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Describe the motion of the object in Graph B of figure 11-2.

kupik [55]

Answer : The motion of the object is, non-uniform

Explanation :

Uniform motion : It is defined as the movement of an object along a straight line with constant speed. It travels equal distances in equal time interval.

The average speed of an object is similar to the actual speed of an object. The distance-time graph shows a straight line.

Non-uniform motion : It is defined as the movement of an object along a straight line with variable speed. It travels unequal distances in equal time interval.

The average speed of an object is different to the actual speed of an object. The distance-time graph shows a curved line.

The given graph B is a non-uniform motion graph.

Hence, the motion of the object is, non-uniform

4 0

3 years ago

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A street light is at the top of a 19 foot tall pole. A woman 5.25 feet tall walks towards the pole with a speed of 7ft/sec along

bezimeni [28]

Answer:

The tip of her shadow is moving at the speed of 9.66 ft/sec

Explanation:

The height of the street light = 19 feet

The height of the woman = 5.25 feet

Distance between the woman and the base of the pole, x = 35 ft

The speed of the woman towards the pole, dx/dt = 7ft/sec

The distance from the base of the streetlight to the tip of the woman's shadow = y

The distance from the woman to the tip of her shadow = y - x

The diagram illustrating this description is shown below

Using similar triangle:

$\begin{gathered} \frac{19}{5.25}=\text{ }\frac{y}{y-x} \\ 19(y-x)\text{ = 5.25y} \\ 19y-19x\text{ = 5.25y} \\ 19y-5.25y\text{ = 19x} \\ 13.75y\text{ = 19x} \\ y\text{ = }\frac{19}{13.75}x \\ y\text{ = }1.38x \end{gathered}$

Find the derivative of both sides with respect to time, t

$\begin{gathered} \frac{dy}{dt}=\text{ 1.38}\frac{dx}{dt} \\ \frac{dy}{dt}=\text{ 1.38(7)} \\ \frac{dy}{dt}\text{ = }9.66\text{ ft/sec} \end{gathered}$

The tip of her shadow is moving at the speed of 9.66 ft/sec

7 0

1 year ago

A 0.100-kg stone rests on a frictionless, horizontal surface. A bullet of mass 6.00 g, travel- ing horizontally at 350 m>s, s

mariarad [96]

Answer:

Magnitude is 25.8 m/s and direction is $35.5^{o}$

Explanation:

From the law of conservation of linear momentum

$m_{b}v_{ib}+ m_{s}v_{is}= m_{b}v_{fb}+ m_{b}v_{fs}$ where $m_{b}$ and $m_{s}$ are masses of bullet and stones respectively, $v_{ib}$ and $v_{is}$ are the initial velocities of bullet and stone respectively, $v_{fb}$ and $v_{fs}$ are the final velocities of bullet and stone respectively

Substituting 6g=0.006 Kg for mass of bullet, 0.1Kg for mass of stone, 350 m/s for initial velocity of bullet and 250 m/s as final velocity for stone but initial velocity of stone is zero

$0.006 Kg *350 m/s (\hat i)+0.1 Kg*0=0.006 Kg* 250 m/s (\hat j)+0.1*v_{fs}$