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Neko [114]
4 years ago
13

How does insulation in your home affect the transfer of thermal energy?

Physics
1 answer:
Llana [10]4 years ago
4 0
Heat loss through walls can be reduced using wall insulation
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A stone is thrown vertically upward with a speed of 18 m/s. (a) How long does it take the stone to reach a height of 11 m? (b) h
bagirrra123 [75]

Answer:

a) It takes the stone 0.7743 s to reach a height of 11 m for the first time on its way up and 2.899 s to reach again that height on its way down.

b) At t = 0.7743 s the velocity is 10.41 m/s and at t = 2.899 s the velocity is -10.41 m/s.

c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down.

Please, see the attached figures and the explanation for a description of the figures.

Explanation:

Hi there!

The equations for the height and velocity of the stone are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

v = velocity at time t

a) Let´s calculate the time it takes the stone to reach a height of 11 m. The origin of the frame of reference is at the throwing point so that y0 = 0:

y = y0 + v0 · t + 1/2 · g · t²        

11 m = 18 m/s · t - 1/2 · 9.8 m/s² · t²    

0 = -4.9 m/s² · t² + 18 m/s · t - 11 m

Solving the quadratic equation:

t = 0.7743 s and t = 2.899 s

(Notice that I have used more significant figures to avoid error by rounding)

The stone will be two times at a height of 11 m, one on its way up (at 0.7743 s) and one on its way down  (at 2.899 s). Then, it takes the stone 0.7743 s to reach a height of  11 m for the first time.

b)  Let´s use the equation of velocity:

v = v0 + g · t

at t = 0.77443 s

v = 18 m/s - 9.8 m/s² · 0.77443 s

v = 10.41 m/s

at t = 2.899 s

v = 18 m/s - 9.8 m/s² · 2.899 s

v = - 10.41 m/s

(Both velocities have to be of the same magnitude but of different sign, that´s why I haven´t rounded the time.)

c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down. On its way up, the velocity is 10.41 m/s and on its way down it is -10.41 m/s.

Figures

The functions to plot are the following:

height in function of time (figure 1, x-axis: time. y-axis: height)

y = -4.9t² + 18t

velocity in function of time (figure 2, x-axis: time. y-axis velocity)

v = -9.8t + 18

Acceleration in function of time (figure 3, x-axis: time. y-axis: acceleration)

a = -9.8

5 0
4 years ago
A wave a strong undertow. What kind of wave is it
shutvik [7]
Hey there! 

<span>Answer: </span>A swell

A wave which creates a strong undertow is called a swell. These types of waves are also known as <span>surface gravity waves. 

Thank you!</span>
8 0
3 years ago
A smooth wooden block is placed on a smooth wooden tabletop. You find that you must exert a force of 22.0 N to keep this 7.00 kg
xxMikexx [17]

Answer:

B . 68.7 N

Explanation:

Given in the question that;

Force = 22 N

Mass = 7.0 kg

Velocity = 4.0 m/s

Gravitational force is the weight of wooden block , calculated as mass times acceleration due to gravity.

F= m* a

F= 7 * 9.81

F= 68.67 N

F= 68.7 N

7 0
3 years ago
Suppose a 2662 kg giant seal is placed on a scale and produces a 20.0 cm compres- sion. If the seal and spring system are set in
Yuki888 [10]

Answer:

  0.8976 seconds

Explanation:

The period of oscillation for the simple harmonic motion can be found using the formula ...

  T = 2π√(d/g)

where d is the displacement of the spring due to the attached weight, and g is the acceleration due to gravity.

__

For d = 0.20 meters, the period is ...

  T = 2π√(0.20/9.8) ≈ 0.8976 . . . . seconds

_____

<em>Additional comment</em>

The formula for the oscillator period is usually seen as ...

  T = 2π√(m/k)

where m is the mass in the system and k is the spring constant. The value of the spring constant is calculated from ...

  k = mg/d

Using that in the formula, we find it simplifies to ...

  T=2\pi\sqrt{\dfrac{m}{k}}=2\pi\sqrt{\dfrac{m}{\left(\dfrac{mg}{d}\right)}}=2\pi\sqrt{\dfrac{d}{g}}

5 0
2 years ago
An object with a mass of m = 3.85 kg is suspended at rest between the ceiling and the floor by two thin vertical ropes.
Aleksandr-060686 [28]

The tension in the upper rope is determined as 50.53 N.

<h3>Tension in the upper rope</h3>

The tension in the upper rope is calculated as follows;

T(u) = T(d)+ mg

where;

  • T(u) is tension in upper rope
  • T(d) is tension in lower rope

T(u) = 12.8 N + 3.85(9.8)

T(u) = 50.53 N

Thus, the tension in the upper rope is determined as 50.53 N.

Learn more about tension here: brainly.com/question/918617

#SPJ1

6 0
2 years ago
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