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3 years ago

Which of the following statements is true of chords? Select all that apply.

1 answer:

5 0

All of those are true, except the one about the radius.

Both ends of a chord have to be on the circle, but one end
of a radius is at the center, so a radius can't be a chord.

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Gravity is affected by (2 points)

Tanya [424]

Answer:

Mass and distance

Explanation:

5 0

3 years ago

A 3.90 kg block is in equilibrium on an incline of 31.0◦. The acceleration of gravity is 9.81 m/s2 . What is Fn of the incline o

storchak [24]

Answer:

Explanation:

The sum of the pore along the plane is expressed according to Newton's law

Fn-Ff = ma

Fn is the moving force

Ff = nR = frictional force

m is the Mass

a is the acceleration

Substitute the given values

Fn - nR = ma

Fn - tan31°(mgcostheta) =3.9(9.8)

Fn - tan31(3.9(9.8)cos31) = 3.9(9.8)

Fn - tan31(38.22cos31)= 38.22

Fn - 32.76tan31 = 38.22

Fn-19.68 = 38.22

Fn = 38.22+19.68.

Fn = 57.90N

Hence Fn (moving force) of the inclined block is 57.90

4 0

3 years ago

Which equation best describes the law of conservation of momentum?

NeX [460]

The law of conservation of momentum<span> states that for two objects colliding in an isolated system, the total </span>momentum<span> before and after the collision is equal. Momentum should be conserved. Hope this answers the question. Have a nice day.</span>

8 0

3 years ago

Read 2 more answers

You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

$I_i$ = Initial moment of inertia = 9.3 kgm²

$I_f$ = Final moment of inertia = 5.1 kgm²

$\omega_i$ = Initial angular speed = 1.8 rev/s

$\omega_f$ = Final angular speed

As the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s$

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J$

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

6 0

3 years ago

Did you know that your answer automatically gets reported if you have "idk" anywhere in the answer?

Maurinko [17]

Answer:

rlly??

Explanation:

4 0

3 years ago

Read 2 more answers