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ohaa [14]
3 years ago
9

When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees

upon landing to reduce the force of the impact. A 75-kg man just before contact with the ground has a speed of 4.6 m/s. (a) In a stiff-legged landing he comes to a halt in 2.1 ms. Find the average net force that acts on him during this time. N (b) When he bends his knees, he comes to a halt in 0.09 s. Find the average force now. N (c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of these forces, find the force of the ground on the man in parts (a) and (b). stiff legged landing N bent legged landing

Physics
1 answer:
dolphi86 [110]3 years ago
6 0

Answer:

a) 1.725*10^5 N

b) 3.83*10^3 N

c) i) 173.24 kN

c) ii) 4.57 kN

Explanation:

See the attachment for calculations

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Net force of 200 newtons is applied to a wagon for 3 seconds
AveGali [126]

Momentum of the wagon increases by (200 x 3)

                                                     = 600 newton-sec

                                                     = 600 kg-m/sec
 
6 0
4 years ago
If potential energy is at point A is 12 joules, what is the potential energy at B ?
deff fn [24]

Answer:

incomplete information

Explanation:

Where is diagram? no point A and point B given

7 0
3 years ago
How do you rationalize the tension being used in Tennis Racket strings using the concept of impulse and momentum?
zheka24 [161]

Answer:

The momentum, ΔP, and therefore, kinetic energy given to the ball in a serve is the result of the product of the tension force, 'F', in the string and the time of contact, Δt, between the ball and the string

ΔP = F × Δt

Explanation:

The impulse, ΔP, is the produce of the force, 'F', applied to a body for a given period of time, Δt', that gives motion to the body, and it is equal to the change of momentum of the body

ΔP = F × Δt

The momentum, 'P', of a body is the product of the mass, 'm', of the body and its velocity, 'v'

P = m × v

Tension is the axial pulling force of a string

T = Axial Force, F_{axial}

The tension used in Tennis Racket strings is between 40 to 65 lbs.

When high tension is used in the string, the string is taut, and the contact duration between the Racket string and the ball is minimal, and the player needs to use more force to obtain a high momentum, and therefore, energy in the ball, which reduces control, and increase stress, as force is more emphasized

When low tension is used in the string, the Tennis Racket strings are more elastic. During a serve, the ball pushes the strings further back into the racket, such that the ball spends more time in contact with the string, (Δt is larger), and therefore, the impulse, F·Δt = ΔP, given to the ball is larger, therefore, the ball has a larger change in momentum, and therefore more energy in the intended direction.

However, a very slackened string will increase the increase area and time (large Δt) of contact of the ball and the racket such that the force given to the ball, F = ΔP/(large Δt) is reduced and therefore reduce the likelihood of gaining points from a serve against an opponent with a much forceful return of a serve.

3 0
3 years ago
According to the directions on a can of frozen orange juice concentrate, 1 can of concentrate is to be mixed with 3 cans of wate
WARRIOR [948]

Answer:

25

Explanation:

Given:

1 can of concentrate requires 3 cans of water

Now,

Total ounces in 200 6-ounce cans = 1200 ounces

also,

for 1 can of concentrate requires 3 cans of water

thus,

for 12 ounces can water can required = 3 × 12 ounces = 36 ounces of cans

Thus,

total ounce of juice per can = 12 + 36 = 48 ounces per can

therefore,

the number of 12-ounce cans required  are = \frac{\textup{Ttoal ounces}}{\textup{Ounces per can}}

or

= \frac{\textup{1200}}{\textup{48}}

or

the number of 12-ounce cans required  are = 25

8 0
3 years ago
You push your couch a distance of 3.9 m across the living room floor with a horizontal force of 220.0 n. the force of friction i
Mrac [35]

The general formula to calculate the work is:

W=Fd \cos \theta

where F is the force, d is the displacement of the couch, and \theta is the angle between the direction of the force and the displacement. Let's apply this formula to the different parts of the problem.


(a) Work done by you: in this case, the force applied is parallel to the displacement of the couch, so \theta=0^{\circ} and \cos \theta=1, therefore the work is just equal to the product between the horizontal force you apply to push the couch and the distance the couch has been moved:

W=Fd=(220.0 N)(3.9 m)=858 J


(b) work done by the frictional force: the frictional force has opposite direction to the displacement, therefore \theta=180^{\circ} and \cos \theta=-1. Therefore, we must include a negative sign when we calculate the work done by the frictional force:

W=-Fd=-(144.0 N)(3.9 m)=-561.6 J


(c) The work done by gravity is zero. In fact, gravity (which points downwards) is perpendicular to the displacement of the couch (which is horizontal), therefore \theta=90^{\circ} and \cos \theta=0: this means

W=0.


(d) Work done by the net force:

The net force is the difference between the horizontal force applied by you and the frictional force:

F=220 N-144 N=76 N

And the net force is in the same direction of the displacement, so \theta=0^{\circ} and \cos \theta=1 and the work done is

W=Fd=(76 N)(3.9 m)=296.4 J


4 0
4 years ago
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