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katrin2010 [14]
3 years ago
14

If it requires 5.0 j of work to stretch a particular spring by 1.8 cm from its equilibrium length, how much more work will be re

quired to stretch it an additional 3.7 cm ?
Physics
1 answer:
ANEK [815]3 years ago
5 0
15.277.. j. I did the problem using a proportion. an additional 3.7m to the current 1.8 cm=5.5cm.

Therefore, 5.0 j/1.8cm=x/5.5cm
You might be interested in
Relative formula mass of CuCO3
Elodia [21]

AnMolar mass of CuCO3 = 123.5549 g/mol

This compound is also known as Copper(II) Carbonate.

Convert grams CuCO3 to moles  or  moles CuCO3 to grams

Molecular weight calculation:

63.546 + 12.0107 + 15.9994*3

Percent composition by element

Element   Symbol   Atomic Mass   # of Atoms   Mass Percent

Copper Cu 63.546 1 51.431%

Carbon C 12.0107 1 9.721%

Oxygen O 15.9994 3 38.848%

Explanation:

5 0
2 years ago
Help<br> A. 12,240<br> B. 6,120
nata0808 [166]

Answer:

A. 12,240.

Explanation:

1,530 times 8.0 = 12,240.

hope this helped :)

3 0
2 years ago
An automobile of mass 1.46 cross times to the power of blank 10 cubed kg rounds a curve of radius 25.0 m with a velocity of 15.0
Vsevolod [243]

The centripetal force exerted on the automobile while rounding the curve is 1.31\times10^4 N

<u>Explanation:</u>

given that

Mass\ of\ the\ automobile\ m  =1.46\times 10^3 kg\\radius\ of\ the\ curve\ r =25 m\\velocity\ of\ the\ automobile\ v=15m/s\\

Objects moving around a circular track will experience centripetal force towards the center of the circular track.

centripetal\ force=mv^2/r\\=1.46\times10^3\times15^2/25\\=1.46\times 10^3\times 225/15\\=1.31\times 10^4 N

4 0
3 years ago
What type of force are you exerting when you stomp your foot on the ground
madam [21]

Answer:

stomp???, frictional force

Explanation:

this allow ur foot to be stomp

3 0
3 years ago
Read 2 more answers
What must be the distance between point charge 91 = 20 MC and point charge 02 -45,0 AC for the electrostatic force between them
Fynjy0 [20]

Answer:

Explanation:

Q₁ = 20 X 10⁻⁶ C

Q₂ = -45 X 10⁻⁶ C

d is required distance.

F = Force between the = 7 N

F = k x Q₁ X Q₂ / d²

d² = k x Q₁ X Q₂ / F = 20 X 45 X 10⁻¹² X 9 X 10⁹  /7

=1.157

d = 1.075 m

7 0
3 years ago
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