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ratelena [41]
2 years ago
15

According to the law of conservation of energy, which statement must be true?

Physics
1 answer:
Nitella [24]2 years ago
5 0

Answer:

B. A system cannot take in additional matter.

Explanation: The total amount of energy in the universe remains constant, it can merely change from one form to another.

Hope it helps you:)))

have a good day

You might be interested in
Se o Universo tem uma idade definida, de aproximadamente 13,7 bilhões de anos, o que existiu antes do big-bang?
UkoKoshka [18]
Ninguém sabe exatamente
4 0
3 years ago
which of the following refers to the motion of an object falling straight down with only the force of gravity acting on it​
MrRissso [65]

Answer:

Since you haven't provided any choices, then the answer is "Free Fall Motion."

Explanation:

In order to learn more about the answer, let's discuss what free fall motion is.

Free Fall- In Physics, this refers to any body motion that is acted upon solely by <u>"gravity."</u> The acceleration in free fall is always downward and there's the absence of other forces. Take note that the<em> acceleration should be the same and is independent of the object's mass. </em>This acceleration is called "acceleration due to gravity."

Gravity- This refers to the force that pulls any object towards the center of the earth.

<u>Examples of Objects in Free Fall Motion</u>

1. A ball dropped at the top of a building.

2.  Dropping a coin from a table.

The ball and the coin are both in free fall motion because they are being pulled by gravity towards the earth. Their acceleration is also constant and there are no other forces acting upon them.

6 0
3 years ago
An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m
Aneli [31]

Answer:

\displaystyle y_m=3.65m

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of v_o and \theta\\ as the initial speed and angle, then we have

\displaystyle v_x=v_o\ cos\theta

\displaystyle v_y=v_o\ sin\theta-gt

\displaystyle x=v_o\ cos\theta\ t

\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}

If we want to know the maximum height reached by the object, we find the value of t when v_y becomes zero, because the object stops going up and starts going down

\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt

Solving for t

\displaystyle t=\frac{v_o\ sin\theta }{g}

Then we replace that value into y, to find the maximum height

\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2

Operating and simplifying

\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}

We have

\displaystyle v_o=20\ m/s,\ \theta=25^o

The maximum height is

\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}

\displaystyle y_m=3.65m

7 0
3 years ago
A spinning wheel has a rotational inertia of 2 kg•m². It has an angular velocity of 6.0 rad/s. An average counterclockwise torqu
ira [324]

Answer:

-20.0 kg m^2/s

Explanation:

The angular momentum of an object in rotation is given by

L=I \omega

where

I is the moment of inertia

\omega is the angular speed

In this problem, initially we have

I=2 kg m^2 is the moment of inertia of the wheel

\omega_i = 6.0 rad/s is the initial angular speed

So the initial angular momentum is

L_i = I\omega_i = (2)(6.0)=12 kg m^2/s

Later, a counterclockwise torque of

\tau=-5.0 Nm is applied

So the angular acceleration of the wheel is:

\alpha = \frac{\tau}{I}=\frac{-5.0}{2}=-2.5 rad/s^2 in the direction opposite to the initial rotation.

As a result, the final angular velocity of the wheel will be:

\omega_f = \omega_i + \alpha t

where

t = 4.0 is the time interval

Solving,

\omega_f = +6.0 +(-2.5)(4.0)=-4 rad/s

which means that now the wheel is rotating in the counterclockwise direction.

Therefore, the new angular momentum of the wheel is:

L_f = I\omega_f =(2)(-4.0)=-8.0 kg m^2/s

So, the change in angular momentum is:

\Delta L=L_f - L_i = (-8.0-(12))=-20.0 kg m/s^2

7 0
3 years ago
A 20 kg truck drives in a circle of radius 4 m at 10m/s. What is the
Tems11 [23]
  • Mass=m=20kg
  • Radius=r=4m
  • Velocity=10m/s=v

We know

\boxed{\sf F_c=\dfrac{mv^2}{r}}

\\ \sf\longmapsto F_c=\dfrac{20(10)^2}{4}

\\ \sf\longmapsto F_c=\dfrac{20(100)}{4}

\\ \sf\longmapsto F_c=\dfrac{2000}{4}

\\ \sf\longmapsto F_c=500N

6 0
3 years ago
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