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skelet666 [1.2K]

2 years ago

11

Waves are observed passing under a dock. Wave crests are 8.0 meters apart. The time for a complete wave to pass by is 4.0 second

s. The markings on the post submerged in water indicate that the water level fluctuates from a trough at 6.0 meters to a crest at 9.0 meters. What is the amplitude of the wave?

1 answer:

Anit [1.1K]2 years ago

3 0

The amplitude of a wave can be obtained by measuring the distance from the resting position of the wave to its crest. The resting position is half of the distance from the crest to the trough. Given that the distance between the crest and the trough is 3 meters, the amplitude should be half of that, which is 1.5 meters.

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What type of circuit is illustrated?

Reil [10]

Answer:

I beleive its B

Explanation:

If not then A but I'm positive its B

6 0

2 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

2 years ago

The sensor in the torso of a crash test dummy records the magnitude and direction of the net force acting on the dummy.If the du

Sunny_sXe [5.5K]

Here in crash test the two forces are acting on the dummy in two different directions

As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.

So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors

so we can say

$F_{net} = \sqrt{F_1^2 + F_2^2}$

here given that

$F_1 = 130.0 N$

$F_2 = 4500.0 N$

now we will plug in all data in the above equation

$F_{net} = \sqrt{4500^2 + 130^}$

$F_{net} = 4501.9 N$

so it will have net force 4501.9 N which will be reported by sensor

4 0

2 years ago

Read 2 more answers

A book that weighs 19 Newtons sits on a table. With what force

iVinArrow [24]

Answer:

We know there's two forces acting on a book while it sits on a table:the force of gravity pulling it down, and the normal force of the table acting upward on the book. The book isn't accelerating while it sits there. That's because the weight of the book is being counteracted by the normal force of the table.

Explanation:

There are two forces acting upon the book. One force - the Earth's gravitational pull - exerts a downward force. The other force - the push of the table on the book (sometimes referred to as a normal force) - pushes upward on the book.

5 0

2 years ago

A transformer has two sets of coils, the primary with N1 = 160 turns and the secondary with N2 = 1400 turns. The input rms volta

vovikov84 [41]

To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.

So things our data are given by

$N_1 = 160$

$N_2 = 1400$

$\Delta V_{1rms} = 62V$

PART A) Since it is a system in equilibrium the relationship between the two transformers would be given by

$\frac{N_1}{N_2} = \frac{\Delta V_{1rms}}{\Delta V_{2rms}}$

So the voltage for transformer 2 would be given by,

$\Delta V_{2rms} = \frac{N_2}{N_1} \Delta V_{1rms}$

PART B) To express the number value we proceed to replace with the previously given values, that is to say

$\Delta V_{2rms} = \frac{N_1}{N_2} \Delta V_{1rms}$

$\Delta V_{2rms} = \frac{1400}{160} 62V$

$\Delta V_{2rms} = 1446.66V$

7 0

2 years ago