A particular HeNe laser beam at 633 nm has a spot size of 0.8 mm. Assuming a

"My distance to the center of the earth is about 4000 miles when I am on the surface. If I go to a height of 8000 miles above th

lord [1]

Given,

Distance from the surface to the center of the earth, d=4000 miles

Distance from the center to you at a height of 8000 miles, a= 8000+4000=12000 miles

The gravitational force acting on a person at the surface is equal to his weight.

From Newton's Universal Law of Gravitation, the gravitational force is

$F=\frac{G\times M\times m}{r^2}$

Where G is the gravitational constant, M is the mass of the earth, m is the mass of the object/person, r is the distance between the center of the earth and the object/person

At the surface, this force is equal to the weight of the person, W=mg

i.e.

$F_s=\frac{G\times M\times m}{d^2}=W$

On substituting the of d,

$W=\frac{\text{GMm}}{4000^2}$

At a height of 8000 miles from the surface, the gravitational force is equal to,

$F_a=\frac{GMm}{12000^2}$

On dividing the above two equations,

$\frac{F_a}{W}=\frac{4000^2^{}}{12000^2}=\frac{1}{9}$

Therefore,

$F_a=\frac{1}{9}W$

Therefore at a height of 8000 miles above the surface of the earth, the force of gravity becomes 1/9 time your weight.

5 0

10 months ago

Two loudspeakers are placed side by side and driven by the same source at 500 Hz. A listener is positioned in front of the two s

Oliga [24]

Answer:

0.68 m

Explanation:

We know that the speed of sound in air is a product of frequency and wavelength. Taking speed of sound in air as 340 m/s

V=frequency*wavelength

Then wavelength is given by 350/500=0.68 m

Therefore, to repeat constructive interference at the listener's ear, a distance of 0.68 m is needed

4 0

2 years ago

If a planet has a non-circular orbit around a star, as the planet moves closer towards a star its orbital speed ….

makkiz [27]

I think the answer you’re looking for is
D)
-I hope this helps! Enjoy the rest of your day

3 0

3 years ago

A spherical balloon is made from a material whose mass is 3.30 kg. The thickness of the material is negligible compared to the 1

stellarik [79]

Answer:

563712.04903 Pa

Explanation:

m = Mass of material = 3.3 kg

r = Radius of sphere = 1.25 m

v = Volume of balloon = $\frac{4}{3}\pi r^3$

M = Molar mass of helium = $4.0026\times 10^{-3}\ kg/mol$

$\rho$ = Density of surrounding air = $1.19\ kg/m^3$

R = Gas constant = 8.314 J/mol K

T = Temperature = 345 K

Weight of balloon + Weight of helium = Weight of air displaced

$mg+m_{He}g=\rho vg\\\Rightarrow m_{He}=\rho vg-m\\\Rightarrow m_{He}=1.19\times \frac{4}{3}\pi 1.25^3-3.3\\\Rightarrow m_{He}=6.4356\ kg$

Mass of helium is 6.4356 kg

Moles of helium

$n=\frac{m}{M}\\\Rightarrow n=\frac{6.4356}{4.0026\times 10^{-3}}\\\Rightarrow n=1607.85489$

Ideal gas law

$P=\frac{nRT}{v}\\\Rightarrow P=\frac{1607.85489\times 8.314\times 345}{\frac{4}{3}\pi 1.25^3}\\\Rightarrow P=563712.04903\ Pa$

The absolute pressure of the Helium gas is 563712.04903 Pa

3 0

2 years ago

A 100 kg box as showsn above is being pulled along the x axis by a student. the box slides across a rough surface, and its posit

Anit [1.1K]

Answer:

a) 2 m/s

b) i) $K.E = 50 (1.5t^2 + 2) ^2\\$

ii) $F = 3tm$

Explanation:

The function for distance is $x = 0.5t ^3 + 2t$

We know that:

Velocity = $v= \frac{d}{dt} x$

Acceleration = $a= \frac{d}{dt}v$

To find speed at time t = 0, we derivate the distance function:

$x = 0.5 t^3 + 2t\\v= x' = 1.5t^2 + 2$

Substitute t = 0 in velocity function:

$v = 1.5t^2 + 2\\v(0) = 1.5 (0) + 2\\v(0) = 2$

Velocity at t = 0 will be 2 m/s.

To find the function for Kinetic Energy of the box at any time, t.

$Kinetic \ Energy = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times 100 \times (1.5t^2 + 2) ^2\\\\K.E = 50 (1.5t^2 + 2) ^2\\$

We know that $Force = mass \times acceleration$

$a = v'(t) = 1.5t^2 + 2\\a = 3t$

$F = m \times a\\F= m \times 3t\\F = 3tm$

6 0

3 years ago