PHYSICS COLLEGE

M = 30.3kg
M = 40.17kg 9
R = 0.5m
G = 6. 67x10^11
F ?​

Answers

Answer 1
Answer:

Answer:

m¹=30.3kg

m²=40.17kg

R=0.5m

G=6.67*10¹¹

F=Gm¹m²/R²

=160.68

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Which are electromagnetic waves? check all that apply.earthquake waves
infrared waves
ocean waves
radio waves
untraviolet waves

Answers

Since electromagnetic waves do not require a medium for their transmission, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

What are electromagnetic waves?

Electromagnetic waves or radiations are waves which occur as a result of the interaction between the electric and magnetic fields.

Electromagnetic waves do not require a material medium for their transmission and as such can travel through a vacuum.

Some examples of electromagnetic waves are radio waves, ultraviolet waves, microwaves, infrared waves etc.

Therefore, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

Learn more about electromagnetic waves at: brainly.com/question/25847009
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An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ.Find an expression for the angular velocity ω in terms of g, L and angle θ.

Answers

Answer:

The expression would be ω = \sqrt{(g)/(L sin 0 ) }

Explanation:

Given that ω is the angular velocity

g is the acceleration due to gravity

L is the length

θ is the angle of downward tilt

For an object we compare the horizontal and vertical component of the forces acting on the body;

For vertical component

T sinθ = mg............1

For the horizontal component

T cos θ = (mv^(2) )/(R) .............2

R is our radius and is  = L cos θ

v = ωR

substituting into equation 2 we have

T cos θ = m(ωR)^(2) /R

T cos θ=m(ω)^(2)R ..................3

Now comparing the vertical and the horizontal component we have;

equation 1 divided by equation 3 we have

T sin θ /T cos θ = mg / m(ω)^(2)R

Tan θ = g / (ω)^(2)R............4

Making ω the subject formula we have;

(ω)^(2) = g/  R Tan θ

But R = L cos θ and Tan θ = sin θ/ cosθ

putting into equation 4 we have;

(ω)^(2) = g /[( L cos θ) x( sin θ/ cosθ)]

(ω)^(2) = g/ L sinθ

ω = \sqrt{(g)/(L sin 0 ) }

Therefor the expression for the angular velocity ω in terms of g, L and angle θ would be ω = \sqrt{(g)/(L sin 0 ) }
Let g, r, L and T are  gravity, radius, length, and angle of string w/r/t vertical, respectively. 
Then
ω²r = rg/Lcos T
ω² = g/L cos T
ω = √(g / L cos T) 

If a small child swallowed a safety pin, whywould an X-ray photograph clearly show the
location of the pin?

Answers

Answer:

yes

Explanation:

it is in the body system

Answer:

it would show clearly because it is a metal piece in the body.

At what partial pressure are argon atoms expected to have a free travel of approximately 5 µm, if the gas is at a temperature of 400 K? The cross section of collision, σ, or Argon is 0.28 nm2Ar molar mass is 39.9 g/mole

Answers

Answer:

2790 Pa

Explanation:

Given wavelength λ= 5μm

temperature T= 400 K

cross section of collision σ= 0.28 nm^2

molar mass = 39.9 g/mole

pressure = P= (RT)/(√(2)N_A\sigma\lambda )

putting values we get

=(8.314*400)/(√(2)*6.022*10^(23)*0.28*10^(-18)*5*10^(-6) )

⇒P = 2790 J/m^3

the partial pressure are argon atoms expected= 2790 Pa

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in meters, t in seconds, and c and b are positive constants. What are the units of (a) constant c and (b) constant b? Find a formula in terms of c, b, and t of the (c) velocity v and (d) acceleration a. (e) At what time t ≥ 0 does the particle reach its maximum x value?

Answers

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=(2c)/(3b).

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=(dx)/(dt)\n=(d)/(dx)(ct^2-bt^3)\n=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = (dv)/(dt)\n=(d)/(dt)(2ct-3bt^2)\n=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left ((dx)/(dt)\right )_(t=t_o)=0.
  2. \rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.

Applying both these conditions,

\rm \left ( (dx)/(dt)\right )_(t=t_o)=0\n2ct_o-3bt_o^2=0\nt_o(2c-3bt_o)=0\nt_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = (2c)/(3b).

For \rm t_o = 0,

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = (2c)/(3b),

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6b\cdot (2c)/(3b)=2c-4c=-2c.

Here,

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.

Thus, the particle reach its maximum x value at time \rm t_o = (2c)/(3b).

wo 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength

Answers

Complete question:

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:

a) the midpoint between the two rings?

b) the center of the left ring?

Answer:

a) the electric field strength at the midpoint between the two rings is 0

b) the electric field strength at the center of the left ring is 2712.44 N/C

Explanation:

Given;

distance between the two rings, d = 25 cm = 0.25 m

diameter of each ring, d = 10 cm = 0.1 m

radius of each ring, r = (0.1)/(2) = 0.05 \ m

the charge on each ring, q = 20 nC

Electric field strength for a ring with radius r and distance x from the center of the ring is given as;

E = (kxQ)/((x^2 +r^2)^(3/2))

The electric field strength at the midpoint;

the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m

E = (kxQ)/((x^2 +r^2)^(3/2)) \n\nE = (8.99*10^(9)*0.125*20*10^(-9))/((0.125^2 + 0.05^2)^(3/2)) \n\nE = 9210.5 \ N/C

E_(left) = 9210.5 \ N/C

The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;

E_(right) = -9210.5 \ N/C

The electric field strength at the midpoint;

E_(mid) = E_(left) + E_(right)\n\nE_(mid) = 9210.5 \ N/C - 9210.5 \ N/C\n\nE_(mid) = 0

(b)

The distance from the right ring to center of the left ring, x = 0.25 m.

E = (KxQ)/((x^2 +r^2)^(3/2)) \n\nE = (8.99*10^(9) *0.25*20*10^(-9))/((0.25^2 + 0.05^2)^(3/2)) \n\nE = 2712.44 \ N/C
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