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Zigmanuir [339]
3 years ago
10

In February 1955, a paratrooper fell 370 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 1.2×10⁵N.
What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?
Physics
1 answer:
Nonamiya [84]3 years ago
7 0

Answer:

(a) 1.11 m

(b) 4760 kgm/s  

Explanation:

height of plane (h) = 370 m

velocity (v) = 56 m/s

mass (m) = 85 kg

force = 1.2 x 10^{5} N = 120,000 N

(a) We can get the minimum depth of snow from the equation below

   force x depth = kinetic energy on impact

 f x d = 0.5 x m x v^{2}

 120000 x d = 0.5 x 85 x 56^{2}

 d= (0.5 x 85 x 56^{2}) ÷ 120000 = 1.11 m

(b) the magnitude of impulse is equal to the momentum of the paratrooper and his gear

 = m x v

= 85 x 56 = 4760 kgm/s  

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An audience of 2250 fills a concert hall of volume 32000 m^3. If there were no ventilation, by how much would the temperature of
Nonamiya [84]
246 and 64 minutes later
7 0
3 years ago
Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the
sdas [7]

Answer:

The new distance is     d = 0.447 d₀

Explanation:

The electric out is given by Coulomb's Law

         F = k q₁ q₂ / r²

This electric force is in balance with tension.

We reduce the charge of sphere B to 1/5 of its initial value (q_{B}=q₂ = q₂ / 5) than new distance (d = n d₀)

dat

     q₁ = q_{A}

     q₂ = q_{B}

     r = d₀

In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

         F = k q₁ q₂ / d₀²

         F = k q₁ (q₂ / 5) / (n d₀)²

         .k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)

          5 n² = 1

          n = √ 1/5

          n = 0.447

The new distance is

         d = 0.447 d₀

6 0
3 years ago
An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f
sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

f=\frac{v_s}{2L}         (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

f=\frac{343m/s}{2(0.47228m)}=362.36Hz      

Closed tube:

f'=\frac{v_s}{4L'}

L': length of the closed tube = 0.702821m

f'=\frac{343m/s}{4(0.702821m)}=122.00Hz

Next, you use the following formula for the beat frequency:

f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0
3 years ago
How many types of crust are there? List the names and one detail about each type
valina [46]

Answer:

There are two different types of crust: thin oceanic crust that underlies the ocean basins, and thicker continental crust that underlies the continents. These two different types of crust are made up of different types of rock.

Explanation:

There ya go !

:) hoping this helped ya out

5 0
3 years ago
Una partícula se mueve en el plano XY efectúa un desplazamiento mientras actúa sobre ella una fuerza constante. X= (4i + 3j) m,
dsp73

Answer:

a) La magnitud del desplazamiento es de 5 m

La magnitud de la fuerza es 20 N

b) El trabajo realizado por la fuerza es de 100 J

c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}

Lo que da;

\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}

Lo que da;

\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N

b) El trabajo, W, realizado por la fuerza = Fuerza, F × Distancia, X

∴ Ancho = 20 N × 5 m = 100 N · m = 100 J

c) La dirección de la fuerza viene dada por la siguiente fórmula;

tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right )  = 38.9^{\circ}

La dirección del plano viene dada por la siguiente fórmula;

tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right )  = 38.9^{\circ}

Por tanto, el ángulo entre la fuerza y el plano = 0 °

La fuerza actúa a lo largo del plano.

6 0
3 years ago
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