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3 years ago

7

Four boxes are sliding with constant speed before each box experiences and unbalanced force of 8 N. Which box would experience t

he greatest acceleration when the unbalanced force is applied? Answer choices: A) The box with a mass of 2 g B) The box with a mass of 4g C) The box with a mass of 6 g D)The box with a mass of 8 g

1 answer:

neonofarm [45]3 years ago

5 0

Answer:

d

Explanation:

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A student moves a box across the floor by exerting 56.7 N of force and doing 195 J of

Paha777 [63]

Answer:

A. 3.4 m

Explanation:

Given the following data;

Force = 56.7N

Workdone = 195J

To find the distance

Workdone is given by the formula;

$Workdone = force * distance$

Making "distance" the subject of formula, we have;

$Distance = \frac {workdone}{force}$

Substituting into the equation, we have;

$Distance = \frac {195}{56.7}$

Distance = 3.4 meters.

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3 years ago

Identify two types of simple machines in this compund machine.

sleet_krkn [62]

the answer would be "screw and lever!", thats the answer because the middle is made into a rod of metal rapped in metal to screw into the cork, and the lever part is the two handles on the side to lift it out of the bottle.

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3 years ago

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A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

jeyben [28]

Answer:

$V_0$

Explanation:

Given that, the range covered by the sphere, $M$ , when released by the robot from the height, $H$ , with the horizontal speed $V_0$ is $D$ as shown in the figure.

The initial velocity in the vertical direction is $0$ .

Let $g$ be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. $V_0$ will remain constant throughout the projectile motion.

So, if the time of flight is $t$ , then

$D=V_0t\; \cdots (i)$

Now, from the equation of motion

$s=ut+\frac 1 2 at^2\;\cdots (ii)$

Where $s$ is the displacement is the direction of force, $u$ is the initial velocity, $a$ is the constant acceleration and $t$ is time.

Here, $s= -H, u=0,$ and $a=-g$ (negative sign is for taking the sigh convention positive in $+y$ direction as shown in the figure.)

So, from equation (ii),

$-H=0\times t +\frac 1 2 (-g)t^2$

$\Rightarrow H=\frac 1 2 gt^2$

$\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)$

Similarly, for the launched height $2H$ , the new time of flight, $t'$ , is

$t'=\sqrt {\frac {4H}{g}}$

From equation (iii), we have

$\Rightarrow t'=\sqrt 2 t\;\cdots (iv)$

Now, the spheres may be launched at speed $V_0$ or $2V_0$ .

Let, the distance covered in the $x-$ direction be $D_1$ for $V_0$ and $D_2$ for $2V_0$ , we have

$D_1=V_0t'$

$D_1=V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_1=\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_1=1.41 D$ (approximately)

This is in the $3$ points range as given in the figure.

Similarly, $D_2=2V_0t'$

$D_2=2V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_2=2\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_2=2.82 D$ (approximately)

This is out of range, so there is no point for $2V_0$ .

Hence, students must choose the speed $V_0$ to launch the sphere to get the maximum number of points.

7 0

3 years ago

In a circus act, when the tumbling lever is measured, the measurements show that the effort arm is 8 meters long and the resista

Yuki888 [10]

The statement the mechanical advantage of the lever is 16 is False.

<h3>Mechanical Advantage</h3>

Mechanical advantage MA = d/D where

d = distance moved by effort and
D = distance moved by load

Given that the effort arm is 8 meters long, d = distance moved by effort = 8 m.

Also, given that the resistance arm is 2 meters long, D = distance moved by load = 2 m.

<h3>Calculating the mechanical advantage</h3>

So, substituting the values of the variables into the equation, we have

MA = d/D

MA = 8 m/2 m

MA = 4

Since MA = 4, so, the stament is False.

So, the statement the mechanical advantage of the lever is 16 is False.

Learn more about mechanical advantage here:

brainly.com/question/13779480

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2 years ago

Definition for compression and rarefraction

s344n2d4d5 [400]

Compression and rarefaction are two phenomenon occurs in longitudunal wave!

when there is denser particle gathering in that wave , there we called it compression and the rarer part of particles is rarefaction !

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