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blsea [12.9K]
3 years ago
10

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]\\\\

F_{N} = 1.47 \ \ N

4 0
3 years ago
two objects were lifted by a machine. One object had a mass of 2 kilograms, was lifted at a speed of 2m/sec . The other had a ma
Lubov Fominskaja [6]

Object 2 has more kinetic energy

Explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion, and it is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

In this problem, for object 1:

m = 2 kg

v = 2 m/s

So its kinetic energy is

K_1 = \frac{1}{2}(2)(2)^2=4 J

For object 2,

m = 4 kg

v = 3 m/s

So its kinetic energy is

K_2 = \frac{1}{2}(4)(3)^2=18 J

Therefore, object 2 has more kinetic energy.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

5 0
3 years ago
A compound pulley is a type of system. How does a system work?
Nastasia [14]
Here you go fam .......

4 0
2 years ago
A point source emits 25.9 W of sound isotropically. A small microphone intercepts the sound in an area of 0.242 cm2, 590 m from
AnnZ [28]

The solution is in the attachment

7 0
3 years ago
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