Which Diagram represents summer?

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

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

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0

3 years ago

Explain why ice floats in water. Explain why ice floats in water. Steam is more dense than water. Ice is the same density as wat

Pani-rosa [81]

Hey how's your day going

I hope after I answer that you understand and don't just paste my answer into your assignment!!! (<- read!!!)

Answer \|/

Ice is less dense than water.

Reason why \|/

When water freezes the molecules inside completely stop moving (They still vibrate but don't change their position much). In doing so, they spread out a touch which makes it less dense than liquid water. So ice floats

3 0

3 years ago

HELP ME PLEASE If an object is placed under a force of 20 N, it accelerates at a rate of 5 m/s^2. If the force is increased to 5

lara [203]

Answer:

b

Explanation:

3 0

3 years ago

Read 2 more answers

If a moving car speeds up until it is going twice as fast, how much kinetic energy doe s it have compared with its initial kinet

dezoksy [38]

The kinetic energy of an object of mass m and velocity v is given by
$K= \frac{1}{2} mv^2$

Let's call $v_i$ the initial speed of the car, so that its initial kinetic energy is
$K_i = \frac{1}{2} mv_i^2$
where m is the mass of the car.

The problem says that the car speeds up until its velocity is twice the original one, so
$v_f = 2 v_i$
and by using the new velocity we can calculate the final kinetic energy of the car
$K_f = \frac{1}{2} mv_f^2 = \frac{1}{2}m (2 v_i)^2 = 4 ( \frac{1}{2} mv_i^2)=4 K_i$
so, if the velocity of the car is doubled, the new kinetic energy is 4 times the initial kinetic energy.

6 0

3 years ago

A microphone is attached to a spring that is suspended from the ceiling, as the drawing indicates. Directly below on the floor i

svet-max [94.6K]

Answer:

$0.261\ \text{m}$

Explanation:

$\Delta f$ = Change in frequency = 2.1 Hz

$f$ = Frequency of source of sound = 440 Hz

$v_m$ = Maximum of the microphone

$v$ = Speed of sound = 343 m/s

$T$ = Time period = 2 s

We have the relation

$\Delta f=2f\dfrac{v_m}{v}\\\Rightarrow v_m=\dfrac{\Delta fv}{2f}\\\Rightarrow v_m=\dfrac{2.1\times 343}{2\times 440}\\\Rightarrow v_m=0.8185\ \text{m/s}$

Amplitude is given by

$A=\dfrac{v_m T}{2\pi}\\\Rightarrow A=\dfrac{0.8185\times 2}{2\pi}\\\Rightarrow A=0.261\ \text{m}$

The amplitude of the simple harmonic motion is $0.261\ \text{m}$ .

4 0

2 years ago

Which Diagram represents summer? ​

Which Diagram represents summer?