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3 years ago

Pure science is best defined as the what?

Physics

2 answers:

Lapatulllka [165]3 years ago

7 0

A (continuing search for new knowledge)

WINSTONCH [101]3 years ago

6 0

Answer:

A) continuing search for new knowledge

Explanation:

Science has always been about discovery and gaining knowledge

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This is a cell, which is the basic unit of all life. All organs in human bodies are made of cells and require oxygen to survive.

kotegsom [21]

The two systems that work together to deliver oxygen are D, respiratory and cardiovascular

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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.

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4 years ago

A centrifuge rotor rotating at 10,000 rpm is shut off and is eventually brought to rest by a frictional force of 1.20m n. if the

Pani-rosa [81]

<span>Answer: The moments of inertia are listed on p. 223, and a uniform cylinder through its center is: I = 1/2mr2 so I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2 Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration: t = Ia -1.20 Nm = (0.0120984 kgm2)a a = -99.19 rad/s/s Now we have a kinematics question to solve: wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s w = 0 a = -99.19 rad/s/s Let's find the time first: w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s t = 10.558 s = 10.6 s And the displacement (Angular) Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form s = (u+v)t/2 Which is q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians But the problem wanted revolutions, so let's change the units: q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>

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3 years ago

What is the momentum of a bird with a mass of 0.018kg flying at 15m/s?

Mazyrski [523]

Answer:

0.27 kg-m/s

Explanation:

i believe this is the correct answer

6 0

3 years ago

The flow of electricity can be compared of water in

Anna007 [38]

The flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.

<h3>How we compare the flow of electricity to water?</h3>

Water flowing in pipes is like flowing of electricity in a circuit. A battery is like a pump from where electricity comes and moves in the circuit. Electrons flowing through wires are like water molecules flowing through pipes. So in comparison between water and electricity, both water and electricity are similar to each other in flowing and movement.

So we can conclude that the flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.

Learn more about electricity here: brainly.com/question/776932

#SPJ1

7 0

2 years ago