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3 years ago

A charge of 24 C pass a certain point in a conductor in 10 s. What is the size of current in the conductor?

Physics

2 answers:

arsen [322]3 years ago

8 0

Answer:

2.4A

Explanation:

I = Q/t

I = 24/10

= 2.4A

allsm [11]3 years ago

6 0

Answer:

write the name of any five districts of nepal

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The frequency of the given sound is 1.5khz then how many vibration it is completing in one second ?

kompoz [17]

Answer:

1500 per second.

Explanation:

vibrations = 1.5 kilohertz

1.5×1000=1500

the answer is 1500 per second.

6 0

3 years ago

Read 2 more answers

Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (Let a = 200 N and b = 400 N. Round y

omeli [17]

Answer: R = 346.4N and angle 30° to the horizontal negative axis

Explanation:

To find the resultant force, we need to sum up the forces on the vertical and horizontal axis.

For the horizontal axis;

Rx = -b + acos60

Rx = -400N +200cos60

Rx = -400N +100N

Rx = -300N

For the vertical axis;

Ry = asin60 = 200sin60

Ry = 173.2N

The resultant force R can be given as;

R = √(Rx^2 +Ry^2)

R = √((-300)^2 + 173.2^2)

R = 346.4N

Angle z can be written as

Tanz = Ry/Rx

z = taninverse (Ry/Rx)

z = taninverse (173.2/300)

z = 30°

6 0

3 years ago

Asha walks 15 m west, then 20 m north, then 15 m east. Calculate the distance and displacement

Norma-Jean [14]

Answer:

Asha traveled a distance of 50 m

The displacement is 20 m north.

Explanation:

Distance and Displacement

A moving object constantly travels for some distances at defined periods of time. The total moved distance is the sum of each individual distance the object traveled. It can be written as:

dtotal=d1+d2+d3+...+dn

This sum is obtained independently of the direction the object moves.

The displacement only takes into consideration the initial and final positions of the object while moving. The displacement, unlike distance, is a vectorial magnitude and can even have magnitude zero if the object starts and ends the movement at the same point.

Asha walks 15 m west, 20 m north, and 15 m east. The total distance is

dt=15 m + 20 m + 15 m = 50 m

Asha traveled a distance of 50 m

Note the final position is 20 m above the initial position since she goes 15 m west and 15 m east. Thus the displacement is 20 m north.

5 0

3 years ago

Planck's constant, h, is 6.626 x 10-34 js. the speed of light in a vacuum, c, is 3.00 x 108 m/s. calculate the frequency of a gr

Misha Larkins [42]

Frequency= velocity of light/wave length
Fr= 3×10^8/510×10^-9
Frequwency=5.88×10^14 Hz

4 0

3 years ago

If a proton and an electron are released when they are 6.50×10⁻¹⁰ m apart (typical atomic distances), find the initial accelerat

Cerrena [4.2K]

Answer:

(a) Acceleration of electron= 5.993×10²⁰ m/s²

(b) Acceleration of proton= 3.264×10¹⁷ m/s²

Explanation:

Given Data

distance r= 6.50×10⁻¹⁰ m

Mass of electron Me=9.109×10⁻³¹ kg

Mass of proton Mp=1.673×10⁻²⁷ kg

Charge of electron qe= -e = -1.602×10⁻¹⁹C

Charge of electron qp= e = 1.602×10⁻¹⁹C

To find

(a) Acceleration of electron

(b) Acceleration of proton

Solution

Since the charges are opposite the Coulomb Force is attractive

$F=\frac{1}{4(\pi)Eo }\frac{|qp*qe|}{r^{2} }\\ F=(8.988*10^{9}Nm^{2}/C^{2})*\frac{(1.602*10^{-19})^{2} }{(6.50*10^{-10} )^{2} } \\F=5.46*10^{-10}N$

From Newtons Second Law of motion

F=ma

a=F/m

For (a) Acceleration of electron

$a=F/Me\\a=(5.46*10^{-10} )/9.109*10^{-31}\\ a=5.993*10^{20}m/s^{2}$

For(b) Acceleration of proton

$a=F/Mp\\a=(5.46*10^{-10} )/1.673*10^{-27} \\a=3.264*10^{17}m/s^{2}$

3 0

3 years ago