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2 years ago

Which characterstic is related to kinetic enegry but not potential energy.

Physics

1 answer:

V125BC [204]2 years ago

8 0

Answer:

Explanation:

Kinetic energy is the energy of motion

KE=.5mv^2

>m= mass

>v= velocity (m/s)

PE=mgh

>m= mass

>g= acceleration due to graviry

>h= height

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Which nutrient is the most difficult to break down and, thus, is generally used last for energy in the body?

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Answer:

proteins

Explanation:

protiens is the most difficult nutrient to brek down

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3 years ago

Read 2 more answers

A daring 510-N swimmer dives off a cliff with a running horizontal leap. What must her minimum speed be just as she leaves the t

oee [108]

Answer:

$v_x = 1.26 m/s$

Explanation:

given,

weight of swimmer = 510 N

length of ledge, L = 1.75 m

vertical height of the cliff, h = 9 m

speed of the swimmer = ?

horizontal velocity of the swimmer should be that much it can cross the wedge.

distance = speed x time

d = v_x × t

1.75 = v_x × t ........(1)

now,time taken by the swimmer to cover 9 m

initial vertical velocity of the swimmer is zero.

using equation of motion for time calculation

$s = ut +\dfrac{1}{2}gt^2$

$9= 0+\dfrac{1}{2}\times 9.8\times t^2$

t² = 1.938

t = 1.39 s

same time will be taken to cover horizontal distance.

now, from equation 1

1.75 = v_x × 1.39

$v_x = 1.26 m/s$

horizontal speed of the swimmer is equal to 1.26 m/s

4 0

3 years ago

Which of the following is equal to 1W? A.1 J B.1 J/s C.1 J·s D.1 N·m

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B. Extra text to get to 20 characters.

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2 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 ( $\frac{1}{3}V(R)$ )