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2 years ago

1. What is wave motion

1 answer:

8 0

A wave is basically propagation of disturbances—that is, deviations from a state of rest or equilibrium—from place to place in a regular and organized way. Most familiar are surface waves on water, but both sound and light travel as wavelike disturbances, and the motion of all subatomic particles exhibits wavelike properties.

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Write a collision scenario here. If you choose your own collision, you can have neither, one, or both of the objects break. Be s

OleMash [197]

Answer:

My scenario would be A Car vs. a guard rail on a road. You have a car that is coming down a Highway at a speed of 43 Mph Miles per hour (69.2018 Kmh)

And it hits a steel guardrail and the car smashes in at the front and the guardrail is only bent while the car has the bumper and the hood along with the headlights and windshield along with the passenger side window break.

Explanation:

This is caused by so much force reacting from one object to another but also depends on molecular density.

5 0

3 years ago

While John is traveling along a straight interstate highway, he notices that the mile marker reads 249 km. John travels until he

erastova [34]

Answer:

Explanation:

Displacement can be displayed as a vector, this because it has magnitud and direction. Because of this, we can think John's Resultant Displacement as the join of this two vectors.

The First Vector is from the 249 Km Marker to the 141 Km Marker, which give us a Vector with a Magnitude equals to 108 Km.

The Second Vector goes from 141 Km Marker to the 174 Km Marker, which give us a Vector with a Magnitude equals to 33 Km.

However is important to know the direction for each Vector, we notice that John was traveling on one direction and then he returned. This makes our Vector to have a different direction, and this means difference signs. Difference signs means substraction. So, the Third Vector will be:

Third Vector = 108 Km - 33 Km

Third Vector = 75 Km

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4 years ago

A shipping pallet holds 10 boxes. Each box holds 300 parts of differenttypes. The part weight is normally distributed with a mea

Nat2105 [25]

Answer:

Explanation:

300

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3 years ago

A 3.80-m-long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building

Virty [35]

Answer:

Explanation:

The question is ;

3.80 m-long, 500. kg steel beam extends horizontally from the point

where it has been bolted to the framework of a new building under

construction. A 67.0 kg construction worker stands at the far end of

the beam. What is the magnitude of the torque about the point where the beam is bolted into place?

Solution:

Here two torques are in action

a) torque T1 due to weight of worker at the edge of the beam - at center of the beam

b) torque T2 due to weight of the uniform beam - at the point where the beam is bolted

So we first calculate the torque produced due to weight of the worker;

We can see that;

Distance of worker from the center of the beam = 1.9m

Mass of the worker = 67 kg

Value of g= 9.8 m/sec2

T1=force × distance from point of rotation

Here force is weight of the worker which is = mass × g=67×9.8=656.6 N

So the torque is

T1= 656.6×1.9=12225.892 Nm

T1 = 12225.892 Nm

Now torque by the beam itself ;

Length of the beam from its center point to the bolt point = 1.9 m

Weight force of beam acting at center point= mass× g= 500×9.8=4900 N

Torque T2 at bolt point by the beam weight = weight of beam × length of beam from its center to the bolt point = 4900×1.9=9310 Nm

T2=9310Nm

So total torque = T1+T2= 12225.892+9310=21565.892 Nm

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3 years ago

A charge Q is transferred from an initially uncharged plastic ball to an identical ball 24 cm away.The force of attraction is th

xxMikexx [17]

Answer:

The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons

Explanation:

Given;

magnitude of the attractive force, F = 17 mN = 0.017 N

distance between the two objects, r = 24 cm = 0.24 m

The attractive force is given by Coulomb's law;

$F = \frac{1}{4\pi \epsilon _0} \times \frac{Q^2}{r^2} = \frac{kQ^2}{r^2} \\\\Q^2 = \frac{Fr^2}{k} \\\\Q = \sqrt{ \frac{Fr^2}{k}} \\\\Q = \sqrt{ \frac{(0.017)(0.24)^2}{9\times 10^9}} \\\\Q = 3.298 \times 10^{-7} \ C$

The charge of 1 electron = 1.602 x 10⁻¹⁹ C

n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷

$n = \frac{3.298 \times 10^{-7}}{1.602 \times 10^{-19}} = 2.06 \times 10^{12} \ electrons$

Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons

4 0

3 years ago