All categories

3 years ago

How did newton use creativity and logic in his approach to investigating light?

Physics

2 answers:

Mekhanik [1.2K]3 years ago

7 0

He used is logic to form a structure that could hold light a light bulb.

Fed [463]3 years ago

4 0

He tested light in two different ways to see which was more of his idea he wanted to approach. newton used his own logic and creativity to seeing how lights are used against a prism.

You might be interested in

Mass is transferred from a normal star in a close binary system toward a white dwarf. The material that is transferred to the wh

Anestetic [448]

Answer:

maelstrom

Explanation:

A whirlpool like this is called a maelstrom.

Hope this Helps!! Have an Awesome Day!! (-:

7 0

3 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago

A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if th

aivan3 [116]

Answer:

Part a)

$a = 3.65 m/s^2$

Part b)

$t = 9.5 s$

Part c)

$v_f = 55.1 m/s$

Explanation:

Part a)

As we know that it starts from rest and moves on runway by total distance 165 m

so we will have

$v_f^2 - v_i^2 = 2ad$

$v_f^2 - 0 = 2(a)(165)$

$v_f = 125 km/h = 34.7 m/s$

now we have

$a = 3.65 m/s^2$

Part b)

Now for take off time we will have

$v_f - v_i = at$

$34.7 - 0 = 3.65 t$

$t = 9.5 s$

Part c)

$v_f = v_i + at$

$v_f = 0 + (3.65)(15.1)$

$v_f = 55.1 m/s$

7 0

3 years ago

When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is referred to as

Dafna11 [192]

When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is called "load triangle". Because of a motorcycle's size and weight<span> and the fact that it has only two wheels, how to carry extra load is very important. One has to make sure that they are keeping the weight low and close to the middle of the motorcycle and keep the load evenly from side to side. Heavier items should be in the "load triangle".</span><span> </span>

3 0

3 years ago

How do you know the speed of an electromagnetic wave in a vacuum?

Ivenika [448]

The speed of electromagnetic waves in a vacuum is the same as the speed of light. It can be measured by finding the frequency and wavelength of two different waves, and then by that correlation, the speed of the waveform.
Hope this helps you (:

7 0

4 years ago