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4 years ago

A student practicing for a track meet ran 250 m in 30 seconds. What was her average speed?

Physics

2 answers:

artcher [175]4 years ago

5 0

8.33 meters/sec. time = 30 sec. 30 sec. = 8.33 meters/sec.

sweet-ann [11.9K]4 years ago

4 0

Answer:

8.33 meters/sec.

time = 30 sec. 30 sec. = 8.33 meters/sec.

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a uniform metre rule of mass 100g balances at the 40cm mark when a mass X is placed at the 10cm mark what is the value of x.pls

Blababa [14]

Answer:

The fulcrum of the metre stick is at the 40 cm mark

100 g * 10 cm = 1000 g-cm clockwise torque

x * 30 cm = 1000 gm-cm = counterclockwise torque for balance

X = 1000 / (40 -10) = 1000 / 30 = 33.3 gm at 10 cm to balance

3 0

3 years ago

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:

The output power is greater in the interval from 1.0 s to 2.5 s

Explanation:

Physical Power

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

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4 years ago

Which objects can be electrically polarized?

SIZIF [17.4K]

neutral objects is correct

5 0

3 years ago

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What would best suit data that describes how a part relates to the whole 1. Line graph2. Bar graph 3. Circle graph 4. Scientific

Lerok [7]

The best option for visually representing how one part of data relates to a whole would likely be a circle graph. This will allow for the entirety of a data set to be seen at once, as well as providing an easy way to compare and contrast the different parts and how they relate to one another.

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4 years ago

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i

bija089 [108]

Answer:

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