All categories

3 years ago

Why do you need to apply a force in order to get the box to move

Mathematics

1 answer:

Readme [11.4K]3 years ago

4 0

Answer:

The reason one must apply a force to a box in order for the box to move is to overcome the friction generated by the box against the plane the box is sitting.

The box at rest has zero friction against the surface on which it sits. Once a force is applied, a box's movement is resisted by the friction created from contact with the surface.

In order for the box to move, one must generate more force than is being generated by the friction of the surface onto the box, allowing the box to move.

Cheers.

You might be interested in

Name where u from<br>age <br>favorite food

mestny [16]

Name: Ash
Where I’m from: New York
Favorite food: Chicken

7 0

3 years ago

Solve the following equation for x: 6(4x+ 1) = 3(x + 8) + 3.

salantis [7]

Answer:

x = -1/7 Or 0.143 (3 dec)

Step-by-step explanation:

6(4x + 5) = 3(x + 8) + 3

= 24x + 30 = 3x + 24 + 3

24x + 30 = 3x + 27

21x + 30 = 27

21x = -3

21x/21 = -3/21

x = -1/7

Or 0.143 (3 dec)

3 0

3 years ago

Read 2 more answers

I need help on this question plz and thank you <br> (Question 5 and if you can number 3 as well)

Alex Ar [27]

H = -2r + s/πr

C = 3A-a-b

If you need step by step comment

7 0

3 years ago

A baseball is thrown into the air with an upward velocity of 30 ft/s. its initial height was 6 ft, and its maximum height is 20.

marissa [1.9K]

Check the picture below.

where is the -16t² coming from? that's Earth's gravity pull in feet.

$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{30}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{6}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+30t+6 \\\\[-0.35em] ~\dotfill$

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+30}t\stackrel{\stackrel{c}{\downarrow }}{+6} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)$

$\bf \left(-\cfrac{30}{2(-16)}~~,~~6-\cfrac{30^2}{4(-16)} \right)\implies \left( \cfrac{30}{32}~,~6+\cfrac{225}{16} \right)\implies \left(\cfrac{15}{16}~,~\cfrac{321}{16} \right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{\stackrel{\textit{how many}}{\textit{seconds it took}}}{0.9375}~~,~~\stackrel{\stackrel{\textit{how many feet}}{\textit{up it went}}}{20.0625})~\hfill$

7 0

3 years ago

Read 2 more answers

How do you solve the slope and y-intercept of a graph?

valkas [14]

Answer:

The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept. The y-intercept of this line is the value of y at the point where the line crosses the y axis.

5 0

3 years ago

Read 2 more answers