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3 years ago

WARNING IF YOU CANT ANSWER IT DONT EVEN ANSWER OR IM REPORTING AND POSTING YOUR ANSWER ONLINE!!!!! (⌐■_■)☜(⌒▽⌒)☞

Physics

2 answers:

likoan [24]3 years ago

5 0

Answer:

false

Explanation:

andrezito [222]3 years ago

3 0

Answer:

the answer is false :)

Explanation:

Science is based on theories and tests, if someone has a theory they will test and if it is correct yippie! and if it isn't they will redo the test over until they get it correct or unless it is unsolvable.

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What is the mass moment of inertia of a 20kg sphere with a radius of 0.2m about a point on the sphere's perimeter

Kobotan [32]

Answer:

I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).

The moment of inertia about the center of a sphere is 2 / 5 M R^2.

By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is I = 2/5 M R^2 + M R^2 = 7/5 M R^2

I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2

7 0

2 years ago

A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient

gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0

3 years ago

What’s the opposite of an electron

Anna007 [38]

As its charge, proton -a positive charged molecule at the center of an atom- is the opposite of the electron -the particle which is orbiting the center of an atom.

7 0

4 years ago

g A bowling ball with a mass of 3.86 kg and a radius of 0.161 m starts from rest at a height of 2.5 m and rolls down a 48.4 o sl

Fynjy0 [20]

Answer:

$v=1.5m/s$

Explanation:

The gravitational potential energy gets transformed into translational and rotational kinetic energy, so we can write $mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}$ . Since $v=r\omega$ (the ball rolls without slipping) and for a solid sphere $I=\frac{2mr^2}{5}$ , we have:

$mgh=\frac{mv^2}{2}+\frac{2mr^2\omega^2}{2*5}=\frac{mv^2}{2}+\frac{mv^2}{5}=\frac{7mv^2}{10}$

So our translational speed will be:

$v=\sqrt{\frac{10gh}{7}}=\sqrt{\frac{10(9.8m/s^2)(0.161m)}{7}}=1.5m/s$

6 0

3 years ago

Electronic configuration of. <br>1)Fe. <br>2)Fe++ <br>3)Fe+++

mezya [45]

Answer:

The electronic configuration of Fe2+ is 1s2 2s2 2p6 3s2 3p6 3d6 and Fe3+ is 1s2 2s2 2p6 3s2 3p6 3d5. Fe2+ contains 2 fewer electrons compared to the electronic configuration of Fe.

7 0

3 years ago