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Amanda [17]
4 years ago
13

The u.s army’s parachuting team, the Golden Knights, are on a routine

Physics
1 answer:
quester [9]4 years ago
7 0

The average force is -2600 N

Explanation:

First of all, we need to calculate the acceleration of the man during the collision, which is given by the suvat equation:

v^2-u^2=2as

where:

v = 0 is his final velocity (he comes to a stop)

u = 4.0 m/s is the initial velocity

a is the acceleration

s = 0.20 m is the distance covered

Solving for a,

a=\frac{v^2-u^2}{2s}=\frac{0-4.0^2}{2(0.20)}=-40 m/s^2

The negative sign indicates that it is a deceleration.

Now we can find the average force on the man by using Newton's second law of motion:

F=ma

where

m = 65 kg is the mass

a=-40 m/s^2

And substituting,

F=(65)(-40)=-2600 N

where the negative sign indicates the force is in the direction opposite to the motion.

Learn more about force and Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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Now let’s apply the work–energy theorem to a more complex, multistep problem. In a pile driver, a steel hammerhead with mass 200
andrew11 [14]

Answer:

a) v = 7.67

b) n = 81562 N

Explanation:

Given:-

- The mass of hammer-head, m = 200 kg

- The height at from which hammer head drops, s12 = 3.00 m

- The amount of distance the I-beam is hammered, s23 = 7.40 cm

- The resistive force by contact of hammer-head and I-beam, F = 60.0 N

Find:-

(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

Solution:-

- We will consider the hammer head as our system and apply the conservation of energy principle because during the journey of hammer-head up till just before it hits the I-beam there are no external forces acting on the system:

                                   ΔK.E = ΔP.E

                                  K_2 - K_1 = P_1- P_2

Where,  K_2: Kinetic energy of hammer head as it hits the I-beam

             K_1: Initial kinetic energy of hammer head ( = 0 ) ... rest

             P_2: Gravitational potential energy of hammer head as it hits the I-beam. (Datum = 0)

             P_1: Initial gravitational potential energy of hammer head      

- The expression simplifies to:

                                K_2 = P_1

Where,                     0.5*m*v2^2 = m*g*s12

                                v2 = √(2*g*s12) = √(2*9.81*3)

                                v2 = 7.67 m/s

- For the complete journey we see that there are fictitious force due to contact between hammer-head and I-beam the system is no longer conserved. All the kinetic energy is used to drive the I-beam down by distance s23. We will apply work energy principle on the system:

                               Wnet = ( P_3 - P_1 ) + W_friction

                               Wnet = m*g*s13 + F*s23

                               n*s23 = m*g*s13 + F*s23

Where,    n: average force the hammerhead exerts on the I-beam.

               s13 = s12 + s23

Hence,

                             n = m*g*( s12/s23 + 1) + F

                             n = 200*9.81*(3/0.074 + 1) + 60

                             n = 81562 N

                               

                                                   

6 0
3 years ago
Calculate the force of an object that has a mass of 10kg and an acceleration of 4m/s²
ICE Princess25 [194]
The answer is A-40N.
8 0
3 years ago
To properly record a measurements, you must record which of the following?
serg [7]
I'm not too sure about this but unit is definitely one of the answers. If "number" refers to (for example) the "6" in 6 inches then it should also count. I think "symbol" could also just be unit so I'm not sure about that.
5 0
4 years ago
A horizontal spring with a spring constant 100 N/m is compressed 20cm and used to launch a 2.5kg box across a frictionless, hori
Verdich [7]

Answer:

   d = 0.544 m

Explanation:

To solve this problem we must work in two parts: one when the surface has no friction and the other when the surface has friction

Let's start with the part without rubbing, let's find the speed that the box reaches., For this we use the conservation of mechanical energy in two points: maximum compression and when the box is free (spring without compression)

Initial, maximum compression

    Em₀ = Ke = ½ k x²

Final, free box without compressing the spring

    Em_{f} = K = ½ m v²

    Emo = Em_{f}

    ½ k x² = ½ m v²

    v = √ (k / m) x

Let's reduce the SI units measures

    x = 20 cm (1m / 100cm) = 0.20 m

    v = √ (100 / 2.5) 0.20

    v = 1,265 m / s

Let's work the second part, where there is friction. In this part the work of the friction force is equal to the change of mechanical energy

   W_{fr} = ΔEm = Em_{f} - Em₀

   W_{fr} = - fr d

Final point. Stopped box

   Em_{f} = 0

Starting point, starting the rough surface

   Em₀ = K = ½ m v²

With Newton's second law we find the force of friction

    fr = μ N

    N-W = 0  

   N = W = mg

   fr = μ mg

   -μ m g d = 0 - ½ m v²

   d = ½ v² / (μ g)

Let's calculate

   d = ½ 1,265² / (0.15 9.8)

   d = 0.544 m

6 0
3 years ago
hich muscle fibers are best suited for activities that involve lifting large, heavy objects for a short period of time? cardiac
Temka [501]

Answer:

Dead lifting uses tho muscle fundamentals

Explanation:

6 0
4 years ago
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