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3 years ago

The frequency of a wave is 560 Hz. What is it’s period

Physics

1 answer:

Crazy boy [7]3 years ago

4 0

The answer would be, "1/560 seconds".

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F(x) = 3x^2+5x-14<br> Find f(-9)

Llana [10]

Answer:

f(-9) = 184

Explanation:

f(x)=3x²+5x-14

f(-9)= 3(-9)² +5(-9)-14 Order of Operations : Exponents

= 3(81)+5(-9)-14

= 243+5(-9)-14

= 243-45-14

= 198-14

f(-9)= 184

Hope this helps :)

8 0

3 years ago

A nonconducting sphere has radius R = 2.81 cm and uniformly distributed charge q = +2.35 fC. Take the electric potential at the

Sladkaya [172]

Answer:

(a). The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$ .

(b). The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$ .

Explanation:

Given that,

Radius of sphere R=2.81 cm

Charge = +2.35 fC

Potential at center of sphere

$V = 0$

(a). We need to calculate the potential at a distance r = 1.60 cm

Using formula of potential difference

$V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr$

$V_{r}-0=-\int_{0}^{r}{\dfrac{qr}{4\pi\epsilon_{0}R^3}}dr$

$V_{r}=-(\dfrac{qr^2}{8\pi\epsilon_{0}R^3})_{0}^{1.60\times10^{-2}}$

$V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})$

$V_{r}=-0.00012190\ V$

$V_{r}=-1.219\times10^{-4}\ V$

The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$ .

(b). We need to calculate the potential at a distance r = R

Using formula of potential difference

$V_{R}=-\dfrac{2.35\times10^{-15}}{8\pi\times8.85\times10^{-12}\times2.81\times10^{-2}}$

$V_{R}=-0.0003759\ V$

$V_{R}=-3.759\times10^{-4}\ V$

The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$ .

Hence, This is the required solution.

5 0

3 years ago

What is the magnitude of the displacement of a car that travels half a lap along a circle of radius of 150m?

Lisa [10]

It would be equal to 2r = 2* 150 = 300m

6 0

3 years ago

The relatively small, rocky bodies generally found orbiting between mars and jupiter are known as _____. meteoroids satellites c

adelina 88 [10]

The answer that is being described above is the ASTEROIDS. The one that we see floating between Mars and Jupiter is what we call the Asteroid Belt. The asteroid belt comprises of different rocky bodies and they also orbit within the solar system. Hope this helps.

6 0

2 years ago

Read 2 more answers

A proton is placed in an electric field of intensity 700 n/

kakasveta [241]

F=ma
F=QE = 1.602e-19C*700N/C = 1.1214e-16N
1.1214e-16N = ma = 1.6726e-27kg * a
a = 6.702e10 m/s² along the direction of the field line

8 0

2 years ago