If the released energy of one earthquake is 100 times that of another, how much greater is its magnitude on the richter scale?

what factors go into decisions about changing what materials should be used when building a products

Answer:

you need to consider the use for the product, how brittle the materials are, how they react to certain things, the cost of the materials, the durability and flexibility of the materials, and how easy to obtain the materials are as well as how they would work and how they would hold

6 0

3 years ago

a boy throw a ball upward from the top of a cliff 73m high.Calculate the time in which ball will fall on the ground and the velo

LekaFEV [45]

The ball rises for v/g seconds; which equals 14.7/9.8=1.5 seconds . After this time, it’s height will be:

h(t)=g/2(1.5)²+14.7(1.5)

=-4.9 x 2.25 + 22.05

=11.025m

The ball then falls for 49+11.025=60.025m, which takes:

g/2t²=60.025

t²=12.25

t=3.5 secs

Total time: 1.5+3.5=5 seconds

3 0

3 years ago

The centre of mass of a metre rule is at the 50cm mark. state what is meant by Centre of mass

Tanya [424]

Answer:

Centre of mass of any body is a point where all mass of a body is supposed to be concentrated

it lies in geometrical centre....

3 0

2 years ago

Which of the following best describes what alveolar are

Zolol [24]

Alveoli are tiny balloon shaped structures and are the smallest passageway in the respiratory system. The alveoli are only one cell thick, allowing the relatively easy passage of oxygen and carbon dioxide (CO2) between the alveoli and blood vessels called capillaries.

6 0

3 years ago

Read 2 more answers

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

2 years ago