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olganol [36]
4 years ago
9

You lift a 25-kg child 0.80 m, slowly carry him 10 m to the playroom, and finally set him back down 0.80 m onto the playroom flo

or. What work do you do on the child for each part of the trip and for the whole trip
Physics
1 answer:
maksim [4K]4 years ago
6 0

To solve this problem we will apply the work theorem which is expressed as the force applied to displace a body. Considering that body strength is equivalent to weight, we will make the following considerations

\text{Mass of the child} = m = 25kg

\text{Acceleration due to gravity} = g = 9.81m/s^2

\text{Height lifted} = h = 0.80m (Upward)

Work done to upward the object

W = mgh

W = (25)(9.81)(0.8)

W = 196.2J

Horizontal Force applied while carrying 10m,

F = 0N

W = 0J

Height descended in setting the child down

h' = -0.8m (Downwoard)

W = mgh'

W = (25)(9.81)(-0.80)

W = -196.2J

For full time, assuming that the total value of work is always expressed in terms of its symbol, it would be zero, since at first it performs the same work that is later complemented in a negative way.

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A crude approximation for the x component of velocity in an incompressible laminar boundary layer is a linear variation from u =
slega [8]

Answer:

2.5 * 10^-3

Explanation:

<u>solution:</u>

The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:

<em>δu/δx+δv/δy=0</em>

so that:  

<em>δv/δy= -δu/δx</em>

Now, since u = Uy/δ, where δ = cx^1/2, we have that:

<em>u=U*y/cx^1/2</em>

and we obtain:  

<em>δv/δy=U*y/2cx^3/2</em>

The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):  

v=∫δv/δy(dy)=U*y/4cx^1/2

 =y/x*(U*y/4cx^1/2)

 =u*y/4x

which is exactly what we needed to demonstrate.  

Also, using u = U*y/δ in the last equation we can obtain:  

v/U=u*y/4*U*x

     =y^2/4*δ*x

which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:

(v/U)_max=δ^2/4δx

                =δ/4x

                =2.5 * 10^-3

7 0
3 years ago
During normal beating, a heart creates a maximum 3.95-mV potential across 0.305 m of a person’s chest, creating a 0.75-Hz electr
WINSTONCH [101]

Answer:

E = 0.0130 V/m.

Explanation:

The electric field is related to the potential difference as follows:

E = \frac{\Delta V}{d}

<u>Where:</u>

E: is electric field

ΔV: is the potential difference = 3.95 mV  

d: is the distance of a person's chest = 0.305 m

Then, the electric field is:

E = \frac{\Delta V}{d} = \frac{3.95 \cdot 10^{-3} V}{0.305 m} = 0.0130 V/m

Therefore, the maximum electric field created is 0.0130 V/m.

I hope it helps you!

7 0
3 years ago
What is quantum computing?
Natali5045456 [20]
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4 0
3 years ago
Consider the following True/False statements:
Ainat [17]

Answer:

6) False

7) True

8) False

9) False

10) False

11) True

12) True

13) True

14) True

Explanation:

The spacing between two energy levels in an atom shows the energy difference between them. Clearly, B has a greater value of ∆E compared to A. This implies that the wavelength emitted by B is greater than A while B will emit fewer, more energetic photons.

When atoms jump from lower to higher energy levels, photons are absorbed. The kinetic energy of the incident photon determines the frequency, wavelength and colour of light emitted by the atom.

The energy level to which an atom is excited is determined by the kinetic energy of the incident electron. As the voltage increases, the kinetic energy of the electron increases, the further the atom is from the source of free electrons, the greater the required kinetic energy of free electron. When electrons are excited to higher energy levels, they must return to ground state.

4 0
3 years ago
I really don’t know the answer for this
EleoNora [17]
The correct answer is the reverse wave I took the test
6 0
3 years ago
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