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The position of a particle changes linearly with time, i.e. as one power of t, as given by the following: h(t) = ( 4.1 t + 5.5 ) meters. Find the speed of the particle, in meters per second.

Answers

Answer 1

Answer:

Answer:

v = 4.1 m / s

Explanation:

Velocity is defined by the relation

v = $(dx)/(dt)$

we perform the derivative

v = 4.1 m / s

Another way to find this magnitude is to see that the velocity on the slope of a graph of h vs t

v = $(\Delta x)/(\Delta t)$

Δx = v Δdt + x₀

h= 4.1 t + 5.5

v = 4.1 m / s

x₀ = 5.5 m

Answer 2

Answer:

The Speed of a Particle is 4.1 meters per second.

The position of a particle can be represented by a linear equation of the form h(t) = (at + b) where a and b are constants.

In this case, the equation is h(t) = (4.1t + 5.5).

To find the speed of the particle, we can take the derivative of the position equation with respect to time.

The derivative of h(t) is the rate of change of position with respect to time, which represents the velocity of the particle.

In this case, the derivative is 4.1 meters per second.

Therefore, the speed of the particle is 4.1 meters per second.

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A white blood cell has a diameter of approximately 12 micrometers or 0.012 um a model represents its diameter as 24 um what ratio of model size
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A 50.0-g hard-boiled egg moves on the end of a spring with force constant k = 25.0 N/m. It is released with an amplitude 0.300 m. A damping force F_x = -bv acts on the egg. After it oscillates for 5.00 s, the amplitude of the motion has decreased to 0.100 m.Calculate the magnitude of the damping coefficient b.Express the magnitude of the damping coefficient numerically in kilograms per second, to three significant figures.

Answers

Answer

given,

mass of the boiled egg = 50 g = 0.05 Kg

spring constant, k = 25 N/m

initial Amplitude, A₁ = 0.3 m

final amplitude, A₂ = 0.1 m

time, t = 5 s

considering the amplitude of damped harmonic oscillation to calculate damping factor.

$A_2 = A_1 e^{-((b)/(2m))t}$

b is the damping factor and t is the time.

$e^{-((b)/(2m))t}=(A_2)/(A_1)$

on solving

$b= (2m)/(t)ln((A_1)/(A_2))$

inserting all the given values

$b= (2* 0.05)/(5)ln((0.3)/(0.1))$

b = 0.0219 Kg/s

damping coefficient in three significant figure

b = 0.022 Kg/s

What causes rain? a.air becomes filled with water vapor
b.water vapor condenses on dust particles
c. dust particles can no longer support water droplets

Answers

Answer:

the awnser is A

Explanation:

Brainliest Please

If R = 12 cm, M = 430 g, and m = 60 g , find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.)

Answers

Answer:

Explanation:

Given

Radius of Pulley r=12 cm

mass of block m=60 gm

mass of Pulley M=430 gm

Block descend h=50 cm

Applying Conservation of Energy

Potential Energy of block convert to rotational Energy of pulley and kinetic energy of block

i.e.

$mgh=(1)/(2)I\omega ^2+(1)/(2)mv^2$

where I=moment of inertia

$I=mr^2$

and for rolling $\omega =(v)/(r)$

$mgh=(1)/(2)Mv^2+(1)/(2)mv^2$

$v^2=(2mgh)/(m+M)$

$v=\sqrt{(2mgh)/(m+M)}$

$v=\sqrt{(2* 60* 9.8* 0.5)/(430+60)}$

$v=\sqrt{(60* 9.8)/(490)}$

$v=√(1.2)$

$v=1.095 m/s$

Convert 56km/h to m/s.

Answers

Explanation:

15.556 metres per second

Which one of the following statements concerning the Stefan-Boltzmann equation is correct? The equation can be used to calculate the power absorbed by any surface. The equation applies only to perfect radiators. The equation applies only to perfect absorbers. The equation is valid with any temperature units. The equation describes the transport of thermal energy by conduction.

Answers

"The equation can be used to calculate the power absorbed by any surface" statement concerning the Stefan-Boltzmann equation is correct.

Answer: Option A

Explanation:

According to Stefan Boltzmann equation, the power radiated by black body radiation source is directly proportionate to the fourth power of temperature of the source. So the radiation transferred is absorbed by another surface and that absorbed power will also be equal to the fourth power of the temperature. So the equation describes the relation of net radiation loss with the change in temperature from hotter temperature to cooler temperature surface.

$P=e \sigma A\left(T^(4)-T_(c)^(4)\right)$

So this law is application for calculating power absorbed by any surface.

One of the primary visible emissions from a distant planet occurs at 425 nm. Calculate the energy of a mole of photons of this emission.]

Answers

Answer:

Explanation:

Given

Wavelength of incoming light $\lambda =425\ nm$

We know

$speed\ of\ wave=frequency* wavelength$

$frequency=(speed)/(wavelength)$

$\mu =(3* 10^8)/(425* 10^(-9))$

$\mu =7.058* 10^(14)\ Hz$

Energy associated with this frequency

$E=h\mu$

where h=Planck's constant

$E=6.626* 10^(-34)* 7.058* 10^(14)$

$E=46.76* 10^(-20)\ Hz$

Energy of one mole of Photon $=N_a* E$

$=6.022* 10^(23)* 46.76* 10^(-20)$

$=281.58* 10^(3)$

$=281.58\ kJ$

Final answer:

To calculate the energy of a mole of photons of the emission at 425 nm, use the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Convert the wavelength to meters, substitute the values into the equation, and calculate to find the energy of a single photon. Multiply this by Avogadro's number to find the energy of a mole of photons.

Explanation:

To calculate the energy of a mole of photons of the emission at 425 nm, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength (in meters).

Converting the wavelength to meters, we have 425 nm = 425 x 10^-9 m.

Substituting the values into the equation, we get E = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m). Calculating this gives us the energy of a single photon of this emission. To find the energy of a mole of photons, we can multiply this value by Avogadro's number (6.02 x 10^23 photons/mol).

Therefore, the energy of a mole of photons of this emission is (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m) x (6.02 x 10^23 photons/mol).

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