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2 years ago

if the spring balance with the iron ball is relaesed what reading will the spring balance record and wht

Physics

1 answer:

Assoli18 [71]2 years ago

4 0

Answer:

when it is released the reading will be zero because gravitational force acts on it you can also say as free fall body

Explanation:

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In a series circuit of two resistors, the resistors have the same __________; in a parallel circuit of two resistors, the resist

bekas [8.4K]

Answer:

current, voltage

Explanation:

Resistors can both be connected in series and parallel configuration. when a resistor is connected serially, the same amount of current flows through each resistor element but the voltage across each resistor is determined by the magnitude of the resistance of each resistor. However, in a parallel connection, the same voltage is applied across each resistor but different current.

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4 years ago

Which of the following is an example of a push?

Alex777 [14]

Throwing a baseball.

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3 years ago

Read 2 more answers

Water runs through a plumbing with a flow of 0.750m3/s and arrives to every exit of a fountain. At what speed will the water com

Lubov Fominskaja [6]

Divide the flow rate (0.750 m³/s) by the cross-sectional area of each pipe:

diameter = 40 mm ==> area = <em>π</em> (0.04 m)² ≈ 0.00503 m²

diameter = 120 mm ==> area = <em>π</em> (0.12 m)² ≈ 0.0452 m²

Then the speed at the end of the 40 mm pipe is

(0.750 m³/s) / (0.00503 m²) ≈ 149.208 m/s ≈ 149 m/s

(0.750 m³/s) / (0.0452 m²) ≈ 16.579 m/s ≈ 16.6 m/s

7 0

3 years ago

Two point charges are brought closer together, increasing the force between them by a factor of 20. By what factor was their sep

olga55 [171]

The separation between them is $\frac{r}{\sqrt{20} }$

Concept :

If the force increases, distance between charges must decrease. Force is indirectly proportional to the distance squared.

Given,

Two point charges are brought closer together, increasing the force between them by a factor of 20.

Original force is

F = $\frac{kq_{1} q_{2} }{r^{2} }$ -------- ( 1 )

Here, $q_{1} , q_{2}$ are charges and r is the distance between them

New force F' = $\frac{kq_{1q_{2} } }{r'^{2} }$ ----------- (2 )

Divide ( 1 ) and ( 2 )

$\frac{F'}{F}$ = $\frac{\frac{kq_{1}q_{2} }{r'^{2} } }{\frac{kq_{1}_{2} }{r^{2} } }$

20 = $\frac{r^{2} }{r'^{2} }$

r' = $\frac{r}{\sqrt{20} }$

Given that force between them are increasing and therefore distance between them decrease by $\frac{r}{\sqrt{20} }$

Learn more about two point charges here : brainly.com/question/24206363

#SPJ4

8 0

2 years ago

In a double-slit experiment, the third-order maximum for light of wavelength 490 nm is located 15 mm from the central bright spo

jeka94

Answer:

The distance from the central bright spot are $156.8\times10^{-3}\ mm$ and $142.9\times10^{-3}\ mm$ .

Explanation:

Given that,

Wavelength = 490 nm

Distance y= 15 mm

Length L=1.6 m

New wavelength = 670

We need to calculate the distance from the central bright spot

Using formula of distance

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

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

Put the value into the formula

$d=\dfrac{3\times490\times10^{-9}\times1.6}{15\times10^{-3}}$

$d=156.8\times10^{-6}\ m$

$d=156.8\times10^{-3}\ mm$

We need to calculate the distance from the central bright spot for new wavelength

Using formula of distance

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

Put the value into the formula

$d=\dfrac{2\times670\times10^{-9}\times1.6}{15\times10^{-3}}$

$d=142.9\times10^{-6} m$

$d=142.9\times10^{-3}\ mm$

Hence, The distance from the central bright spot are $156.8\times10^{-3}\ mm$ and $142.9\times10^{-3}\ mm$ .

8 0

3 years ago