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3 years ago

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A textbook is dropped from a second floor window and falls to the ground. What happens to the acceleration as the book falls to

the ground
-the acceleration increases
-the acceleration decreases
-the acceleration remains constant *

1 answer:

Sav [38]3 years ago

6 0

Answer:

The correct option is;

The acceleration remains constant

Explanation:

The acceleration is due to the force of gravitational attraction between the text book and the Earth

According to Newton's law of gravitation, there is an attractive force between all objects given by the following relation;

F = G×M₁×m₂/r²

Where;

G, M₁, m₂, and r are constant such that we have;

G×M₁/r² = Constant = The acceleraton due to gravity, g

F = g×m₂

So the acceleration of the textbook as it is being attracted by the force of gravity towards the ground (Earth) is remains constant.

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Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$

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3 years ago

The rectangular boat shown below has base dimensions 10.0 cm × 8.0 cm. Each cube has a mass of 40 g, and the liquid in the tank

Paladinen [302]

When boat is sunk into the liquid the net buoyancy on the boat is counterbalanced by weight of the boat

So here weight of the boat = Buoyancy force

let say boat is sunk by distance "h"

now we can say

$F_b = \rho * V * g$

$F_b = 1000*0.10 * 0.08 * h * 9.8$

now by above force balance equation we can write

$m*g = F_b$

$0.040 * 9.8 = 1000 * 0.10 * 0.08 * h * 9.8$

$0.040 = 8h$

$h = 5 * 10^{-3} m$

so boat will sunk by total 5 mm distance

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3 years ago

HELP!! Why is motion not considered a force?

Liono4ka [1.6K]

Answer:

There cannot be a force without motion and if there is no motion, then there is no force acting.

Explanation:

A force is a vector that causes an object with mass to accelerate.

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densk [106]

Answer:

$y = 24+64\cdot e^{-150\cdot t}$

Explanation:

Let solve the differential equation by separating corresponding variables:

$\int\limits^t_0\, dt = -\frac{1}{150} \int\limits^y_{y_{o}} \frac{dy}{y-24}$

The solution of this equation is:

$t = -\frac{1}{150}\cdot (\ln|y-24|-\ln |y_{o}-24|)$

The explicit form of the temperature as a function of time is:

$\ln |y-24|=-150\cdot t + \ln |y_{o}-24|$

$y-24 = C\cdot e^{-150\cdot t}$

The value of the integration constant is:

$C = 64$

The complete expression is:

$y = 24+64\cdot e^{-150\cdot t}$

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