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Neko [114]
3 years ago
5

Define the si units for length mass time and temperature

Physics
1 answer:
Sladkaya [172]3 years ago
7 0

For this case we have the following measures in the international system:

  • <em>For the length: </em>

It is a unit that is used to measure the length of any object: Meter.

  • <em>For the mass: </em>

The mass in the international system is denoted by: kilogram. It is used to have an approximate idea of the weight of objects.

  • <em>For the time: </em>

The measure in the international system is: Seconds. It is used to know how much time elapses in an event to be studied.

  • <em>For the temperature: </em>

The measure in the international system is: Kelvin. It is used to get an idea of how cold or hot an object, process, place, person can be.

Answer:

Length: meter

Mass: Kilogram

Time: Second

Temperature: Kelvin

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lord [1]

Answer:

The force is 272.73 newtons

Explanation:

We're going to use impulse-momentum theorem that states impulse is the change on the linear momentum this is:

\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i} (1)

Impulse is also defined as average force times the time the force is applied:

\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t) (2)

By (2) on (1):

\overrightarrow{F}_{avg}(\varDelta t)= \overrightarrow{p}_{f}-\overrightarrow{p}_{i}

solving for \overrightarrow{F}_{avg}:

\overrightarrow{F}_{avg}=\frac{\overrightarrow{p}_{f}-\overrightarrow{p}_{i}}{\varDelta t} (3)

We already know Δt is equal to 0.22 s, all we should do now is to find \overrightarrow{p}_{f}-\overrightarrow{p}_{i} and put on (3) (\overrightarrow{p_{i}} the initial momentum and \overrightarrow{p_{f}} the final momentum). Linear momentum is defined as \overrightarrow{p}=m\overrightarrow{v} , using that on (3):

\varDelta\overrightarrow{p}=m \overrightarrow{v_{f}}-m \overrightarrow{v_{i}} (4)

Velocity (v) are vectors so direction matters, if positive direction is the right direction and negative direction left \overrightarrow{v_{i}}=+3\, \frac{m}{s} and \overrightarrow{v_{f}}=-3\, \frac{m}{s} so (4) becomes:

\varDelta\overrightarrow{p}=m(-3\frac{m}{s}- (+3\frac{m}{s}))=-(10kg)(6\frac{m}{s})

\varDelta\overrightarrow{p}=-60\, \frac{mkg}{s} (5)

Using (5) on (3):

\overrightarrow{F}_{avg}=\frac{-60\, \frac{mkg}{s}}{0.22s}

F_{avg}=272.73N

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4 years ago
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When discussing distances between objects in the solar system, which term do you use?
Bond [772]
I think it's (ASTRONOMICAL UNIT)

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Answer:

<h2>C. </h2>

Explanation:

<h3>#CARRY ON LEARNING</h3><h3>#MARK ON LEARNING</h3><h3>#HELPING HAND</h3>
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3 years ago
Is it possible for the momentum of a system consisting of two carts on a low-friction track to be zero even if both carts are mo
tamaranim1 [39]

Answer:

Yes, if the carts are travelling into opposite directions

Explanation:

The total momentum of a two carts system is the sum of the momenta of the individual carts:

p=p_1 +p_2 = m_1 v_1 + m_2 v_2

where

m1, m2 are the masses of the two carts

v1, v2 are the velocities of the two carts

In order to have a total momentum of zero, we must have

m_1 v_1 + m_2 v_2 = 0\\m_1 v_1 = - m_2 v_2 (1)

Let's remind that velocity (and so, momentum as well) is a vector quantity: this means that it has a direction, so when summing together the momenta, we must also consider the sign of the velocity, depending on its direction.

Therefore, if the two carts are moving in opposite directions, the signs of the two velocities will be opposite. For example, we can have

v_1 > 0 \\v_2 < 0

This means that the condition in eq.(1) can be satisfied, provided that the two carts are travelling into opposite directions.

8 0
3 years ago
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