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3 years ago

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A small dog is trained to jump straight up a distance of 1.2m. How much kinetic energy does the 7.2kg dog need to jump this high

? (The acceleration due to gravity is 9.8m/s2.) Please show work & and explain in simple terms...

2 answers:

Bas_tet [7]3 years ago

5 0

Remember that since energy is conserved, the kinetic energy it takes to jump to a height of 1.2 m is just the same as the potential energy difference between ground level and the 1.2 m height.
Potential energy = (mass)(gravity)(height) = (7.2 kg)(9.81 m/s^2)(1.2 m) = 84.76 J
Therefore, the kinetic energy required for the dog to jump to a 1.2-m height is 84.76 Joules.

lesantik [10]3 years ago

3 0

The answer is A. 85 J
This is because we rounded 84.7 to 85

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The radius of the rod is half the diameter: $r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m$ , so the cross-sectional area is

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The equation that gives the change in resistance as a function of the temperature is

$R(T)=R_0 (1+\alpha(T-T_0))$

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$R_0=0.81 \Omega$ is the resistance at the original temperature (20.0°C)

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$T_0 = 20^{\circ}$

Solving the formula for $\alpha$ , we find

$\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C$

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