I NEED HELP WITH THE QUESTION ABOVEEE

A kangaroo has a maximum gravitational potential energy during one jump of 770 J

KengaRu [80]

Answer:

k = 44000 N/m

Explanation:

Given the following data;

Maximum gravitational potential energy = 770 J

Elastic potential energy = 14% of 770 J = 14/100 * 770 = 107.8 J

Extension, x = 42 - 35 = 7cm to meters = 7/100 = 0.07 m

To find the spring constant, k;

The elastic potential energy of an object is given by the formula;

$E.P.E = \frac {1}{2}kx^{2}$

Substituting into the equation, we have;

$107.8 = \frac {1}{2}*k*0.07^{2}$

$107.8 = \frac {1}{2}*k*0.0049$

Cross-multiplying, we have;

$215.6 = k*0.0049$

$k = \frac {215.6}{0.0049}$

k = 44000 N/m

5 0

3 years ago

There are pluses and minuses to being a competitive person.<br> True <br> False

melomori [17]

True, because being a competitive person means that you like challenging yourself to do more things, but if you’re really competitive you might challenge yourself too hard, and might hurt yourself, which will make you not want to really compete with anything anymore

4 0

2 years ago

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a rugby player passes the ball 5.34 m across the field, where it is caught at the same height as it left his hand. at what angle

MakcuM [25]

Answer:

$31.035^{\circ}$

Explanation:

x = Displacement in x direction = 5.34 m

t = Time taken to travel the displacement

y = Displacement in y direction = 0

u = Initial velocity of ball = 7.7 m/s

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Displacement in x direction is given by

$x=u\cos\theta t\\\Rightarrow t=\dfrac{5.34}{7.7 \cos\theta}$

Displacement in y direction is given by

$y=u\sin\theta t-\dfrac{1}{2}gt^2\\\Rightarrow 0=7.7\sin\theta \dfrac{5.34}{7.7\cos\theta}-\dfrac{1}{2}\times 9.81 (\dfrac{5.34}{7.7\cos\theta})^2\\\Rightarrow 0=7.7\sin\theta-4.905\times \dfrac{5.34}{7.7\cos\theta}\\\Rightarrow 0=7.7^2\sin\theta \cos\theta-4.905\times 5.34\\\Rightarrow 0=7.7^2\dfrac{\sin2\theta}{2}-4.905\times 5.34\\\Rightarrow 0=7.7^2\sin2\theta-4.905\times5.34\times 2\\\Rightarrow \sin2\theta=\dfrac{4.905\times 5.34\times 2}{7.7^2}\\\Rightarrow 2\theta=\sin^{-1}\dfrac{4.905\times 5.34\times 2}{7.7^2}$