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san4es73 [151]
2 years ago
11

Give an example of a situation in which there is a force and a displacement, but the force does no work. explain why it does no

work
Physics
2 answers:
makkiz [27]2 years ago
6 0
Work is equal to force times distance times cosine of the angle between the force and the path vector making it a scalar or dot product. Work is independent of the path taken. It refers to positive change in KE or negative change in PE.
 So for example if you lift a box from ground to waist height then, you did positive work over the distance you lifted it. If you pick it up to that same height, then put it back done on the ground at same place, then no work is done. 

Hope this makes sense. Any questions please ask. Thank You!
harkovskaia [24]2 years ago
6 0
Say you have an object such as a block on a flat surface with no friction to resist it while being displaced. The block moves but no work is done.
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3. Hahn determined that barium (atomic number 56) was one of the elements created when a uranium atom (atomic number 92) split .
creativ13 [48]

According to a periodic table, Krypton was created during the fission of Uranium.

<h3>What is the atomic number?</h3>

<em>Atomic</em> number is a characteristic associated with an element and indicates its number of protons, when a fision occurs, the total number protons is conserved.

Thus, the fission of uranium is led by two elements with <em>atomic</em> numbers 56 and 36. According to a periodic table, those <em>atomic</em> numbers are associated to elements Barium (Ba) and Krypton (Kr), respectively.

According to a periodic table, Krypton was created during the fission of Uranium. \blacksquare

To learn more on fission, we kindly invite to check this verified question:  brainly.com/question/6572079

8 0
1 year ago
Eric drops a 2.20 kg water balloon that falls a distance of 45.08 m off the top of a
Marianna [84]

Answer:

972 J

Explanation:

At the bottom, all the gravitational potential energy was converted into kinetic energy. If you calculate the GPE, its value will be the same that the KE at the bottom. The GPE can be calculated this way:

GPE = mass×gravity×heigth

GPE = 2.2×9.8×45.08 ≈ 972

4 0
2 years ago
An elevator manufacturing company is stress-testing a new elevator in an airless test shaft. The elevator is traveling at an unk
devlian [24]

Answer:

3.192 m/s

Explanation:

t = Time taken = 0.900 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 1.1 meters

a = Acceleration due to gravity = 9.81 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1.1-\frac{1}{2}\times 9.81\times 0.9^2}{0.9}\\\Rightarrow u=-3.192\ m/s

Velocity of the elevator when it snapped is 3.192 m/s

4 0
3 years ago
Consider the free-body diagram. If you want the box to move, the force applied while dragging must be greater than the
VLD [36.1K]

You would want it to be greater than D. friction force

It needs be greater than the friction applied to it.

6 0
2 years ago
Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug
Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

F_D=\dfrac{1}{2}\rho C_DA v^2

C_D=Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

Now by putting the values

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}

\dfrac{F_D_1}{F_D_2}=0.444

Percentage Change in the drag force

\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100

\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100

\Delta F=\dfrac{1-0.444}{0.444}\times 100

ΔF=125.22 %

Therefore force will increase by 125.22  %.

3 0
3 years ago
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