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katrin2010 [14]
3 years ago
8

The coefficient of linear expansion of iron is 1.0 × 10−5 per C◦ . The surface area of an iron cube, with an edge length of 5.0

cm, will increase by what amount if it is heated from 10◦ C to 60◦ C?
Physics
1 answer:
Allushta [10]3 years ago
7 0
The linear expansion of iron is a = 1× 10∧ -5 /°C
Length edge is 5cm
Initial surface area of cube A = 6L ∧²
 = 150cm∧²
The edge length of cube after being heated from 
10°C to 60°C is
L = L[I +a dt] where the dt is = 60-10 = 50°C
5.0025cm
A surface area of the cube after heated is 
A' = 6L' ∧²
= 150.1500375 cm∧²
The increased area is = A' -A = 0.15cm∧²
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A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show
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Answer:

vf = 3.27[m/s]

Explanation:

In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.

With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.

With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.

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a = 3.58 [m/s^2]

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x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]

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3 years ago
You throw a ball upward from ground level with initial upward speed v0. What is the max height of the trajectory?
Inga [223]

Answer:

The max height of the ball is y = -1/2 (v0²/g).

It takes the ball t = -2 · v0/g to hit the ground.

The speed of the ball when it hits the ground is v = -v0.

Explanation:

The height and velocity of the ball is given by the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

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v = velocity at time t

When the ball is at max height, the velocity is 0. So, let´s find the time at which the velocity of the ball is 0.

v = v0 + g · t

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Now, replacing t =  -v0/g in the equation of height, we will obtain the maximum height:

y = y0 + v0 · t + 1/2 · g · t²   (y0 = 0 because the origin of the frame of reference is located on the ground)

y = v0 · t + 1/2 · g · t²

Replacing t:

y = v0 · (-v0/g) + 1/2 · g ·  (-v0/g)²

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Since the acceleration of the ball is always the same, the time it takes the ball to impact the ground will be twice the time it takes to reach its max height, t = -2 v0/g.

However, let´s calculate that time knowing that at that time the height is 0:

y = y0 + v0 · t + 1/2 · g · t²

0 =  v0 · t + 1/2 · g · t²

0 = t · ( v0 + 1/2 · g · t)

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Let´s use the equation of velocity at final time (t = -2 · v0/g):

v = v0 + g · t

v = v0 + g · ( -2 · v0/g)

v = v0 - 2· v0

v = -v0

The speed of the ball when it hits the ground is v = -v0.

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