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1 year ago

What is the period of a water wave with a frequency of 0.5Hz

Physics

1 answer:

stiv31 [10]1 year ago

8 0

Answer:

a wave with a time period of 2 seconds has a frequency of 1 ÷ 2 = 0.5 Hz. A sound wave has a time period of 0.0001 seconds.

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1.15) What distance will 650 joules of work move a box weighing 50 newtons?

son4ous [18]

Answer:

The answer is 13 however make sure if they ask for a certain measurement like meter answer it by saying 13 meters.

Explanation:

This basically turns into basic algebra if you know the formula for work. The formula for work is W=F*d

Here are the variables that you know 650J=50N*d so you need d.

All you do is divide 650J by 50N and you get a total of 13 (meters since I don't know what they want you to put it in).

6 0

2 years ago

An electron that has an instantaneous velocity of ???? = 2.0 × 106 m ???? ???? + 3.0 × 106 m ???? ???? is moving through the uni

Setler79 [48]

Explanation:

It is given that,

Velocity of the electron, $v=(2\times 10^6i+3\times 10^6j)\ m/s$

Magnetic field, $B=(0.030i-0.15j)\ T$

Charge of electron, $q_e=-1.6\times 10^{-19}\ C$

(a) Let $F_e$ is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :

$F_e=q_e(v\times B)$

$F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)]$

$F_e=-1.6\times 10^{-19}\times (-390000)(k)$

$F_e=6.24\times 10^{-14}k\ N$

(b) The charge of electron, $q_p=1.6\times 10^{-19}\ C$

The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.

8 0

2 years ago

Winner of five swimming gold medals at the 1988 olympics

Fed [463]

Matt Biondi..?
(I don’t know if it’s right, sorry if it is wrong)
:)

6 0

2 years ago

A very long straight wire has charge per unit length 1.44×10-10C/m.

4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m

Explanation:

We have given charge per unit length, that is liner charge density $\lambda =1.44\times 10^{-10}C/m$

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

$E=\frac{\lambda }{2\pi \epsilon _0r}$ , here $\lambda$ is linear charge density and r is distance of the point where we have to find the electric field

So $2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}$

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

3 0

2 years ago

Identify the positive and negative about trade blocs.

ohaa [14]

Answer:

if this is on odyssey ware then i can edit my answer to help you

Explanation:

5 0

2 years ago