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klemol [59]
3 years ago
10

What is the common formula for work? Assume that W is the work, F is a constant force, 656-06-01-00-00_files/i0060000.jpgv is th

e change in velocity, and d is the displacement.
Physics
1 answer:
klio [65]3 years ago
6 0
The common formula for work is:
Work = force x displacement
W = F x d
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garik1379 [7]
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5 0
3 years ago
The sun's energy begins as what form of energy?
blagie [28]

Answer:

nuclear energy.............

4 0
3 years ago
If it is fixed at C and subjected to the horizontal 60-lblb force acting on the handle of the pipe wrench at its end, determine
pickupchik [31]

Answer:

τ = 132.773 lb/in² = 132.773 psi

Explanation:

b = 12 in

F = 60 lb

D = 3.90 in (outer diameter)  ⇒ R = D/2 = 3.90 in/2 = 1.95 in

d = 3.65 in (inner diameter)  ⇒ r = d/2 = 3.65 in/2 = 1.825 in

We can see the pic shown in order to understand the question.

Then we get

Mt = b*F*Sin 30°

⇒ Mt = 12 in*60 lb*(0.5) = 360 lb-in

Now we find ωt as follows

ωt = π*(R⁴ - r⁴)/(2R)

⇒ ωt = π*((1.95 in)⁴ - (1.825 in)⁴)/(2*1.95 in)

⇒ ωt = 2.7114 in³

then the principal stresses in the pipe at point A is

τ = Mt/ωt ⇒ τ = (360 lb-in)/(2.7114 in³)

⇒ τ = 132.773 lb/in² = 132.773 psi

7 0
3 years ago
A 2kg water balloon is flying at a rate of 4m/s^2. With what force will it hit its target?
blondinia [14]

Explanation:

F=m×a

m=2kg

a=4m/s^2

F=2kg×4m/s^2

F=8N

6 0
3 years ago
In Millikan's oil drop experiment an oil drop is at rest between two large plates separated by 1.3 cm when the potential differe
sweet-ann [11.9K]

Answer:

Mass of the oil drop, m=3.01\times 10^{-15}\ kg

Explanation:

Potential difference between the plates, V = 400 V

Separation between plates, d = 1.3 cm = 0.013 m

If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

qE=mg

m=\dfrac{qE}{g}

q=6e, e is the charge on electron

E is the electric field, E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m

m=\dfrac{6\times 1.6\times 10^{-19}\times 30769.23}{9.8}

m=3.01\times 10^{-15}\ kg

So, the mass of the oil drop is 3.01\times 10^{-15}\ kg. Hence, this is the required solution.

5 0
3 years ago
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