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3 years ago

True or False An object will slow down and stop as long as no forces act upon it

Physics

2 answers:

kolezko [41]3 years ago

6 0

The answer is false <3

My name is Ann [436]3 years ago

5 0

Answer:

false

Explanation:

Understand Newton’s first law of motion. Experience suggests that an object at rest will remain at rest if left alone, and that an object in motion tends to slow down and stop unless some effort is made to keep it moving. An object will go on forever if there were no forces to act upon it. In space, a rocket will go forever at the speed it projects.

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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

The diagram shows four paths from point A to point B.

iVinArrow [24]

Path 2.

Displacement is the direction and magnitude of an object from its starting point, so path 2 is the direct route you would need to take to find direction and magnitude.

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2 years ago

Write a question you would like to see on a quiz to calculate NET FORCE.

Pie

Answer:

How do you find the net force?

Explanation:

i think this is want you asked for?

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onsider the free body diagram. If the sum of the tension forces is equal to the force of gravity, which description BEST applies

cluponka [151]

The answer is C) A girl hangs by both hands, motionless, from a trapeze.

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3 years ago

Which example identifies a change in motion that produces acceleration?

kifflom [539]

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<span>A ball moving at a constant speed around a circular track produces acceleration. </span>

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3 years ago

Read 2 more answers