All categories

2 years ago

Why are there multiple versions of the scientific method?

Physics

1 answer:

gavmur [86]2 years ago

4 0

Answer:

it's because some versions have more steps and others have less

You might be interested in

How will a current change if the resistance of a circuit remains constant while the voltage across the circuit decreases to half

beks73 [17]

Answer:

1. The current will drop to half of its original value.

Explanation:

The problem can be solved by using Ohm's law:

$V=RI$

where

V is the voltage across the circuit

R is the resistance of the circuit

I is the current

We can rewrite it as

$I=\frac{V}{R}$

In this problem, we have:

- the resistance of the circuit remains the same: R' = R

- the voltage is decreased to half of its original value: $V'=\frac{V}{2}$

So, the new current will be

$I'=\frac{V'}{R'}=\frac{V/2}{R}=\frac{1}{2}\frac{V}{R}=\frac{I}{2}$

so, the current will drop to half of its original value.

4 0

2 years ago

Prima problema este urgent

Aleonysh [2.5K]

Answerana alyom kint gahda amshi

Explanation:

5 0

2 years ago

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

5 0

2 years ago

you ride your bike for a distance of 30km. you travel ata a speed of 0.75km/minute.how many minutes does it take

mihalych1998 [28]

Distance for which the bike is ridden = 30 km
Speed at which the bike is driven = 0.75 km/minute
Let us assume the number of minutes taken to travel the distance of 30 km = x
Now we already know the formula of speed can be written as
Speed = Distance traveled/ Time taken
0.75 = 30/x
0.75x = 30
x = 30/0.75
= 40 minutes
So the time taken for riding a distance of 30 km will be 40 minutes. I hope this procedure is simple enough for you to understand.

7 0

2 years ago

Read 2 more answers

A block of ice of mass 4.30 kg is placed against a horizontal spring that has a force constant k = 250 N/m and is compressed a d

OleMash [197]

Answer:

W = 0.060 J

v_2 = 0.18 m/s

Explanation:

solution:

for the spring:

W = 1/2*k*x_1^2 - 1/2*k*x_2^2

x_1 = -0.025 m and x_2 = 0

W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2

W = 0.060 J

the work-energy theorem,

W_tot = K_2 - K_1 = ΔK

with K = 1/2*m*v^2

v_2 = √2*W/m

v_2 = 0.18 m/s

8 0

2 years ago