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tester [92]
3 years ago
14

Bernie, whose mass is 70.0 kg, leaves a ski jump with a velocity of 21.0 m/s.What is Bernie’s momentum (kg*m/s) as he leaves the

ski jump?
Physics
1 answer:
zheka24 [161]3 years ago
6 0
Momentum = mass * velocity
Momentum = 70 * 21
Momentum = 1,470

Therefore, Bernie’s momentum was 1,470kg*m/s as he leaves the ski jump. :)
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what is the angular speed w of the system immediately after the collision in terms f the sstem parameters and I
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Answer: hello some part of your question is missing attached below is the missing detail

answer :

<em>w</em>f = M( v cos∅ )D / I

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The Angular speed <em>wf  </em>of the system after collision in terms of the system parameters  and I can be expressed as

considering angular momentum conservation

Li = Lf

M( v cos∅ ) D = ( ML^2 / 3  + mD^2 ) <em>w</em>f

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<em>w</em>f = M( v cos∅ )D / I

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Reading Check: (Describe) how pressure changes as the velocity of a fluid<br> increases.
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A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = \frac{I}{nqA}

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

<em>First let's calculate the number of free electrons per cubic meter (n)</em>

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

<em>Substitute the values of Nₐ, ρ and M into equation (ii) as follows;</em>

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

<em>Now let's calculate the drift electron</em>

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

<em>Substitute these values into equation (i) as follows;</em>

v = \frac{I}{nqA}

v = \frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0
3 years ago
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