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Anastaziya [24]
4 years ago
9

2. An airplane travels 335

Physics
1 answer:
Mama L [17]4 years ago
3 0

Answer: The answer is 716.8 km/hr

Explanation:

The main idea here is the convert the given units to km and hr. Here's the conversion factors.

1mi= 1.6km

1hr= 60min

We have to multiply 335miles by 1.60km/mi in order to get the answer in km then again we have to multiply 45 minutes by 1hr/60min in order to get the answer in hr.

The last step is to calculate the speed using the formula : v=d/t

=537.6km/0.75hr

=716.8km/hr

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A 2.43ug particle moves at 1.97 x 108 m/s. What is its momentum?
Ira Lisetskai [31]

Answer:

0.48 kgm/s

Explanation:

m = mass of the particle = 2.43 μg = 2.43 x 10⁻⁶ x 10⁻³ kg = 2.43 x 10⁻⁹ kg

v  = velocity of the particle = 1.97 x 10⁸ m/s

p = momentum of the particle

momentum of the particle is given as

p = m v

inserting the values

p = (2.43\times 10^{-9})(1.97\times 10^{8})

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For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

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andrew11 [14]

Answer:

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For this to occur, the surface incident must be smooth and without rough patterns on it.

A path way with rough rocks, small patch of soil and rough logs will give off diffused reflection

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