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ehidna [41]
3 years ago
9

Help would be greatly appreciated, will give brainliest!

Physics
2 answers:
Fed [463]3 years ago
3 0

According to our principle, when an object is slowing down, the acceleration is in the opposite direction as the velocity. Thus, this object has a negative acceleration. ... When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object also has a negative acceleration.

ad-work [718]3 years ago
3 0

I don't know of a standard 4-step procedure for solving these problems, other than the standard procedure for solving ANY technical problem:

-- Be sure to understand the question; know exactly what it's asking for.

-- Be sure to understand exactly what information the question GIVES you.

-- Go to your toolbox:  Gather all the formulas, equations, and laws you know that involve the information that's given and the answer that's to be found.

-- Pick the tools that build the best bridge between the given information and whatever has to be found.

= = = = = = = = =

Here is a good, useful, helpful, basic toolbox that everyone should carry around with them.

For the statics, dynamics, and kinematics topics that you listed, there are a few basic, fundamental tools that must be in everybody's toolbox.  You really should memorize these.  They'll solve at least half of all the distance and motion problems you'll ever encounter:

-- Distance covered = (mileage reading at the end of the trip) - (reading at the beginning)

-- Displacement = straight-line distance and direction from the start-place to the end-place

-- Speed = (distance covered) ÷ (time to cover the distance)

-- Change in speed = (speed at the end) - (speed at the beginning)

-- Velocity = (displacement and direction) ÷ (time for the trip)

-- Acceleration = (change in speed and direction of the change) ÷ (time for the change)  

-- Distance covered from 'rest', with acceleration =

(1/2)·(acceleration)·(time squared)

= = = = =

Newton's laws & conservation:

-- Things keep moving in a straight line at constant speed, unless something makes them speed up, slow down, or curve.

-- Acceleration = (force) ÷ (mass)

-- Force ALWAYS comes in pairs.  For every force, there's always another force, equal to the first one and in the opposite direction.  Always.

-- Mass is conserved.  It never appears or disappears.  There's always the same amount after as there was before.  If more shows up, it had to come from somewhere.  If some disappears, it went somewhere, and you can find it if you look for it.

-- "Momentum" = (mass) · (speed)

-- Momentum is conserved.  It never appears or disappears.  There's always the same amount after as there was before.  If more shows up, it had to come from somewhere.  If some disappears, it went somewhere, and you can find it if you look for it.

-- Potential energy = (mass) · (gravity) · (height)

-- Kinetic energy = (1/2) · (mass) · (speed squared)

-- Energy is conserved.  It never just appears or disappears.  There's always the same amount after as there was before.  If more shows up, it had to come from somewhere.  If some disappears, it went somewhere, and you can find it if you look for it.

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flat sheet is in the shape of a rectangle with sides of lengths 0.400 mm and 0.600 mm. The sheet is immersed in a uniform electr
Ksenya-84 [330]

Answer:

6.29591\times 10^{-6}\ N/C^2

Explanation:

Flux is given by

\phi=EAcos\theta

A = Area

A=0.4\times 10^{-3}\times 0.6\times 10^{-3}

E = Electric field = 76.7 N/C

Angle is given by

\theta=90-20\\\Rightarrow \theta=70^{\circ}

\phi=76.7\times 0.4\times 10^{-3}\times 0.6\times 10^{-3}\times cos70\\\Rightarrow \phi=6.29591\times 10^{-6}\ N/C^2

The flux through the sheet is 6.29591\times 10^{-6}\ N/C^2

6 0
3 years ago
Describe what happens to the moving boat when the oars are out of the water and the forward thrust is zero
Tatiana [17]

Answer:

The boat won't be able to move if the oars were out and there was no thruster. If there was a flow of the water then yes there would be a moving boat.

6 0
3 years ago
You are in Paris, 60 m up in the Eiffel Tower. If you throw a euro downward at a velocity of 2.0 m/s, how long would it take the
kondor19780726 [428]

Answer:

t = 3.29 seconds

Explanation:

It is given that,

Height of the Eiffel tower is 60 m

Initial speed of a euro, u = 2 m/s

It will move under the action of gravity in the downward direction. Firstly, we can find the final velocity as follows :

v^2-u^2=2ad\\\\v=\sqrt{u^2+2ad} \\\\v=\sqrt{(2)^2+2\times 9.81\times 60} \\\\v=34.36\ m/s

Let t is the time taken by the euro to hit the ground. It can be calculated as :

v=u+at\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{34.36-2}{9.81}\\\\t=3.29\ s

Hence, it will take 3.29 seconds to hit the ground.

4 0
3 years ago
What is the total amount of kinetic and potential energy in a system ?
Solnce55 [7]

Answer:

Its the sum of the potential energy and the kinetic energy

7 0
3 years ago
You are exploring a planet and drop a small rock from the edge of a cliff. In coordinates where the +y direction is downward and
Lelu [443]

Answer:

value of the acceleration of gravity on the planet is 5.00 m/s²

Explanation:

The problem is similar to a free fall exercise, with another gravity value, the expression they give us is the following:

       y-yo = ½ gₐ t²       (1)

They tell us that they make a squared time graph with the variation of the distance, it is appropriate to clarify this in a method to linearize a curve, which is plotted the nonlinear axis to the power that is raised, specifically, the linearization of a curve The square is plotted against the other variable.

  Let's continue our analysis, as we have a linear equation, write the equation of the line.

     

        y1 = m x1 + b       (2)

where  “y1” the dependent variable, “x1” the independent variable, “m” the slope and “b” the short point

In this case as the stone is released its initial velocity is zero which implies that b = 0,

We plot on the “y” axis the time squared “t²” and on the horizontal axis we place “y-yo”.  To better see the relationship we rewrite equation 1 with this form

        t² = 2 /gₐ  (y-yo)

 

With the two expressions written in the same way, let's relate the terms one by one

        y1 = t²

        x1 = (y-yo)

        m = 2/gap

        b= 0

We substitute and calculate

        m = 2/gp

        gₐ = 2/m

        gₐ = 2/ 0.400

        gₐ = 5.00 m / s²

This is the value of the acceleration of gravity on the planet, note that the decimals are to keep the figures significant

6 0
3 years ago
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