Refraction occurs when a wave

Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo

ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

a)

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

$F_{net}=ma$

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

$a=9.50 m/s^2$ (acceleration)

So, the net force is

$F_{net}=(94.0)(9.50)=893 N$

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

b)

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

$a=9.50 m/s^2$ (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

$v=0+(9.50)(0.890)=8.5 m/s$

c)

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

$W=K_f - K_i = \frac{1}{2}mv^2-0$

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

$K_i = 0 J$ is the initial kinetic energy

So the work done is

$W=\frac{1}{2}(94.0)(8.5)^2=3396 J$

The power expended is given by

$P=\frac{W}{t}$

where

t = 0.890 s is the time elapsed

Substituting,

$P=\frac{3396}{0.890}=3816 W$

d)

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

$F_1=893 N$

We know that the average force for the whole race is

$F_{avg}=820 N$

Which can be rewritten as

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}$

And solving for $F_2$ , we find the average force exerted by Bolt on the ground during the second phase:

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N$

The net force exerted by Bolt during the second phase can be written as

$F_{net}=F_2-D$ (1)

where D is the air drag.

The net force can also be rewritten as

$F_{net}=ma$

where

$a=\frac{v-u}{t}$ is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

$a=\frac{12.4-8.5}{8.69}=0.45 m/s^2$

So we can now find the average drag force from (1):

$D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N$

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

$\Delta E=W=Ds$

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

$s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m$

And so,

$\Delta E=Ds=(770.2)(90.6)=69780 J$

e)

The power that Bolt must expend just to voercome the drag force is given by

$P=\frac{\Delta E}{t}$

where

$\Delta E$ is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

$\Delta E=69780 J$

t = 8.69 s is the time elapsed

Substituting,

$P=\frac{69780}{8.69}=8030 W$

And we see that it is about twice larger than the power calculated in part c.

3 0

2 years ago

Increasing which two things would increase gravitational potential energy?

sukhopar [10]

Answer:

Mass and height

Explanation:

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about $9.8 m/s2.$

Which is represented as;

$PE_g=mgh$

$PE_g$ stands for gravitational potantial energy,

m stands for mass of object,

g is the gravitational constant and

h is the height.

Here we see that mass of object and height is directly proportional to the gravitational potential energy.

That means increasing in mass and height will result in increasing gravitational potential energy.

7 0

2 years ago

Hi please, I Have An attachment on Waves, Just two Objective Questions Whoever Answers Will be Marked Brainliest thank you.

kifflom [539]

Answer:

The first answer is W and Z, since they appear to be a period apart. Dont know the second question. I did what I could, hope someone can answer the second.

5 0

2 years ago

3. El tambor de una lavadora que gira a 3 000 revoluciones por minuto (rpm) se acelera uniformemente hasta que alcanza las 6 000

cestrela7 [59]

Answer:

d) α = 1693.5 rad / s² , a = 392.7 m / s² , a_total = α √(R² +1) ,

e) tan θ = a / α

Explanation:

This is an exercise in linear and angular kinematics.

We initialize reduction of all the magnitudes to the SI system

w₀ = 3000 rev / min (2π rad / 1rev) (1min / 60s) = 314.16 rad / s

w = 6000 rev / mi = 628.32 rad / s

θ = 12 rev = 12 rev (2π rad / 1 rev) = 75.398 rad

d) ask for centripetal, tangential and total acceleration.

Let's start by looking for centripetal acceleration, let's use the formula

w² = w₀² + 2 α θ

α = (w²- w₀²) / 2θ

we calculate

α = (628.32²2 - 314.16²) / 2 75.398

α = 1693.5 rad / s²

the quantity is linear and angular are related

the linear or tangential acceleration is

a = α R

where R is the radius of the drum

a = 1693.5 R

Unfortunately you do not give the radius of the drum for a complete calculation, but suppose it is a washing machine drum R = 20 cm = 0.20 m

a = 1693.5 0.20

a = 392.7 m / s²

the total acceleration is

a_total = √(a² + α²)

a_total = √ (α² R² + α²)

a_total = α √(R² +1)

e) The centripetal acceleration is directed towards the center of the movement is radial and its magnitude is constant

Tangential acceleration is tangency to radius and its value varies proportionally radius

the total accelracicon is the result of the vector sum of the two accelerations and their directions given by trigonometry

tan θ = a / α

the angular velocity increases linearly when with centripetal acceleration

8 0

2 years ago

A missle is fired horizontally with an initial velocity of 45 m/s from the top of a building 75 m high.

NARA [144]

The horizontal range of the missile is b) 176 m

Explanation:

The motion of the missile is a projectile motion, so it consists of two independent motions:

- A uniform motion with constant velocity along the horizontal direction

- A uniformly accelerated motion with constant acceleration (equal to the acceleration of gravity) in the vertical-downward direction

To find the time of flight of the missile, we study the vertical motion. We can use the following suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where:

s = 75 m is the vertical displacement of the missile (the height of the building)

$u_y=0$ is the initial vertical velocity (the missile is thrown horizontally)

t is the time of flight

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find the time of flight:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(75)}{9.8}}=3.91 s$

This means that the missile takes 3.91 s to reach the ground.

Now we study the horizontal motion: the missile moves with a constant horizontal velocity of

$v_x = 45 m/s$

Therefore, the distance covered in a time t is

$d=v_x t$

and by substituting t = 3.91 s, we find the horizontal range of the missile:

$d=(45)(3.91)=176 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0

2 years ago