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3 years ago

Why is it difficult to convert miles to yards

2 answers:

6 0

There's really no "why", because it's not difficult at all.

Simply multiply the number of miles by 1,760, and bada boom,
there you have the same distance described in yards.

Lelechka [254]3 years ago

3 0

In today's technological age it would be immensely easy to make such a conversion. There are countless tools on the internet that can do the job for you incredibly quickly and easily. Nevertheless, I'll give you an idea about how it could be done.

$1\quad mile\quad =\quad 1760\quad yards\\ \\ \frac { 1 }{ 1760 } \quad \times \quad 1\quad mile\quad =\quad 1760\quad \times \quad \frac { 1 }{ 1760 } \quad yards\\ \\ \frac { 1 }{ 1760 } \quad miles\quad =\quad 1\quad yard$

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2 years ago

How does natural selection produce change in a population of mice?

sergey [27]

Answer:

The population of the mice will decrease to the faster, stronger, and smarter mice because the weaker will die because of natural selection.

Explanation:

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2 years ago

What is the minimal mass of helium (density 0.18 kg/m3) needed to lift a balloon carrying two people in a basket, if the total m

Sergio039 [100]

Answer:
M[min] = M[basket+people+ balloon, not gas] * Î”R/R[b]
Î”R is the difference in density between the gas inside and surrounding the balloon.
R[b] is the density of gas inside the baloon.
====================================
Let V be the volume of helium required.
Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V
U = 1.225gV newtons
----
Weight of Helium = Volume of Helium * Density of Helium * g
W[h] = 0.18gV N
Net Upward force produced by helium, F = Upthrust - Weight = (1.225-0.18) gV = 1.045gV N  -----

Weight of 260kg = 2549.7 N
Then to lift the whole thing, F > 2549.7
So minimal F would be 2549.7
----
1.045gV = 2549.7
V = 248.8 m^3
Mass of helium required = V * Density of Helium = 248.8 * 0.18 = 44.8kg (3sf)
=====
Let the density of the surroundings be R
Then U-W = (1-0.9)RgV = 0.1RgV
So 0.1RgV = 2549.7 N
V = 2549.7 / 0.1Rg
Assuming that R is again 1.255, V = 2071.7 m^3
Then mass of hot air required = 230.2 * 0.9R = 2340 kg
Notice from this that M = 2549.7/0.9Rg * 0.1R so
M[min] = Weight of basket * (difference in density between balloon's gas and surroundings / density of gas in balloon)
M[min] = M[basket] * Î”R/R[b]

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3 years ago

HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLE

Romashka [77]

$A_x = 5.0$

$A_y = -6.3$

6 0

2 years ago

A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g gla

miss Akunina [59]

Answer:

The ball traveled 0.827 m

Explanation:

Given;

distance between the metal plates of the room, d = 3.1 m

mass of the glass, m = 1.1g

charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

F = qV/d

|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)

|F| = 4.548 x 10⁻³ N

F = - 4.548 x 10⁻³ N

The net horizontal force experienced by this ball is;

$F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N$

The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.

$W = F_{net} *h\\\\W = 15.328 *10^{-3} * h$

W = K.E

$15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m$

Therefore, the ball traveled 0.827 m

4 0

3 years ago