When sunlight strikes the side of a building, what form of energy is it<br> transformed to?

You walk six blocks east along 12th Street, then three blocks west, then one block east, then six blocks east, then seven blocks

Oksana_A [137]

Displacement depends upon the path taken as it is a vector.

From your problem above we would have a total displacement of;
Defining +x direction as east and -x direction as west

6east-3west+1east+6east-7west

6-3+1+6-7=3 blocks east or + x-direction

So even though they walked a total of 17 blocks it ends up only being 3 blocks total in +xdirection that was travelled by displacement.

Any questions please ask.

5 0

3 years ago

PLEASE HLEP I WILL GIVE BRAINLIEST TO CORRECT ANSWER!!!!!!

Hoochie [10]

Answer:

75 km/h

Explanation:

Speed = Distance divided by Time.

The train went a distance of 300 km in 4 hours. Therefore, the speed of the train is: 300 divided by 4 = 75 km/h

8 0

3 years ago

Read 2 more answers

How do the discovery of gravity and the invention of electronic satellites most likely relate to the processes of scientific inv

Leya [2.2K]

Answer:

c

Explanation:

because i said so

5 0

3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago

Ex 10: My dog runs at 6 m/s for 18 meters. How long did she run for?

Hunter-Best [27]

She ran for 3s

Put 18/6 because in order to find how long she ran for you need to divide the distance by the meters ran, once you do that you will get 3.

7 0

3 years ago

When sunlight strikes the side of a building, what form of energy is it transformed to?