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3 years ago

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What is the net force of this object if it is 400 N up and 600 N down?

1 answer:

Marianna [84]3 years ago

3 0

Hello!

what is the net force of this object if it is 400 N up and 600 N down?

In this case, as the meanings of the vectors are opposite, one of the forces receives the negative signal and thus a piece of the other vector is drawn, as shown in the figure. And to know the module of the resultant force, if it subtracts the modules of the forces.

$F_{Net} = F_{grav} - F_{air}$

$F_{Net} = 600 - 400$

$\boxed{\boxed{F_{Net} = 200N}}\end{array}}\qquad\checkmark$

Answer:

200N

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I Hope this helps, greetings ... Dexteright02! =)

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Answer:

Explanation:

Part A

- Work function ( ∅ ) is the minimum energy required by the photon to knock an electron out of the metal surface. That is the portion of energy of a photon transferred to an electron so that it can escape the metal.

- We can mathematically express it:

∅ = Ep - Ek

= $\frac{h*c}{lamba} - 0.5*m_e*v^2_e\\$

Where,

Planck's constant ( h ) = 6.6261*10^-34

Speed of light ( c ) = 3*10^8 m/s

mass of an electron ( m_e ) = 9.1094*10^-31 kg

Given:-

Incident light's wavelength ( λ ) = 250*10^-9 m

The maximum speed o electron ( v_e ) = 4*10^5 m/s

Solution:-

- Plug the values into the expression derived before:

∅ = $\frac{(6.6261*10^-^3^4)*(3*10^8)}{250*10^-^9} \\\\$

∅ = $7.22257*10^-^1^9 J * \frac{6.242*10^1^8 eV}{J}$

∅ = 4.508 eV ... Answer

Part B

- 2.5% of the energy emitted by a 100-W light bulb was visible light with wavelength ( λ ) = 500*10^-9 m. In 1.0 min the amount of energy harbored by a stream of photons are:

$E_p = n_p*\frac{h*c}{lambda} = P*t*e$

Where,

Planck's constant ( h ) = 6.6261*10^-34

Speed of light ( c ) = 3*10^8 m/s

Given:-

visible light's wavelength ( λ ) = 500*10^-9 m

Power of light bulb ( P ) = 100 W

Time taken ( t ) = 1.0 min = 60 s

Portion of energy as light ( e ) = 0.025

Solution:-

- Plug the values into the expression derived before:

$n_p = \frac{(lambda)*(P)*(t)*(e)}{h*c} \\\\n_p = \frac{(500*10^-^9)*(100)*(60)*(0.025)}{(6.6261*10^-^3^4)*(3*10^8)} \\$

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Part C

- A blood vessel of radius ( r ) with length ( L ) carries blood with viscosity ( μ ). The pressure drop ( ΔP ) in the blood vessel was witnessed.

- Pressure loss ( ΔP ) in a cylindrical blood vessel is given by the Darcy's equation given below:

$\frac{dP}{p} = f*\frac{L}{D}*\frac{v^2}{2}$

Where,

ρ: Density of blood

f: Friction factor

D: Diameter of vessel

v: Average velocity

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$v = \frac{Q}{\pi*r^2 }$

Given:-

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$v = \frac{3.6361*10^-^1^2}{\pi*(10*10^-^6)^2 }$

v = 0.01157 m/s

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$Q = \frac{\pi*(2.5)*(12*10^-^6)^4 }{8*(0.0027)*(10^-^6)}$

Q = 7.54*10^-12 m^3 / s

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