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4 years ago

It takes brooke 10minutes to run 1 mile what is her speed in miles per minutes

1 answer:

hjlf4 years ago

6 0

If you divide miles by minutes, the answer will have units of
miles per minute, which is exactly what you want.

(1 mile) / (10 minutes) = 1/10 mile/minute = 0.1 mile per minute

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The world energy consumption was about 6*10^22 J. How much area must a parallel plate capacitor need to store this energy Assume

Nastasia [14]

Answer:

$A = 5 \times 10^{32} m^2$

Explanation:

As we know that the energy stored in the capacitor is given as

$Q = \frac{1}{2}CV^2$

here we know that

$Q = 6 \times 10^{22} J$

also we know that

$V = 5 Volts$

now we have

$6 \times 10^{22} = \frac{1}{2}C(5^2)$

$C = 4.8 \times 10^{21} F$

now we know the formula of capacitance

$C = \frac{\epsilon_0 A}{d}$

$4.8 \times 10^{21} = \frac{(8.85 \times 10^{-12})(A)}{1}$

$A = 5 \times 10^{32} m^2$

3 0

3 years ago

A 5kg wheel rolls off a flat roof of a 50 m tall building at 12m/s.

Olegator [25]

Explanation:

a) Given in the y direction (taking down to be positive):

Δy = 50 m

v₀ = 0 m/s

a = 10 m/s²

Find: t

Δy = v₀ t + ½ at²

50 m = (0 m/s) t + ½ (10 m/s²) t²

t = 3.2 s

b) Given in the x direction:

v₀ = 12 m/s

a = 0 m/s²

t = 3.2 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (12 m/s) (3.2 s) + ½ (0 m/s²) (3.2 s)²

Δx = 38 m

6 0

3 years ago

Reply quick pls

Sholpan [36]

The ball was moving for 0.8seconds
This is because 4/5 x 1 = 0.8

Hope this helps and good luckkkk :)

4 0

3 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

When a star goes from the main sequence to the red giant phase, it becomes larger rather than smaller. This is true because grav

GuDViN [60]

I believe this is electron degeneracy, because the star is essentially having too many reactions too fast and collapses in on itself eventually.

4 0

3 years ago