Which of the following layers in the earth has the highest density?

Physics

1 answer:

Ray Of Light [21]3 years ago

5 0

If we are being specific, the inner core has the highest density, but if not then the core in general

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The angle turned through the flywheel of a generator during a time interval t is given by theta = at + bt^3 -ct^4, where a, b, a

Alenkinab [10]

Answer:

α = $6bt - 12ct^2$

Explanation:

The angle, θ, has been given as:

θ = $at + bt^3 -ct^4$

To obtain angular acceleration, α, we have to differentiate the angle twice, with respect to time, t.

Differentiating once will yield the angular velocity, ω:

ω = dθ/dt = $a + 3bt^2 - 4ct^3$

This can then be differentiated to obtain angular acceleration:

α = dω/dt = $6bt - 12ct^2$

5 0

3 years ago

A 50.0 g toy car is released from rest on a frictionless track with a vertical loop of radius R (loop-the-loop). The initial hei

Mariana [72]

Answer:

the speed of the car at the top of the vertical loop $v_{top} = 2.0 \sqrt{gR \ \ }$

the magnitude of the normal force acting on the car at the top of the vertical loop $F_{N} = 1.47 \ \ N$

Explanation:

Using the law of conservation of energy ;

$mgh = mg (2R) + \frac{1}{2}mv^2_{top}\\\\mg ( 4.00 \ R) = mg (2R) + \frac{1}{2}mv^2_{top}\\\\g(4.00 \ R) = g (2R) + \frac{1}{2}v^2 _{top}\\\\v_{top} = \sqrt{2g(4.00R - 2R)}\\\\v_{top} = \sqrt{2g(4.00-2)R$

$v_{top} = 2.0 \sqrt{gR \ \ }$

The magnitude of the normal force acting on the car at the top of the vertical loop can be calculated as:

$F_{N} = \frac{mv^2_{top}}{R} \ - mg\\\\F_{N} = \frac{m(2.0 \sqrt{gR})^2}{R} \ - mg\\\\F_{N} = [(2.0^2-1]mg\\\\F_{N} = [(2.0)^2 -1) (50*10^{-3} \ kg)(9.8 \ m/s^2]\\\\$