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omeli [17]
3 years ago
15

True or false: Everything in the universe works with harmony because of balanced energy

Physics
2 answers:
Eduardwww [97]3 years ago
7 0

Answer:

Explanation:

True

valentinak56 [21]3 years ago
7 0

Answer:

True..................

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An object is accelerating if it is moving____. circle all that apply. ​
tino4ka555 [31]

Answer:

With changing speed and/or in a circle

3 0
4 years ago
What is the mechanical advantage of the machine shown below? a 5 b 4 c 3 d 2
Alex

Answer:

Explanation: Mechanical advantage is a measure of the force amplification achieved by using a tool, mechanical device or machine system. The device preserves the input power and simply trades off forces against movement to obtain a desired amplification in the output force.

4 0
3 years ago
An athlete is working out in the weight room. He steadily holds 50 kilograms above his head for 10 seconds. Which statement is t
postnew [5]

Answer:

A. The athlete isn’t doing any work because he doesn’t move the weight.

Explanation:

We must remember the definition of work, which says that work is equal to the product of mass by the distance displaced. In this case, the athlete only does work when he lifts the weight from the ground to the point where he holds the weight suspended.

So when he's holding the weight, he doesn't do any work.

3 0
3 years ago
A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the
Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

E=\dfrac{Q}{4\pi\epsilon_0 r^2}

Substitute numerical values:

E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

#SPJ4

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2 years ago
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