Qué órganos forman parte del sistema nervioso

A force of 20 N is exerted on a box with a mass of 15 kg. if friction exerts a force of 4 N on the box, at what rate does the bo

MakcuM [25]

Answer:

1.06 metres per second squared

Explanation:

since friction acts against foward force

20 N - 4 N = 16 N

use Newtons 2nd law F=ma Solve for a:

a= F÷m

= 16 ÷ 15

= 1.06 metres per second squared

3 0

3 years ago

Who submits the federal budget every year?

artcher [175]

C) The president submits the federal budget every year.
Hope this helps you!

8 0

3 years ago

Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative

SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

Explanation:

Given there are three blocks of masses $M_{a}$ , $M_{b}$ and $M_{c}$ (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *a ................(equation 1)

Also it is given that $M_{a}$ does not move with respect to $M_{c}$ , which gives tension T is exerted on pulley by $M_{a}$ only, Hence tension T is

T = $M_{a}$ *a ..........(equation 2)

There is also also tension exerted by $M_{b}$ . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = $M_{b}$ $\sqrt{a^{2} +g^{2} }$ ................(equation 3)

From equation 2 and 3, we get

$M_{a}$ *a = $M_{b}$ $\sqrt{a^{2} +g^{2} }$

Squaring both sides we get

$M_{a} ^{2}$ * $a^{2}$ = $M_{b} ^{2}$ * ( $a^{2}$ + $g^{2}$ )

$M_{a} ^{2}$ * $a^{2}$ = ( $M_{b} ^{2}$ * $a^{2}$ )+ ( $M_{b} ^{2}$ * $g^{2}$ )

( $M_{a} ^{2}$ - $M_{b} ^{2}$ ) * $a^{2}$ = $M_{b} ^{2}$ * $g^{2}$

$a^{2}$ = $M_{b} ^{2}$ * $g^{2}$ /( $M_{a} ^{2}$ - $M_{b} ^{2}$ )

Taking square root on both sides, we get acceleration a

a = $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ )

Hence substituting the value of a in equation 1, we get

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

3 0

3 years ago

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

vovikov84 [41]

Gravitational force is given by, $F= G\frac{mM}{R^{2}}$

Where, m and M are the masses of the objects, R is the distance between them and G gravitational constant.

Gravitational force of the star on planet 1, $F_{1}= G\frac{m_{1}M}{R^{2}}$

Gravitational force of the star on planet 2, $F_{2}= G\frac{3m_{1}M}{(3R)^{2}}$

Ratio, $\frac{F_{1}}{F_{2}}= \frac{\frac{Gm_{1}M}{R^{2}}}{\frac{G3m_{1}M}{(3R)^{2}}}$

$\frac{F_{1}}{F_{2}}= \frac{3}{1}$

Therefore, the gravitational force of the star on the planet 1 is three times that on planet 2.

6 0

3 years ago

Read 2 more answers

Direct current, DC, flows in a _________ direction while alternating current, AC, the direction of flow ___________. A) single,

kramer

The flow of Direct current (DC) is constant and flows in one direction. Most digital electronics make use of DC. Alternating current (AC) periodically flows in reverse and is mostly used to deliver power to houses, buildings and the like. With that alone, you can already rule out A, C and D.

The answer would then be B. constant, periodically reversing.

4 0

3 years ago

Read 2 more answers