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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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Describe a situation when you might travel a high velocity but with low acceleration
Umnica [9.8K]
Answer:
One situation could be when someone is sitting in a car that is travelling at a constant speed of 150mph

Explanation:
Let’s review the terms:
Velocity: the speed of something in a given direction
Acceleration: increase in the rate or speed of something

The car is travelling at a high speed but the speed is constant, so it isn’t increasing. In other words, the car the person is sitting in is travelling in in a high velocity but with no/low acceleration.

I hope this helps! Please comment if you have any questions.
4 0
3 years ago
An object of mass m is traveling in a circle with centripetal force f c. If the velocity of the object is v, what is the radius
Gnom [1K]

Answer:

r = mv^2/ Fc

Explanation:

Got it right

7 0
3 years ago
A tuning fork of frequency 254 Hz and an open orang pipe of slightly lower frequency are at 15oC. When
katovenus [111]

The temperature of the air in the open orang pipe has been altered by 18.73° C

The frequency of an open orang pipe is estimated by using the formula:

\mathbf{f = \dfrac{v}{2L}}

Then, the combination of the frequency of the tuning fork and the open orang pipe is:

\mathbf{254 - \dfrac{v}{2L} }

These combinations of frequency produce 4 beats per sound.

i.e.

\mathbf{254 - \dfrac{v}{2L}   =4}

\mathbf{ \dfrac{v}{2L} = 254-4 }

\mathbf{ \dfrac{v}{2L} = 250 ----(1)}

When it is altered, the beats first diminish and increase again by 4.

i.e.

\mathbf{ \dfrac{v'}{2L} = 254+4 }

\mathbf{ \dfrac{v'}{2L} = 258 --- (2) }

If we equate both equations (1) and (2) together, we have:

\mathbf{\dfrac{v'}{v}= \dfrac{258}{250}}

However, from our previous knowledge, we understand that the velocity of an object varies directly proportional to the square root of its temperature.

Hence;

  • when the temperature of the pipe  = unknown ???
  • the temperature of the open orang pipe = 15

∴

\implies \mathbf{\sqrt{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \dfrac{258}{250}}

By squaring both sides, we have:

\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \Big (\dfrac{258}{250}\Big )^2}

\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)= \Big (\dfrac{66564}{62500}\Big )}

\implies \mathbf{\Big(\dfrac{273 + T}{288}\Big)= \Big (1.065024\Big )}

\implies \mathbf{273 +T =306.726912  }

T = 306.726912 - 273

T ≅ 33.73 ° C

∴

The change in temperature ΔT = 33.73° C - 15° C

The change in temperature ΔT = 18.73° C

Learn more about wave frequency here:

brainly.com/question/14316711?referrer=searchResults

4 0
3 years ago
Which of these would NOT transfer and change electrical energy into another type of energy in a closed circuit? A) fan B) air C)
Nitella [24]
The answer is B) air.
7 0
4 years ago
Read 2 more answers
A magnet in the form of a cylindrical rod has a length of 7.30 cm and a diameter of 1.5 cm. It has a uniform magnetization of 5.
yulyashka [42]

Answer:

Magnetic dipole moment is 0.0683 J/T.

Explanation:

It is given that,

Length of the rod, l = 7.3 cm = 0.073 m

Diameter of the cylinder, d = 1.5 cm = 0.015 m

Magnetization, M=5.3\times 10^3\ A/m

The dipole moment per unit volume is called the magnetization of a magnet. Mathematically, it is given by :

M=\dfrac{\mu}{V}

\mu=M\times \pi r^2\times l

Where

r is the radius of rod, r = 0.0075 m

\mu=5.3\times 10^3\ A/m\times \pi (0.0075)^2\times 0.073\ m

\mu=0.0683\ J/T

So, its magnetic dipole moment is 0.0683 J/T. Hence, this is the required solution.

6 0
3 years ago
Read 2 more answers
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