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3 years ago

How many significant figures should be written in the sum 4.65g +9.032g+580.0078+540.439g?

Physics

1 answer:

Ne4ueva [31]3 years ago

7 0

Answer: 1.13 X 10^3 g or 1130 g

Explanation: When you add up everything, you will get 1134.1288 g = 1.11341218 X 10^3. But while adding significant figures, we always look for the least significant figures (here 4.65 has least sig. figures i.e. 3) and take that as a reference to how many significant figures we should have in the answer. (Pretty hard to explain!) Hope you got it!!

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3 years ago

Find the electric force acting on an alpha particle in a horizontal electric field of 600N/C

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Since an alpha particle has 2 protons and no negative particles (electrons) to balance the net charge, its charge is
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4 years ago

A block of mass 0.404 0.404 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result, the spring

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To solve this problem it is necessary to apply the concepts related to the Force from Hook's law as well as the definition of the period provided by the same definition.

We know that the Force can be defined as

$F = xk \rightarrow mg = kx \Rightarrow k = \frac{mg}{x}$

Where

k = Spring constant

x = Displacement

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At the same time the period of a spring mass system is defined as

$T = 2\pi \sqrt{\frac{m}{k}}$

Where

m = Mass

k = Spring constant

Our values are given as,

m = 0.404kg

x = 0.666m

Replacing to find the value of the Spring constant we have that

$k = \frac{mg}{x}$

$k = \frac{(0.404)(9.8)}{0.666}$

$k = 5.944N/m$

Now using the formula of the period we know that

$T = 2\pi \sqrt{\frac{m}{k}}$

$T = 2\pi \sqrt{\frac{0.404}{5.944}}$

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3 years ago

Answer ASAP and only if you know its correct

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Answer:

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2 years ago

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A horizontal spring with spring constant 750 N/m is attached to a wall. An athlete presses against the free end of the spring, c

ANTONII [103]

Answer:

37.5 N Hard

Explanation:

Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.

Using the expression for hook's law,

F = ke.............. Equation 1

F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.

Given: k = 750 N/m, e = 5.0 cm = 0.05 m

Substitute into equation 1

F = 750(0.05)

F = 37.5 N

Hence the athlete is pushing 37.5 N hard

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3 years ago

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