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iogann1982 [59]
3 years ago
15

A girl is standing still in the middle of a baseball field. The girl catches the baseball and begins to move in the same directi

on as the thrown ball. What type of collision is being described?
Physics
2 answers:
ch4aika [34]3 years ago
7 0

Answer:

inelastic collision

Explanation:

an elastic collision occurs when both total momentum and total kinetic energy are conserved, while during an inelastic collision only the total momentum is conserved, while the total kinetic energy is not. In particular, if two objects stick together after the collision, only the momentum is conserved, so it is an inelastic collision.

In this problem, we have a girl that catches a ball and then continues together with the ball: the two 'objects' (girl and ball) stick together after the collision, so this is an inelastic collision.

marusya05 [52]3 years ago
5 0
A. inelastic, since the girl moves in the same direction as the thrown ball
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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
A bike rider approaches a hill with a speed of 8.5 m/s. The total mass of the bike rider is 91kg. What is the kinetic energy of
Anestetic [448]

a) The kinetic energy (KE) of an object is expressed as the product of half of the mass (m) of the object and the square of its velocity (v²):

KE = \frac{1}{2}m* v^{2}

It is given:

v = 8.5 m/s

m = 91 kg

So:

KE= \frac{1}{2}*91*8.5^{2} =3,287.4J


b) We can calculate height by using the formula for potential energy (PE):
PE = m*g*h

In this case, h is eight, and PE is the same as KE:
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m = 91 kg

g = 9.81 m/s² - gravitational acceleration

h = ? - height


Now, let's replace those:

3,287.4= 91 * 9.81 * h

⇒ h = 3,287.4/(91*9.81) = 3,287.4/892.7 = 3.7 m

3 0
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Plz help!
Strike441 [17]
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Alekssandra [29.7K]

Answer: "B" Changing Position

Great Question!

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m_a_m_a [10]

Answer:

i think c

Explanation:

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7 0
3 years ago
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