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3 years ago

What objects do balanced forces act on?

Physics

1 answer:

leva [86]3 years ago

4 0

Answer: Stationary or constant velocity

Explanation:

Objects with balanced forces acting on them experience no change in motion, or no acceleration. So these objects could either be stationary at rest or have a constant velocity. These include a hanging object, a floating object, an object on a table that doesn't move, and a car moving at a constant 10 mph

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4 years ago

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What is the frequency of an x-ray with wavelength 0.15 nm ? assume that the wave travels in free space?

Roman55 [17]

Let's convert first the wavelength into meters:
$\lambda = 0.15 nm = 0.15 \cdot 10^{-9}m$
Then, we know that the wavelength is related to the frequency by the following equation:
$f = \frac{c}{\lambda}$
Where $c=3 \cdot 10^8 m/s$ is the speed of light in free space. Therefore, putting in the numbers, we find the frequency f:
$f= \frac{3 \cdot 10^8 m/s}{0.15 \cdot 10^{-9}m}=2 \cdot 10^{18}Hz$

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3 years ago

You move a 25 N object 5 meters. If it takes 8 s how much power did you do?

klasskru [66]

Answer:

15.625 watts

Explanation:

Recall that power is defined as the worked performed per unit of time:

Power = Work / time

The work done is Force * distance, so in our case the work is:

Work = 25 M * 5 m = 125 J

Then the power will be:

Power = 125 J / 8 sec = 15.625 watts

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3 years ago

Regina has a very busy schedule and can only exercise every now and then. How do the benefits of occasional exercise compare wit

bixtya [17]

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3 years ago

A sphere of radius 5.15 cm and uniform surface charge density +12.1 µC/m2 exerts an electrostatic force of magnitude 35.9 ✕ 10-3

rosijanka [135]

The radius of the sphere is r=5.15 cm=0.0515 m, and its surface is given by
$A=4 \pi r^2 = 4 \pi (0.0515 m)^2 = 0.033 m^2$

So the total charge on the surface of the sphere is, using the charge density
$\rho=+1.21 \mu C/m^2 = +1.21 \cdot 10^{-6} C/m^2$ :
$Q= \rho A = (+1.21 \cdot 10^{-6} C/m^2)(0.033 m^2)=4.03 \cdot 10^{-8}C$

The electrostatic force between the sphere and the point charge is:
$F=k_e \frac{Qq}{r^2}$
where
ke is the Coulomb's constant
Q is the charge on the sphere
$q=+1.75 \muC = +1.75 \cdot 10^{-6}C$ is the point charge
r is their separation

Re-arranging the equation, we can find the separation between the sphere and the point charge:
$r=\sqrt{ \frac{k_e Q q}{F} }= \sqrt{ \frac{(8.99 \cdot 10^9 Nm^2 C^{-2})(4.03 \cdot 10^{-8} C)(1.75 \cdot 10^{-6}C)}{35.9 \cdot 10^{-3}N} }=0.133 m=13.3 cm$

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3 years ago