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3 years ago

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You are launching a 2 kg potato out of a potato cannon. The cannon is 2.0 m long and is aimed 70 degrees above the horizontal. I

t exerts a 52 N force on the potato. What is the kinetic energy (in J) of the potato as it leaves the muzzle of the potato cannon?

1 answer:

DochEvi [55]3 years ago

8 0

Answer:

Explanation:

The net force on the potatoes is given by:

F= 52 - mgSintheta

F= 52- (2×9.8× Sin70°)

F = 52 -18.4

F= 33.58N

Using Newton's 2nd law

F = ma

a=F/m = 33.58/ 2 = 16.79m/s^2

Using the equation of motion:

V^2= u^2 + 2as

V^2 = 0 + 2× 16.79 x2

V^2 = 67.16

V=sqrt(68.16)

V= 8.195m/s This is the exit velocity of the potatoes

Kinetic energy, K.E = 1/2mv^2

KE= 1/2 × 2 × 8.195^2

KE = 67.16J

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Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

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When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 49.

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Answer:

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When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth.

If the muscle generates a force

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Answer:

<em>a. 4.21 moles</em>

<em>b. 478.6 m/s</em>

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Explanation:

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Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

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