Please guys i need help in this

Physics

1 answer:

JulijaS [17]2 years ago

8 0

Answer:

10 seconds

Explanation:

We have the equation V = at (speed = acceleration x time)

We want to find the time, so can rearrange to T = V/a (time = speed / acceleration).

From the question, we know V is 5 and a is 0.5.

Now we can substitute that into our equation: 5/0.5 = 10.

So the time is 10 seconds.

Hope this helps! Let me know if you have any questions :)

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A spotlight on the ground shines on a wall 15 meters away. A man 1.8 meters tall walks away from the spotlight toward the buildi

Alex17521 [72]

Answer:

$s=(15-x)m$

Explanation:

Let the distance from spotlight to wall be 15m, and distance from the man to the building be $x$ .

#Therefore the height of the shadow as a function of the above is $s=(15-x )m$

Hence, height of the shadow is expressed as s=(15-x)m

#See attached photo for illustration

6 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

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