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Misha Larkins [42]

3 years ago

9

Which of the following equations illustrates the law of conservation of matter?

1 answer:

Anvisha [2.4K]3 years ago

8 0

Answer:

A

Explanation:

the balancing is correct in the first one

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Which of the following IS NOT a part of an electromagnet?

Agata [3.3K]

I'd go with electricity source. Good luck!!

3 0

3 years ago

A circular-motion addict of mass 80 kg rides a Ferris wheel around in a vertical circle of radius 10 m at a constant speed of 6.

bekas [8.4K]

Answer with Explanation:

We are given that

Mass=80 kg

Radius,r=10 m

Speed,v=6.1 m/s

a.Time period,T= $\frac{2\pi r}{v}=\frac{2\pi\times 10}{6.1}=10.3 s$

b. $F_N=mg-\frac{mv^2}{r}$

Substitute the values

$F_N=80\times 9.8-\frac{80\times(6.1)^2}{10}=486 N$

c. $F_N=mg+\frac{mv^2}{r}$

Substitute the values

$F_N=80\times 9.8+\frac{80\times (6.1)^2}{10}=1081.7 N$

7 0

3 years ago

To reduce inductive reactance, what devices are normally placed on transmission or distribution lines?

UkoKoshka [18]

Answer : Capacitors

Explanation : Capacitors are normally placed on transmission or distribution lines when to reduce inductive reactance.

This is because it enhances electromechanical and voltage stability , limit voltage dips at network nodes and reduces the power loss.

So, we can say that inductive reactance normally replace by the capacitors.

4 0

3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

3 years ago

2. A bus drove 8 meters East, then turned to drive 8 meters North, then 2 meters

Sphinxa [80]

Answer:

Long question good luck:)

Explanation:

3 0

3 years ago