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Radda [10]
3 years ago
7

A 26.0 kg child plays on a swing having support ropes that are 2.40 m long. A friend pulls her back until the ropes are 45.0 ∘ f

rom the vertical and releases her from rest. What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing? How fast will she be moving at the bottom of the swing? How much work does the tension in the ropes do as the child swings from the initial position to the bottom?
Physics
1 answer:
Sloan [31]3 years ago
8 0
A)Ep'=mgh=mgl(1-cosa).At the bottom of the swing Ep=0(reference level),so the potential energy as the child is just released is bigger than the potential energy at the bottom of the swing.;B)The speed of the child at the bottom of the swing-->v=√(2gh)=√[2gl(1-cosa)];C)I don't think that the tension does any work.
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A shaving or makeup mirror is designed to magnify your face by a factor of 1.40 when your face is placed 20.0cm in front of it
Dennis_Churaev [7]

Answer:

(a) convex mirror

(b) virtual and magnified

(c) 23.3 cm

Explanation:

The having mirror is convex mirror.

distance of object, u = - 20 cm

magnification, m = 1.4

(a) As the image is magnified and virtual , so the mirror is convex in nature.

(b) The image is virtual and magnified.

(c) Let the distance of image is v.

Use the formula of magnification.

m =-\frac{v}{u}\\1.4=-\frac{v}{-20}\\v =28 cm

Use the mirror equation, let the focal length is f.

\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\frac{1}{f}=\frac{1}{28}+\frac{1}{20}\\\frac{1}{f}=\frac{28+20}{560}\\f=11.67cm

Radius of curvature, R = 2 f = 2 x 11.67 = 23.3 cm

5 0
3 years ago
a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

4 0
3 years ago
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7 0
3 years ago
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iris [78.8K]

Answer:

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Explanation:

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8 0
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