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3 years ago

What is the accepted value of the speed of light through empty space?

Physics

1 answer:

galina1969 [7]3 years ago

8 0

Answer:

299,792 kilometers per second

Explanation:

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Review 1: A plane is located x = 40 miles (horizontally) away from an airport at an altitude of h miles. Radar at the airport de

-Dominant- [34]

Explanation:

Let h is the height of the plane above ground. x is the horizontal distance between the ground and the airport. Let s(t) is the distance between the plane and the airport. So,

$s(t)^2={h^2+x^2}$ ...........(1)

Given, h = 4, x = 40 and s(t) = -20 mph

Differentiate equation (1) wrt t

$2s(t)s'(t)=2x(t)x'(t)$

$x'(t)=\dfrac{s(t)s'(t)}{x(t)}$

When x = 40, $s(t)=\sqrt{40^2+4^2}=40.19\ m$

$x'(t)=\dfrac{-240s(t)}{x(t)}$

$x'(t)=\dfrac{-240\times 40.19}{40}$

$x'(t)=-241.14\ m/s$

So, the speed of the airplane is 241.14 m/s. Hence, this is the required solution.

8 0

3 years ago

ANSWER QUCIK<br> PLS THANK YOU

Reil [10]

Electrical energy in the charger and cable

Chemical energy in the battery of the mobile phone

8 0

2 years ago

Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli

Masja [62]

Answer:

Not possible

Explanation:

$E_{cl}$ = longitudinal modulus of elasticity = 35 Gpa

$E_{ct}$ = transverse modulus of elasticity = 5.17 Gpa

$E_m$ = Epoxy modulus of elasticity = 3.4 Gpa

$V_{\rho l}$ = Volume fraction of fibre (longitudinal)

$V_{\rho t}$ = Volume fraction of fibre (transvers)

$E_f$ = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

$E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764$

Transverse modulus of elasticity is given by

$E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148$