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vaieri [72.5K]
3 years ago
7

You release a block from the top of a long, slippery inclined plane of length l that makes an angle θ with the horizontal. The m

agnitude of the block's acceleration is gsin(θ).
1- For an x axis pointing down the incline and having its origin at the release position, derive an expression for the potential energy of the block-Earth system as a function of x. Suppose that the gravitational potential energy is measured relative to the ground at the bottom of the incline, UG(x=l)=0.
Express your answer in terms of g and the variables m, l, x, and θ.
(U^G=?)
2-
Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline. (Vx,f=?)
Express your answer in terms of g and the variables m, l, and θ.
Physics
1 answer:
Mars2501 [29]3 years ago
4 0

Answer:

potential enrgy   U = m g L sin θ

speed     V = √(2g L sin θ)

Explanation:

The expression for the gravitational potential energy of a body is

                    U = mg Y - mg Yo

Where Y give us a constant initial energy from which the differences are measured, for general simplicity it is selected as zero, Yo= 0

What we find an expression for height, let's use trigonometry

                       sin  θ= Y / L

                       Y = l sin θ

   We substitute in the power energy equation

                     U = m g L sin θ

                         

2. The mechanical energy of the system is conserved, so we will write the mechanical energy at two points the highest and the lowest

Highest           Em = U

Lower              Em = K

                         U = K

                         m g L sin θ = ½ m v²

                         V = √(2g L sin θ)

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A ball is dropped from a height of 20 meters. At what helght does the ball have a velocity of 10 meters/second?
Sunny_sXe [5.5K]

Answer:

5.10 meters.

Explanation:

v²=u²+2gh

or, (10)²=(0)²+2×9.8×h

or, 19.6h=100

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2 years ago
A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that incre
Evgen [1.6K]

Complete Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of 1.20  \ \Omega

Answer:

The current is   I =  0.0007 41 \ A

Explanation:

From the question we are told that

   The  area is  A = 8.00 \ cm^2  = 8.0 *10^{-4} \  m^2

   The initial magnetic field at t_o = 0 \ seconds  is B_i = 0.500 \ T

   The magnetic field at t_1 = 0.99 \ seconds is  B_f  = 1.60 \ T

     The resistance is  R = 1.20  \ \Omega

Generally the induced emf is mathematically represented as

      \epsilon  =  A * \frac{B_f - B_i }{ t_f - t_o }

=>   \epsilon  =  8.0 *10^{-4} * \frac{1.60  - 0.500 }{ 0.99- 0  }

=>   \epsilon  = 0.000889 \ V

Generally the current induced is mathematically represented as

     I = \frac{\epsilon}{R }

=>  I = \frac{0.000889}{ 1.20 }  

=>  I =  0.0007 41 \ A  

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Why do raindrops fall with a constant speed during the later stages of their descents?
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Firstly they have a acceleration downwards due the force downwards due they gravitational field acting on it's mass.
as it falls it gains speed, and as it gains speed the air Resistance which is a upward force actin on the drop increases, eventually the rain drop's upward and downward forces are balanced and hence there is no RESULTANT force therefore no acceleration, so the drops falls in constant speed (terminal verlocity is a better term)

Are you wondering that why is the raindrop still moving given that the forces are balanced? If so according to Newton's 1st law an object will keep moving or Remain at rest until a RESULTANT force acts on it.
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3 years ago
Which example demonstrates constant speed with changing direction?
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-- If there's a tug-of-war going on, and there are 300 freshmen pulling on one
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-- If a lady in the supermarket is pushing her shopping cart up the aisle, and her
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balanced. Then the cart doesn't move at all, and it's just as if nobody is pushing
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From these examples, you can see a few things:

-- There's no such thing as "a balanced force" or "an unbalanced force".
It's a <em><u>group</u> of forces</em> that is either balanced or unbalanced.

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-- When the group of forces on an object is balanced, then the effect on the
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4 0
3 years ago
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