How to use unbalanced forces in a sentence

During her high bar routine from question 2, Gabby Douglas slipped and falls from the high bar, landing on a 10 cm thick gymnast

tankabanditka [31]

a) 122.5 J

b) -122.5 J

c) -1884.6 N

d) -3769.2 N

e) $-753.8 m/s^2$

f) $a=-76.9 g$

Explanation:

a)

The kinetic energy of an object is the energy possessed by the object due to its motion.

Mathematically, it is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

Here, we want to find the kinetic energy of the head just before hitting the mat.

At that instant, the speed is:

v = 7 m/s

The mass of the head is:

m = 5 kg

So, the kinetic energy is

$K=\frac{1}{2}(5)(7)^2=122.5 J$

b)

According to the work-energy theorem, the work done by a force on an object is equal to the change in kinetic energy of the object:

$W=K_f - K_i$

where

W is the work done

$K_f$ is the final kinetic energy

$K_i$ is the initial kinetic energy

In this problem:

$K_i=122.5 J$ is the kinetic energy of the head just before hitting the mat

$K_f=0 J$ is the final kinetic energy (since the head comes to a stop)

So, the work done by the mat is:

$W=0-122.5 = -122.5 J$

The work is negative because the force exerted by the mat is opposite to the direction of motion of the head.

c)

The work exerted by a force on an object is given by

$W=Fd$

where

F is the force applied

d is the displacement of the object

W is the work done

In this problem:

W = -122.5 J is the work done by the mat on the head

d = 6.5 cm = 0.065 m is the displacement of the head (since it deflects the mat by this amount)

So, the average force exerted by the mat on the head is:

$F=\frac{W}{d}=\frac{-122.5}{0.065}=-1884.6 N$

(the negative sign indicates that the force is in direction opposite to the motion of the head)

d)

The force calculated in part d) represents the average force exerted by the mat on the head:

$F_{avg}=-1884.6 N$

We can assume that as the head first hits the mat, the initial force is zero, then increases at a constant rate up to a peak value of $F_{peak}$ , then it decreases again until the head stops.

In this case, the relationship between average force and peak force is:

$F_{avg}=\frac{0+F_{peak}}{2}$

And therefore, the peak impact force exerted by the mat on the head is:

$F_{peak}=2F_{avg}=2(1884.6)=-3769.2 N$

e)

The peak acceleration of the head can be found by using Newton's second law, which states that:

$F=ma$

where

F is the force on the head

m is the mass of the head

a is the acceleration

Here we have:

F = -3769.2 N is the peak force

m = 5 kg is the mass of the head

So, solving for the acceleration, we find:

$a=\frac{F}{m}=\frac{-3769.2}{5}=-753.8 m/s^2$

f)

The value of the acceleration due to gravity is

$g=9.8 m/s^2$

Here we want to express the peak acceleration of the head in terms of the acceleration due to gravity; so we can write:

$a=Ng$

where

$a=-753.8 m/s^2$ is the peak acceleration

N is the ratio between the peak acceleration and the gravity acceleration

Solving for N,

$N=\frac{a}{g}=\frac{-753.8}{9.8}=-76.9$

This means that the peak acceleration can be written as

$a=-76.9 g$

6 0

3 years ago

The caste system is an example of

snow_lady [41]

Answer:

It is example of Closed system

7 0

3 years ago

X-rays cannot pass through Earth's atmosphere. Which of these is the best location to place a telescope used to observe x-rays f

Amiraneli [1.4K]

Mountains, tops of buildings, and high-flying aircraft are all part of Earth's atmosphere, no matter how high they are. On the other hand, space doesn't belong to our atmosphere, it is outside of it. Having this in mind, the best location to place a telescope used to observe x-rays from stars is in space.

7 0

3 years ago

Read 2 more answers

A rod of length L is pivoted about its left end and has a force F applied perpendicular to the other end. The force F is now rem

timurjin [86]

Answer:

$F' = 4F$

Explanation:

When force applied to the end of the rod in perpendicular direction then net torque on the rod is given as

$\tau = L F sin90$

$\tau = LF$

Now another force is applied at mid point of the rod at an angle of 30 degree with the rod

so new value of torque is given as

$\tau = \frac{L}{2}F' sin\theta$

$LF = \frac{L}{2}F' sin30$

$LF = \frac{F'L}{4}$

so we have

$F' = 4F$

3 0

3 years ago

For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant

Scilla [17]

Consider the projectile launched at initial velocity V at angle θ relative to the horizontal.
Neglect wind or aerodynamic resistance.

The initial vertical velocity is Vsinθ.
When the projectile reaches its maximum height of h, its vertical velocity will be zero.
If the time taken to attain maximum height is t, then
0 = Vsinθ - gt
t = (Vsinθ)/g, where g = acceleration due to gravity.

The horizontal component of launch velocity is Vcosθ. This velocity remains constant because aerodynamic resistance is ignored.
The time to travel the horizontal distance D is twice the value of t.
Therefore
D = Vcosθ*[(2Vsinθ)/g]
= (2V²sinθ cosθ)/g
= (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, $\frac{dD}{d \theta} =0$
That is,
$\frac{2V^{2}}{g} cos(2 \theta )=0$

Because $\frac{2V^{2}}{g} \neq 0$ , therefore cos(2θ) = 0.
This is true when 2θ = π/2 => θ = π/4.

It has been shown that the maximum horizontal traveled can be attained when the launch angle is π/4 radians, or 45°.

4 0

4 years ago