Ask question

Ask question

All categories

sleet_krkn [62]

3 years ago

8

Urgent! Which is not an example of work? Question 2 options: pushing a box across the floor picking up a box off the floor raisi

ng a barbell over your head trying to push a rock that never moves

1 answer:

OLga [1]3 years ago

8 0

Answer:

trying to push a rock that never moves

Explanation:

You might be interested in

What is electrical energy

creativ13 [48]

Explanation:

it is energy that flows of electric charge it has the ability to do work or apply force to move an object.

8 0

3 years ago

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

A person holding a lunch bag is moving upward in a hot air balloon at a constant speed of 7.3 m/s . When the balloon is 24 m abo

Kitty [74]

Explanation:

Given that,

Initial speed of the bag, u = 7.3 m/s

Height above ground, s = 24 m

We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :

$v^2=u^2+2as$

$v^2=(7.3)^2+2\times 9.8\times 24$

$v=\sqrt{523.69}$

v = 22.88 m/s

So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.

8 0

3 years ago

Yoga not only builds flexibility, but strength and balance.<br><br> True or False

azamat

Answer:

True

Explanation:

I just know.

3 0

3 years ago

Which of the following represents a possible magnitude for the force of static friction when Xavier applied 72.1 Newtons of forc

lana66690 [7]

The possible magnitude for the force of static friction on the stationary cart is 72.1 N.

The given parameters:

<em>Applied force on the cart, F = 72.1 N</em>

<em />

Based on Newton's second law of motion, the force applied to object is directly proportional to the product of mass and acceleration of the object.

F = ma

Static frictional force is the force resisting the motion of an object at rest.

$\Sigma F = 0\\\\F -F_f = 0$

where;

$F_f$ is the frictional force

$F= F_f \\\\72.1 = F_f\\\\F_f = 72.1\ N$

Thus, the possible magnitude for the force of static friction on the stationary cart is 72.1 N.

Learn more about Newton's second law of motion: brainly.com/question/25307325

8 0

2 years ago