In an energy pyramid, the total energy available for consumption from one trophic level to the next one it is

What is the massof the largest ruby?

alexandr402 [8]

I think the answer is 2283g

4 0

3 years ago

Help me please !!!!!

bezimeni [28]

Answer:

confounding cause they had exposure to many programmes

8 0

3 years ago

Read 2 more answers

Ballistic data obtained on a firing range show that aerodynamic drag reduces the speed of a .44 magnum revolver bullet from 250

m_a_m_a [10]

Answer:

0.363999909622

Explanation:

F = Force

m = Mass = 15.6 g

C = Drag coefficient

ρ = Density of air = 1.21 kg/m³

A = Surface area = $\dfrac{\pi}{4}d^2$

v = Terminal velocity = $v=210\ m/s$

s = Displacement = 150 m

$a=\dfrac{v^2-u^2}{2s}$

Force is given by

F = ma

$F=\dfrac{1}{2}\rho CAv^2\\\Rightarrow ma=\dfrac{1}{2}\rho CAv^2\\\Rightarrow m\dfrac{v^2-u^2}{2s}=\dfrac{1}{2}\rho CAv^2\\\Rightarrow C=2\times m\dfrac{v^2-u^2}{2s}\times\dfrac{1}{\rho Av^2}\\\Rightarrow C=2\times15.6\times 10^{-3}\dfrac{210^2-250^2}{2\times 150}\times\dfrac{1}{1.21\times\dfrac{\pi}{4}\times (11.2\times 10^{-3})^2(210)^2}\\\Rightarrow C=-0.363999909622$

The drag coefficient is 0.363999909622 (ignoring negative sign)

4 0

3 years ago

Help meee please please please

ehidna [41]

A) le ha dado a conocer el nombre del director del festival del río y en la ciudad del sur del mar de la plata en el centro del mar del sur sur del sur y el mar de la sierra del mar de la plata del mar del plata

8 0

2 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

5 0

3 years ago