frac{5(8^2)}{6(9.81)} = 5.44 m

Part b)

If ball is not rolling and just sliding over the hill then in that case

\frac{1}{2}mv^2 = mgh

h = \frac{v^2}{2g}

h = \frac{8^2}{2(9.81)} = 3.16 m

3 0
2 years ago
What is an example of potential energy to kinetic energy?
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3 years ago
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g Estimate the number of photons emitted by the Sun in a second. The power output from the Sun is 4 × 1026 W and assume that th
vagabundo [1.1K]

Answer:

The value is N  =  1.107 *10^{45 }  \ photons    

Explanation:

From the question we are told that

   The  power output from the sun is  P_o =  4 * 10^{26} \  W

   The average wavelength of each photon is  \lambda  = 550 \  nm  =  550 *10^{-9} \  m

Generally the energy of each photon emitted is mathematically represented as

        E_c =  \frac{h * c  }{ \lambda }

Here  h is the Plank's constant with value  h  =  6.62607015 * 10^{-34} J \cdot s

          c is the speed of light with value  c =  3.0 *10^{8} \  m/s

So

       E_c =  \frac{6.62607015 * 10^{-34}  * 3.0 *10^{8}  }{ 550 *10^{-9} }          

=>   E_c =  3.614 *10^{-19} \  J          

Generally the  number of photons emitted by the Sun in a second is mathematically represented as

         N  =  \frac{P }{E_c}

=>      N  =  \frac{4 * 10^{26} }{3.614 *10^{-19}}

=>      N  =  1.107 *10^{45 }  \ photons    

5 0
2 years ago
A type of bacteria lives in the roots of plants. The bacteria helps the plants obtain nutrients. The roots are a moist home, whi
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Answer:

Rhizobium.

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The roots of most of these plants have nitrogen-fixing bacteria known as rhizobium. It helps to change nitrogen which is present in the air into the nitrates which are plants, and that is need to synthesize proteins.

Rhizobium bacteria occupy the root hairs which are present on the plants, they help the plant to grow and keep up with other plants. When the one plant dies, the fixed nitrogen which is present in it will release, and making it available for other plants, and this process helps to fertilize the soil in the environment.

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