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Natali5045456 [20]
3 years ago
5

In 1939 or 1940, Emanuel Zacchini took his human-cannonball act to an extreme. After being shot from a cannon, he soared over th

ree Ferris wheels and into a net. Assume that he is launched with a speed of 26.5 m/s and at an angle of 53.0° and neglect air drag.
a) Treating him as a particle, calculate his clearance over the first wheel.
b) If he reach maximum height over the middle wheel, by how much did he clear it?
c) How far from the cannon should the net's center have been positioned?

Physics
1 answer:
Arlecino [84]3 years ago
8 0
 We first determine the vertex by using the formula,<span>-b/2a = vertex, in order to get the values for the t-coordinate. That is why we got 
</span>
v_y=26.5 sin(53)=21.163v_x=26.5 cos(53)=15.948 
then

let x=0since you are going to land on a 3m tally=-.5(9.8)t^2+ 21.163*t
y=0=-4.9t+21.163t=4.31
vx*4.31= total distance travelled=68.88m
Then for the first wheel, you have 15.948m=vxdetermine the time when he reaches 23 meters, that is

23/15.948=1.44218 sec

substitute t with1.44218 sec, then determine the height.
h(1.44218)=20.329

determine vertex by using a graphing calculatort=2.1594s h=22.85m
using the time value of the vertex, determine horizontal distance travelled
34.438m away from cannon
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(Thought question) At noon On June 21 will the shadow length of 42 degrees north latitude be shorter or longer than it is on Mar
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5 0
3 years ago
A karate master wants to break a board by hitting the board swiftly with his hand. The master's hand has a mass of 0.30 kg, and
slava [35]

Answer:

f=-1380N

Explanation:

A karate master wants to break a board by hitting the board swiftly with his hand. The master's hand has a mass of 0.30 kg, and as it strikes the board, his hand has a velocity of 23.3 m/s. The master contacts the board for 0.0050 seconds

.the concluding part to the question should be

What is the impact force (impulse) on the board?

solution

from the Newton's second law of motion which states that

the rate of change in momentum is directly proportional to the force applied

f=m(v-u)/t

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6 0
4 years ago
Consider your moment of inertia about a vertical axis through the center of your body, both when you are standing straight up wi
jeka94

Answer:

     I₁ / I₂ = 1.43

Explanation:

To find the relationship of the two inertial memits, let's calculate each one, let's start at the moment of inertia with the arms extended

Before starting let's reduce all units to the SI system

       d₁ = 42 in (2.54 10⁻² m / 1 in) = 106.68 10⁻² m

       d₂ = 38 in = 96.52 10⁻² m

The moment of inertia is a scalar quantity for which it can be added, the moment of total inertia would be the moment of inertia of the man (cylinder) plus the moment of inertia of each arm

        I₁ = I_man + 2 I_ arm

Man indicates that we can approximate them to a cylinder where the average diameter is

         d = (d₁ + d₂) / 2

         d = (106.68 + 96.52) 10-2 = 101.6 10⁻² m

The average radius is

         r = d / 2 = 50.8 10⁻² m = 0.508 m

The mass of the trunk is the mass of man minus the masses of each arm.

        M = M_man - 0.2 M_man = 80 (1-0.2)

        M = 64 kg

The moments of inertia are:

A cylinder with respect to a vertical axis:         Ic = ½ M r²

A rod that rotates at the end:                            I_arm = 1/3 m L²

Let us note that the arm rotates with respect to man, but this is at a distance from the axis of rotation of the body, so we must use the parallel axes theorem for the moment of inertia of the arm with respect to e = of the body axis.

           I1 = I_arm + m D²

Where D is the distance from the axis of rotation of the arm to the axis of the body

          D = d / 2 = 101.6 10⁻² /2 = 0.508 m

Let's replace

          I₁ = ½ M r² + 2 [(1/3 m L²) + m D²]

Let's calculate

         I₁ = ½ 64 (0.508)² + 2 [1/3 8 1² + 8 0.508²]

         I₁ = 8.258 + 5.33 + 4.129

         I₁ = 17,717 Kg m² / s²

Now let's calculate the moment of inertia with our arms at our sides, in this case the distance L = 0,

          I₂ = ½ M r² + 2 m D²

          I₂ = ½ 64 0.508² + 2 8 0.508²

          I₂ = 8,258 + 4,129

          I₂ = 12,387 kg m² / s²

The relationship between these two magnitudes is

          I₁ / I₂ = 17,717 /12,387

          I₁ / I₂ = 1.43

3 0
3 years ago
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