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2 years ago

Express force in terms of base units

Physics

1 answer:

MariettaO [177]2 years ago

4 0

<h2><u>Answer:</u></h2>

<u></u>

<u>Force = kgm/s² ⟶ Newton</u>

<h2><u>Explanation:</u></h2>

<u></u>

According to the formula we've learnt,

Force = mass × acceleration.

To find force in terms of base units, we must first identify the base SI units of mass & acceleration.

Base SI unit of mass = kg (kilogram)

Base SI unit of acceleration = m/s² (meter per second squared)

So, the base unit of force is,

Force = mass × acceleration.

Force = kg × m/s²

<u>Force = kgm/s² ⟶ Newton</u>

________________

Hope it helps!

$\mathfrak{Lucazz}$

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Red light of wavelength 651 nm produces photoelectrons from a certain photoemissive material. Green light of wavelength 521 nm p

Mnenie [13.5K]

Answer:

material work function is 0.956 eV

Explanation:

given data

red wavelength 651 nm

green wavelength 521 nm

photo electrons = 1.50 × maximum kinetic energy

to find out

material work function

solution

we know by Einstein photo electric equation that is

for red light

h ( c / λr ) = Ф + kinetic energy

for green light

h ( c / λg ) = Ф + 1.50 × kinetic energy

now from both equation put kinetic energy from red to green

h ( c / λg ) = Ф + 1.50 × (h ( c / λr ) - Ф)

Ф =( hc / 0.50) × ( 1.50/ λr - 1/ λg)

put all value

Ф =( 6.63 × $10^{-34}$ (3 × $10^{8}$ ) / 0.50) × ( 1.50/ λr - 1/ λg)

Ф =( 6.63 × $10^{-34}$ (3 × $10^{8}$ ) / 0.50 ) × ( 1.50/ 651× $10^{-9}$ - 1/ 521 × $10^{-9}$ )

Ф = 1.5305 × $10^{-19}$ J × ( 1ev / 1.6 × $10^{-19}$ J )

Ф = 0.956 eV

material work function is 0.956 eV

4 0

3 years ago

Read 2 more answers

A bystander observes the musicians heading toward each other. When musician #1 is 100 m away, the intensity is 1.24 x 10-8 W/m^2

777dan777 [17]

Explanation:

Given that,

Distance 1, r = 100 m

Intensity, $I_1=1.24\times 10^{-8}\ W/m^2$

If distance 2, r' = 25 m

We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :

$I=\dfrac{P}{4\pi r^2}$ .........(1)

Let I' is the intensity at r'. So,

$I'=\dfrac{P}{4\pi r'^2}$ ............(2)

From equation (1) and (2) :

$I'=\dfrac{Ir}{r'^2}$

$I'=\dfrac{1.24\times 10^{-8}\times 100}{25^2}$

$I'=1.98\times 10^{-9}\ W/m^2$

Intensity level is given by :

$dB=10\ log(\dfrac{I'}{I_o})$ , $I_o=10^{-12}\ W/m^2$

$dB=10\ log(\dfrac{1.98\times 10^{-9}}{10^{-12}})$

dB = 32.96 dB

Hence, this is the required solution.

7 0

3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$