Another word for transformations in science

In a bi-prism experiment the eye-piece was placed at a distance 1.5m from the source. The distance between the virtual sources w

Vladimir [108]

Answer:

λ = 1.4 × 10^(-7) m

Explanation:

We are given;

distance of eye piece from the source;D = 1.5 m

distance between the virtual sources;d = 7.5 × 10^(-4) m

To find the wavelength, we will use the formula for fringe width;

X = λD/d

Where X is fringe width, λ is wavelength, while d and D remain as before.

Now, fringe width = eye-piece distance moved transversely/number of fringes

Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m

Thus,

Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m

Thus;

1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))

λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5

λ = 1.4 × 10^(-7) m

6 0

2 years ago

Using a scale diagram, calculate the resultant force acting on a sailing boat when an easterly wind provides 2, point, 50, k, N,

EastWind [94]

Answer:

F = 3.6 kN, direction is 9.6º to the North - East

Explanation:

The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.

Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.

Wind

X axis

F₁ = 2.50 kN

Tide

cos 30 = F₂ₓ / F₂

sin 30 = F_{2y} / F₂

F₂ₓ = F₂ cos 30

F_{2y} = F₂ sin 30

F₂ₓ = 1.20cos 30 = 1.039 kN

F_{2y} = 1.20 sin 30 = 0.600 kN

the resultant force is

X axis

Fₓ = F₁ₓ + F₂ₓ

Fₓ = 2.50 +1.039

Fₓ = 3,539 kN

F_y = F_{2y}

F_y = 0.600

to find the vector we use the Pythagorean theorem

F = $\sqrt{F_x^2 +F_y^2}$

F = $\sqrt{ 3.539^2 + 0.600^2 }$

F = 3,589 kN

the address is

tan θ = F_y / Fₓ

θ = tan⁻¹ $\frac{F_y}{F_x}$

θ = tan⁻¹ $\frac{0.6}{3.539}$ 0.6 / 3.539

θ = 9.6º

the resultant force to two significant figures is

F = 3.6 kN

the direction is 9.6º to the North - East

7 0

2 years ago

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

2 years ago

PLEASE HELP!! At what temperature will silver have a resistivity that is four times the resistivity of tungsten at room temperat

Svetach [21]

Consider 20 deg.C. as room temperature.

From tables,
Silver has a resistivity of 1.6*10^-8 ohm-m at 20 deg.C, and it increases by 0.0038 ohm-m per deg.K increase.
Therefore if the temperature rise above 20 deg.C is T, then silver will have resistivity of
1.6*10^-8(1 + 0.0038T) ohm-m

At room temperature, the resistivity of tungsten (from tables) is 5.6*10^-8.

The resistivity of silver will be 4 times that of tungsten (at room temperature) when
1.6*10^-8(1 + 0.0038T) = 4*5.6*10^-8
1 + 0.0038T = 14
T = 13/.0038 = 3421 deg.K approx

Answer: 20 + 3421 = 3441 °C

4 0

2 years ago

A glider is gliding through the air at a height of 416 meters with a speed of 45.2 m/s. The glider

julia-pushkina [17]

There's not enough information to find an answer.

I think the idea here is that in descending (416 - 278) = 138 meters,
the glider gives up some gravitational potential energy, which
becomes kinetic energy at the lower altitude. This is all well and
good, but we can't calculate the difference in potential energy
without knowing the mass of the glider.

3 0

2 years ago