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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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The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?
BigorU [14]

1. Frequency: 7.06\cdot 10^{14} Hz

The frequency of a light wave is given by:

f=\frac{c}{\lambda}

where

c=3\cdot 10^{-8} m/s is the speed of light

\lambda is the wavelength of the wave

In this problem, we have light with wavelength

\lambda=425 nm=425\cdot 10^{-9} m

Substituting into the equation, we find the frequency:

f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz


2. Period: 1.42 \cdot 10^{-15}s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

The frequency of this light wave is 7.06\cdot 10^{14} Hz (found in the previous exercise), so the period is:

T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s


4 0
3 years ago
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Natalija [7]

Answer:

0.775 m

Explanation:

As the car collides with the bumper, all the kinetic energy of the car (K) is converted into elastic potential energy of the bumper (U):

U=K\\frac{1}{2}kx^2 = \frac{1}{2}mv^2

where we have

k=2\cdot 10^6 N/m is the spring constant of the bumper

x is the maximum compression of the bumper

m=30\cdot 10^4 kg is the mass of the car

v=2.0 m/s is the speed of the car

Solving for x, we find the maximum compression of the spring:

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(30\cdot 10^4 kg)(2.0 m/s)^2}{2\cdot 10^6 N/m}}=0.775 m

8 0
3 years ago
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Nady [450]

Answer:

244mm

Explanation:

I₁ = 3.35A

I₂ = 6.99A

μ₀ = 4π*10^-7

force per unit length (F/L) = 6.03*10⁻⁵N/m

B = (μ₀ I₁ I₂ )/ 2πr .........equation i

B = F / L ..........equation ii

equating equation i & ii,

F / L = (μ₀ I₁ I₂ )/ 2πr

Note F/L = B = F

F = (μ₀ I₁ I₂ ) / 2πr

2πr*F = (μ₀ I₁ I₂ )

r = (μ₀ I₁ I₂ ) / 2πF

r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵

r = 1.4713*10⁻⁵ / 6.03*10⁻⁵

r = 0.244m = 244mm

The distance between the wires is 244m

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Norma-Jean [14]

vf ^2 = kx^2/m = 56(0.75)^2 / 2.5 = 12.6


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Elena L [17]

Answer:

D. Downshift to allow you to turn more sharply

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