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lara [203]
3 years ago
12

At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force t

hat the pole exerts back on the athlete is given by F(x) = (140 N/m )x - (190 N/m2 )x2 acting over a distance of 0.19 m .How much work is done on the athlete?
Physics
1 answer:
Salsk061 [2.6K]3 years ago
5 0

Answer:

The work done on the athlete is approximately 2.09 J

Explanation:

From the definition of the work done by a variable force:

\displaystyle{\int_{x_i}^{x_j}F(x)dx}

and substituting with the function of our problem:

\displaystyle{\int_{0}^{0.19}(140x-190x^2)dx\approx2.09\mathrm{J}}

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a uniform metre rule is pivoted at its centre and weights of 5 N and 12 N are hung at the 3 cm and 5 cm marks respectively. how
schepotkina [342]

In order to balance the stick on the pivot, the total "moments" must be equal on both sides.  A "moment" is (a weight) x (its distance from the center).

for the 5N weight: Moment = (5N) x (3 cm) = 15 N-cm

for the 12N weight: Moment = (12N) x (5 cm) = 60 N-cm

Sum of the moments trying to pull the stick down on that side = 75 N-cm

Whatever we hang on the other side has to provide a moment of 75 N-cm in the other direction.  We have a 25N weight. Where should we hang it ?  

(25N) x (distance from the pivot) = 75 N-cm

Distance from the pivot = (75 N-cm) / (25 N)

<em>Distance from the pivot =  3 cm </em>

8 0
3 years ago
A hammer slides down a roof that makes a 30.0°angle with the horizontal. What are the magnitudes of the components of the hammer
dlinn [17]
<h2>Answer:7.14ms^{-1},4.125ms^{-1}</h2>

Explanation:

Whenever an object is moving in a 2D frame,its motion can be analysed as if it is travelling in two independent 1D frames.

One of such independent 1D frames are along horizontal and another along vertical.

Let v be the total velocity.

Given that,v=8.25ms^{-1}

We call the horizontal velocity as v_{h} and the vertical velocity as v_{v}.

v_{h}=vCos\alpha

v_{v}=vSin\alpha

where \alpha is the angle between the object and horizontal.

It is given that \alpha =30^{0}

v_{h}=8.25\times Cos(30^{0})=7.14ms^{-1}

v_{v}=8.25\times Sin(30^{0})=4.125ms^{-1}

7 0
3 years ago
Jack rolled a marble down a ramp and recorded the potential energy and kinetic energy of the marble at different positions on th
mote1985 [20]

Answer: D

Height of marble from ground

Explanation:

From the formula of kinetic energy and potential energy,

K.E = 1/2mv^2

While

P.E = mgh

From all the parameters given from the question. You can see that mass is constant, acceleration due to gravity is also constant.

Independent variable must be a value that can varies.

Since Jack rolled a marble down a ramp and recorded the potential energy and kinetic energy of the marble at different positions on the ramp to see the effects on both energies.

This different position must be the height which will produce an effect on the potential and kinetic energy of the marble.

Independent variable always provides an effect for dependent variable. Which are kinetic energy and potential energy in this case.

Height of marble from ground is the right answer.

6 0
3 years ago
If toner particles in a laser printer have a negative charge, then what charge do you think the surface of the paper in the prin
Elina [12.6K]

Answer:

Explanation:

As it moves along, the paper is given a strong negative electrical charge by another corona wire. When the paper moves near the drum, its negative charge attracts the positively charged toner particles away from the drum.

3 0
2 years ago
A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin
gogolik [260]

Answer:

a) 23 m/s

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

       p_{o} = p_{f}  (1)

  • The initial momentum p₀, can be written as follows:

       p_{o} =  m_{1}  * v_{1o} + m_{2}* v_{2o} =   6.0 kg * 5.0 m/s + 2.0 kg * v_{2o}  (2)

  • The final momentum pf, can be written as follows:

        p_{f} = (m_{1} + m_{2} )* v_{f}  = 8.0 kg* (-2.0 m/s)  (3)

  • Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

       v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg}  = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s  (4)

  • This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.
6 0
3 years ago
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