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g100num [7]
3 years ago
6

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...
Physics
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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Answer:

Subcategories: Ice field

Explanation:

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How long would it take for Sofia to walk 300 meters if she is walking at a velocity of 2.5 m/s?
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Answer:

Time=120seconds

Explanation:

S=300m

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t=?

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7 0
3 years ago
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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

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Static electricity can be an alternating current.
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