All categories

3 years ago

Plz help difference between closed system and open system?

Physics

1 answer:

frozen [14]3 years ago

4 0

A closed system is a system that is completely isolated from its environment. The physical universe, as we currently understand it, appears to be a closed system. An open system is a system that has flows of information, energy, and/or matter between the system and its environment, and which adapts to the exchange.

D makes the most sense, but you just have to put two and two together and go with your gut feeling, first cross out the answers that don't make sense (A didnt make sense) and go from there! I hope the little bit above me helped you answer or decide :) Good luck!

You might be interested in

A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

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

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the block, measured in meters per second.

$\Sigma F$ - Horizontal net force, measured in newtons.

$\Delta t$ - Impact time, measured in seconds.

Now we clear the final speed in (2):

$v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $m = 14\,kg$ , $\Sigma F = 98\,N$ and $\Delta t = 1.40\,s$ , then final speed of the ice block is:

$v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}$

$v_{f} = 9.8\,\frac{m}{s}$

The final speed of the block of ice is 9.8 meters per second.

6 0

3 years ago

A detailed explanation of static and dynamic friction . plsssss

Gemiola [76]

Answer:

Static friction is what keeps the box from moving without being pushed, and it must be overcome with a sufficient opposing force before the box will move. Kinetic friction (also referred to as dynamic friction) is the force that resists the relative movement of the surfaces once they're in motionExplanation:

#if you need any questions answered within mins/secs hit me up and I got you

7 0

3 years ago

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

3 years ago

What is an example of an organism that reproduces

RideAnS [48]

Sea anemones; Coral; Starfish; Some non-flowering plants. Strawberry; Onion; Potato. Mushrooms are an example of asexually reproducing organisms.

hoped i helped my fellow brainily user

4 0

3 years ago

In a 2-dimensional coordinate system, the x- and y-axes are typically _____.

dem82 [27]

In a 2-dimensional coordinate system, the x- and y-axes
are typically perpendicular to each other. (C)

7 0

3 years ago