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dsp73
1 year ago
5

It is easier to roll a stone up a sloping road than to lift it vertical upwards because.

Physics
1 answer:
kvasek [131]1 year ago
3 0

Answer:

Because the force one needs to apply is reduced.

Explanation:

  • Work can be expressed in terms of force and distance as follows,

                        Work = Force × distance

  • To be more precise, the selection of these two parameters should be in a manner that the distance is what the line of action of the force traces.
  • It is obvious that for a given amount of work when one parameter is changed the other changes to counteract the change of the first one.
  • Imagine the situation shown in the attached figure. The amount of work that needs to be done during the lifting of the mass to a height of \small h vertically is E = mg\times h = mgh.
  • On the slope, however, one needs to push the object an L distance up to get the object to that h height ultimately.
  • The work he does on the slope is mgh = F\times L and sinceL > h , F < mg.
  • The force that needs to be applied is less than that applied in the vertical lift. Hence it feels easier.             #SPJ4                                                                    

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2 years ago
An 8.2 kg object accelerates at 8.0 m/s^2. What is the acceleration?
ZanzabumX [31]

Answer:

8.0m/s²

Explanation:

meters per second squared is a unit of acceleration so 8.0m/s² is the answer

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How do you state a hypothesis! Branliest
VLD [36.1K]

Answer:

State your hypothesis as concisely, and to the point, as possible. A hypothesis is usually written in a form where it proposes that, if something is done, then something else will occur. Usually, you don't want to state a hypothesis as a question. You believe in something, and you're seeking to prove it.

Explanation:

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3 years ago
Read 2 more answers
A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz
gladu [14]

This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle?  Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa  

Tt is stagnation temperature = ?

T is the static temperature  = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa =  ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 =  Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

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Answer:

Explanation is given

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