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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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The acceleration of the ball after leaving the hand is 9.8 m/s^2 downward

Explanation:

In order to find the acceleration of the ball during its motion, we have to study which forces are acting on it.

After the ball leaves the hand, if we neglect air resistance, there is only one force acting on the ball: the force of gravity, whose magnitude is

F=mg

where m is the mass of the ball and g is the acceleration of gravity (g=9.8 m/s^2), acting in the downward direction.

According to Newton's second law, the acceleration of the ball is given by

a=\frac{\sum F}{m}

where

\sum F is the net force acting on the ball

After the ball leaves the hand, the only force acting on it is the force of gravity, so we can substitute (mg) into the previous equation:

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This means that the acceleration of the ball remains 9.8 m/s^2 downward for the entire motion, after leaving the hand.

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3 years ago
Two blocks are connected by a light weight, flexible cord that passes over a frictionless pulley.Ifm1=2 kg and m2 = 3 kg, and bl
Vera_Pavlovna [14]

Answer:

t = 1.41 sec.

Explanation:

If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.

First, we need to find the value of  acceleration, which is the same for both blocks.

If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:

F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)

⇒ a = (\frac{(m₂-m₁)}({m₁+m₂} * g = g/5 m/s²

Once we got the value of a, we can use for instance this kinematic equation, and solve for t:

Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.

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Energy can be transformed from one form to another. The diagram shows one such process.
densk [106]

Answer:

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A 40.0 -kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 13
Irina18 [472]

The work done by the gravitational force = 0

Given the mass of the box = 40 kg

The box is initially at rest.

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The box is moved only in the horizontal direction by the applied force.

Gravitational force is applied in a direction perpendicular to the applied force. hence it doesn't do any work on the box.

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Answer:

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