Op- amp as integrater

State the laws of vibration of a stringed Instrument

olchik [2.2K]

If the length and linear density are constant, the frequency is directly proportional to the square root of the tension.

7 0

3 years ago

A proton is moving in a circular orbit of radius 12 cm in a uniform 0.31-T magnetic field perpendicular to the velocity of the p

PIT_PIT [208]

Answer:

Explanation:

A proton of charge

q=+1.609×10^-19C

Orbit a radius of 12cm

r=0.12m

Magnetic Field of 0.31T

Angle between velocity and field is 90°

a. Because the magnetic force F supplies the centripetal force Fc.

The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by

F = qvB sin θ

And the centripetal force is given as

Fc=mv²/r

Where m is mass of proton

m=1.673×10^-27kg

Then, F=Fc

qvB sin θ=mv²/r

qBSin90=mv/r

rqB=mv

Then, v=rqB/m

v=0.12×1.609×10^-19×0.31/1.673×10^-23

v=3577692.78m/s

v=3.58×10^6m/s

b. Since,

F=qVBSin90

F=1.609×10^-19×3.58×10^6×0.31

F=1.785×10^-13 N.

6 0

4 years ago

The y component of a vector is 36, and the angle between the vector and the x axis is 27 what is the magnitude of the vector

xz_007 [3.2K]

Answer:

Magnitude of Vector = 79.3

Explanation:

When a vector is resolved into its rectangular components, it forms two vector components. These components are named as x-component and y-component, they are calculated by the following formulae:

x-component of vector = (Magnitude of Vector)(Cos θ)

y-component of vector = (Magnitude of Vector)(Sin θ)

where,

θ = angle of the vector with x-axis = 27°

Therefore, using the values in the equation of y-component, we get:

36 = (Magnitude of Vector)(Sin 27°)

Magnitude of Vector = 36/Sin 27°

<u>Magnitude of Vector = 79.3</u>

3 0

3 years ago

4. If a book falls off the kitchen counter and hits the floor 0.51 seconds later,

Zigmanuir [339]

Answer:

1.2 miles per second

Explanation:

6 0

3 years ago

A 20.0 kg crate sits at rest at the bottom of a 15.0-m-long ramp that is inclined at 34.0° above horizontal. A constant horizont

ankoles [38]

Answer:

987 joules, 3.01s

Explanation:

(A)

from the attached diagram

net force, Fnet, pulling the crate up the ramp is given by

Fnet = FcosФ - WsinФ - Fr

where FcosФ is the component of horizontal force 290N resolved parallel to the plane

WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane

Fr = constant opposing frictional force

Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65

Fnet = 240.421 - 109.602 - 65

Fnet = 65.82N

Work done on the crate up the ramp, W, is given by

W = Fnet × d (distance up the plane)

W = 65.819 × 15

W = 987.285 joules

W = 987 joules (to 3 significant Figures)

(B)

to calculate the time of travel up the ramp

we use the equation of motion

$s = ut + \frac{1}{2}at^{2}$

where s = distance up the plane, 15m

u = Initial velocity of the crate, which is 0 for a body that is initially at rest

a = acceleration up the plane, given by

$a = \frac{Fnet}{m}$

where m = mass of the crate, 20 kg

now, $a = \frac{65.819}{20} \\a = 3.291\frac{m^{2} }{s}$

from, $s = ut + \frac{1}{2}at^{2}$

$15 = 0*t + \frac{1}{2}* 3.291 * t^{2}$

15 = 0 + 1.645 $t^{2}$

15 = 1.645 $t^{2}$

$t = \sqrt{\frac{15}{1.645} }$

$t = 3.019$

t = 3.01s (to 3 sig fig)

7 0

3 years ago

Op- amp as integrater ​

Op- amp as integrater