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A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

3 years ago

Drag the positive or negative feedback loop on the left to each process on the right. terms may be used once, more than once, or

slamgirl [31]

The order of the positive and negative feedback loops are positive, positive, negative, positive, positive, negative.

<h3>What is a feedback loop?</h3>

A system component known as a feedback loop is one in which all or a portion of the output is used as input for subsequent actions. A minimum of four phases comprise each feedback loop. Input is produced in the initial phase. Input is recorded and stored in the subsequent stage. Input is examined in the third stage, and during the fourth, decisions are made using the knowledge from the examination.

Both negative and positive feedback loops are possible. Insofar as they stay within predetermined bounds, negative feedback loops are self-regulating and helpful for sustaining an ideal condition. One of the most well-known examples of a self-regulating negative feedback loop is an old-fashioned home thermostat that turns on or off a furnace using bang-bang control.

To learn more about feedback loop, visit:

brainly.com/question/11312580

#SPJ4

5 0

2 years ago

Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s

Viefleur [7K]

Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density = 0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

= -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level ) = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = $\sqrt{CRT}$ = $\sqrt{1.4*287*200.5 }$ = 283.8 m/s

hence it is a subsonic aircraft

4 0

3 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have: