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schepotkina [342]
3 years ago
10

A heavy boy and a lightweight girl are balanced on a massless seesaw. The boy moves backward, increasing his distance from the p

ivot point. What happens to the seesaw? It is impossible to predict without knowing additional information. The side the boy is sitting on will tilt downward. The side the girl is sitting on will tilt downward. Nothing; the seesaw will remain balanced.
Physics
1 answer:
victus00 [196]3 years ago
6 0

Answer:

The side the boy is sitting on will tilt downward.

Explanation:

According to the law of moments when the same force is applied at a greater distance from the pivot then the effect of moment is greater about that point.

<u>Mathematically momentum is given as:</u>

M=F\times r

where:

F is the applied force at a distance 'r' acting in a direction perpendicular to the line joining the point of application and the hinge.

  • Moment is the rotational effect of the applied force on the body.

<em>When the boy of a heavier mass than the girl was sitting on a balanced see-saw then it is certain that he was closer to the hinge than the girl to balance the turning effect (in case of an unbiased see-saw). When the body moves farther his weight is same but the radial distance from the hinge increases which increases his moment of weight.</em>

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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
A cat climbs 10 m directly up a tree.
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 <span>there is no horizontal displacement if he went straight up 

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6 0
3 years ago
Can someone help me?!!!!!
ladessa [460]

Answer:

magnitude: 21.6; direction: 33.7 degrees

Explanation:

When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have

vector: (-3,-2)

Scalar: -6

so the vector multiplied by the scalar will have components

(-3\cdot (-6), -2 \cdot (-6))=(18,12)

The magnitude is given by Pythagorean's theorem:

m=\sqrt{18^2+12^2}=21.6

and the direction is given by the arctan of the ratio between the y-component and the x-component:

\theta = tan^{-1} (\frac{12}{18})=33.7^{\circ}

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