Magnetic field 0.02 mfrom a wire is 0.1 T. What is the magnitude of the magnetic field of 0.01 m from the same wire?

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

andrew11 [14]

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}$

For second order

$\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

8 0

2 years ago

4) A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

VMariaS [17]

Answer:

(a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

Explanation:

Given that,

A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

Suppose, the distance between 40 yard line and 25 yard line is 20 yard.

(a). We need to calculate their speed during that run

Using formula of speed

$v=\dfrac{d}{t}$

Where. d = distance

t = time

Put the value into the formula

$v=\dfrac{18.288}{2}$

$v=10\ m/s$ during that run

(b). We need to calculate their velocity

Using formula of speed

$v=\dfrac{\Delta d}{\Delta t}$

Put the value into the formula

$v=\dfrac{22.86-36.58}{2}$

$v=-6.86\ m/s$

Negative sign shows the direction of motion.

(c). If they kept running at that velocity for another 1.3 seconds,

We need to calculate the final position

Using formula of position

$d=vt$

Put the value into the formula

$d=6.86\times1.3$

$d=8.91\ m$

Hence, (a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

8 0

3 years ago

Anyone willing to help?

grigory [225]

Answer:

I think it is "A" because the wind systems are created by uneven heating of Earth's surface.It also may help form large global wind patterns.

Explanation:

5 0

2 years ago

Read 2 more answers

Artemon [7]

Alike because they both act on the quarks making up the nucleons and they have very short ranges. The Strong Nuclear Force is an attractive force between protons and neutrons that keep the nucleus together and the Weak Nuclear Force is responsible for the radioactive decay of certain nuclei. Which also makes them very different

7 0

3 years ago

A pendulum is transported from sea-level, where the acceleration due to gravity g is 9.80 m/s2, to the bottom of Death Valley. A

tatyana61 [14]

Answer: the value of g in Death Valley is 10.417 m/s²

Explanation:

Given that;

acceleration due to gravity at the point is g = 9.8 m/s²

Lets say the acceleration due to gravity at the bottom of Death valley is g'

as the period of the pendulum is decreased by 3.00%

T' = 0.97 T

T is the period of the pendulum at sea level and T' is the period of the pendulum at bottom of Death valley

therefore from the relation

T = 2π√(l/g)

g'/g = T²/T'²

g' = (T²/ (0.97T)²)g

g' = 1.063g

g' = 10.417 m/s²

therefore the value of g in Death Valley is 10.417 m/s²

7 0

3 years ago