PHYSICS COLLEGE

What operation do you apply to the position function of a particle to compute the particles velocity

Answers

Answer 1
Answer:

Answer:

the derivative with respect to time

Explanation:

This is an exercise in kinematics, where the velocity is defined as a function of the position of a body of the form

            v = dx/dt

where v is the velocity of the body, x is the position that we assume is a continuous and differentiable function.

The function written in the equation is the derivative with respect to time

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Nitrogen makes up about what percent of a human's body weight?

Answers

Answer:

the answer is 3.3 %

Explanation:

What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1.90 × 103 m/s?

Answers

The mass of a neutron is:
m=1.67 \cdot 10^(-27)kg
Since we know its speed, we can calculate the neutron's momentum:
p=mv=(1.67 \cdot 10^(-27)kg)(1.90 \cdot 10^3 m/s)=3.17 \cdot 10^(-24) kg m/s

The problem says the photon has the same momentum of the neutron, p.  The photon momentum is given by
p= (h)/(\lambda)
where h is the Planck constant, and \lambda is the photon wavelength. If we re-arrange the equation and we use the momentum we found before, we can calculate the photon's wavelength:
\lambda= (h)/(p)= (6.6 \cdot 10^(-34)Js)/(3.17 \cdot 10^(-24) kg m/s)=2.08 \cdot 10^(-10) m

And since we know the photon travels at speed of light c, we can now calculate the photon frequency:
f= (c)/(\lambda)= (3 \cdot 10^8 m/s)/(2.08 \cdot 10^(-10) m)= 1.44 \cdot 10^(18) Hz

A force of 200N acts on a body that moves along a horizontal plane in the same direction of movement. The body moves 30m. What is the work done by that force?

Answers

Work = Force times Distance

Work = 200 x 30

Work = 6000

The work done by a force of 200N on a body that moved 30m is 6000J or 6000 Joules.

A piccolo and a flute can be approximated as cylindrical tubes with both ends open. The lowest fundamental frequency produced by one kind of piccolo is 522.5 Hz, and that produced by one kind of flute is 256.9 Hz. What is the ratio of the piccolo's length to the flute's length?

Answers

Answer:

ratio of the piccolo's length to the flute's length is 0.4916

Explanation:

given data

frequency of piccolo = 522.5 Hz

frequency of flute = 256.9 Hz

to find out

ratio of the piccolo's length to the flute's length

solution

we get here length of tube that is express as

length of tube = velocity of sound ÷ fundamental frequency .......................1

so here ratio of Piccolo length to flute that is

(L\ picco)/(L\ flute) = (f\ flute)/(f \ piccolo)

(l \ piccolo)/(L\ flute) = (256.9)/(522.5)  =  0.4916

so ratio of the piccolo's length to the flute's length is 0.4916

If 2050 J of heat are added to a 150 g object its temperature increases by 15°C.(a) What is the heat capacity of this object?
(b) What is the object's specific heat?

Answers

When an object gets heated by a temperature ΔT energy needed, E = mcΔT

Here energy is given E = 2050 J

Mass of object = 150 g

Change in temperature ΔT  = 15 ^0C = 15 K

a) Heat capacity of an object equal to the ratio of the heat added to (or removed from) an object to the resulting temperature change.

  So heat capacity = E/ΔT = 2050/15 = 136.67 J/K

b) We have E = mcΔT

                    c = (2050)/(150*10^(-3)*15)  = 911.11 J/kgK

 So object's specific heat = 911.11 J/kgK

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 100 cm . a) What is their initial "launch" speed off the ground?b)How long are they in the air?

Answers

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\n\Rightarrow -u^2=2as-v^2\n\Rightarrow u=√(v^2-2as)\n\Rightarrow u=√(0^2-2* -9.81* 1)\n\Rightarrow u=4.45\ m/s

a) The vertical speed when the player leaves the ground is 4.45 m/s

v=u+at\n\Rightarrow t=(v-u)/(a)\n\Rightarrow t=(0-4.45)/(-9.81)\n\Rightarrow t=0.45\ s

Time taken to reach the maximum height is 0.45 seconds

s=ut+(1)/(2)at^2\n\Rightarrow 1=0t+(1)/(2)* 9.81* t^2\n\Rightarrow t=\sqrt{(1* 2)/(9.81)}\n\Rightarrow t=0.45\ s

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds
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