A pendulum with a length of 2 m has a period of 2.8 s. What is the period of a pendulum with a length of 8 m

If an electronin an electron beam experiences a downward force of 2.0x10^-14N while traveling in a magnetic field of 8.3x10^-2T

Anni [7]

Answer:

Explanation:

Given that,

Force is downward I.e negative y-axis

F = -2 × 10^-14 •j N

Magnetic field is westward, +x direction

B = 8.3 × 10^-2 •i T

Charge of an electron

q = 1.6 × 10^-19C

Velocity and it direction?

Force in a magnetic field is given as

F = q(V×B)

Angle between V and B is 270, check attachment

The cross product of velocity and magnetic field

F =qVB•Sin270

2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2

Then,

v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)

v = 1.51 × 10^6 m/s

Direction of the force

Let x be the direction of v

-F•j = v•x × B•i

From cross product

We know that

i×j = k, j×i = -k

j×k =i, k×j = -i

k×i = j, i×k = -j OR -k×i = -j

Comparing -k×i = -j to given problem

We notice that

-F•j = q ( -V•k × B×i)

So, the direction of V is negative z- direction

V = -1.51 × 10^6 •k m/s

6 0

2 years ago

A particle with charge −− 5.90 nCnC is moving in a uniform magnetic field B⃗ =−(B→=−( 1.28 TT )k^k^. The magnetic force on the p

maks197457 [2]

Answer:

Explanation:

Given that,

Charge q=-5.90nC

Magnetic field B= -1.28T k

And the magnetic force

F =−( 3.70×10−7N )i+( 7.60×10−7N )j

Let the velocity be V(xi + yj + zk)

Then, the force is given as

Note i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

The force in a magnetic field is given as

F= q(v×B)

−( 3.70×10−7N )i+( 7.60×10−7N )j =

q(xi + yj + zk) × -1.28k

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28x i×k - 1.28y j×k - 1.28z k×k)

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( 1.28xj - 1.28y i )

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28y i + 1.28x j)

So comparing comparing coefficients

let compare x axis component

-( 3.70×10−7N )i=-1.28qy i

−3.70×10−7N = -1.28qy

y= -3.7×10^-7/-1.28q

y= -3.7×10^-7/-1.28×-5.90×10^-9)

y=-48.99m/s

y≈-49m/s

Let compare y-axisaxis

7.6×10−7N j = 1.28qx j

7.6×10−7N = 1.28qx

x= 7.6×10^-7/1.28q

x= 7.6×10^-7/1.28×-5.90×10^-9

x=-100.64m/s

a. Then, the velocity of the x component is x= -100.64m/s

b. Also, the velocity component of the y axis is y=-49m/s

c. We will compute

V•F

V=-100.64i -49j

F=−( 3.70×10−7 N )i+( 7.60×10−7 N )j

Note

i.j=j.i=0. Also i.i = j.j =1

V•F is

(-100.64i-49j)•−(3.70×10−7N)i+(7.60×10−7 N )j =

3.724×10^-5 - 3.724×10^-5=0

V•F=0

d. Angle between V and F

V•F=|V||F|Cosx

0=|V||F|Cosx

Cosx=0

x= arccos(0)

x=90°

Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other

4 0

3 years ago

A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above

Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed $v_{1}= 2\ m/s$

Speed $v_{2}=7\ m/s$

Pressure in main pipe $P_{1}=2\times10^{5}\ Pa$

(I). We need to calculate the diameter

Using equation of continuity

$Av_{1}=Av_{2}$

$\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}$

$(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7$

$d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}$

$d_{2}=5.345\ cm$

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

$P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})$

$P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)$

$P_{2}=1.28500\times10^{5}\ Pa$

(III). If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}$

$P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})$

Here, $h_{3}=0$

Put the value in the equation

$P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17$

$P_{3}=3.5400\times10^{5}\ Pa$

Hence, This is required solution.

7 0

3 years ago

HELP PLZ HELP ME ANSER THIS WILL GIVE BRAINLIEST

zaharov [31]

Answer:

fertilization

Explanation:

4 0

3 years ago

Suppose that light from a laser with wavelength 633 nm is incident on a thin slit of width 0.500 mm. If the diffracted light pro

coldgirl [10]

Explanation:

It is given that,

Wavelength, $\lambda=633\ nm=633\times 10^{-9}\ m$

Slit width, $a=0.5\ mm=0.0005\ m$

Order, m = 2

If the diffracted light projects onto a screen at distance 1.50 m, L = 1.5 m

For the diffraction of light,

$y=\dfrac{m\lambda L}{a}$

$y=\dfrac{2\times 633\times 10^{-9}\times 1.5}{0.0005}$

y = 0.0037 m

So, the distance from the center of the diffraction pattern to the dark band is 0.0037 meters. Hence, this is the required solution.

3 0

3 years ago