What type of foods do bodyguards eat give 5 examples

What are the Properties of Transverse Waves and Longitudinal wave

jeyben [28]

Answer:

Key terms

TermMeaningTransverse waveOscillations where particles are displaced perpendicular to the wave direction.Longitudinal waveOscillations where particles are displaced parallel to the wave direction

7 0

3 years ago

standing side by side, you and a friend step off a bridge and fall for 1.6s to the water below. your friend goes first, and you

Andrew [12]

(a) The distance will be more than 2.0 meters.

In fact, you starts your fall after your friend has already fallen 2.0 meters. This means that your friend has already accelerated for a while, therefore his velocity will be greater than yours. But this statement will be actually true for the entire fall, since you has some delay, therefore when your friend will hit the water, the separation between you and him will be greater than the initial separation of 2.0 meters.

b) First of all we need to calculate the height of the bridge with respect to the water. We know that you take 1.6 s to fall down, therefore we can use the following equation:

$S=\frac{1}{2}gt^2=\frac{1}{2}(9.81 m/s^2)(1.6s)^2=12.56 m$

We know that your friend will take 1.6 s to falls down. Instead, you start your jump after he has already fallen 2.0 m, therefore after a time given by the equation:

$S=\frac{1}{2}gt^2$

Using S=2.0 m,

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(2.0 m)}{9.81 m/s^2}}=0.64 s$

So we know that you start your fall 0.64 s after your friend. Therefore, now we can find how much did you fall between the moment you started your fall (0.64 s) and the moment your friend hits the water (1.6 s). Using

$t=1.6 s-0.64 s=0.96 s$

we find

$S=\frac{1}{2}gt^2=\frac{1}{2}(9.81 m/s^2)(0.96 s)^2 =4.52 m$

So, when your friend hits the water, you just covered 4.52 m, while he already covered 12.56 m. Therefore, the separation between you and your friend is more than 2 meters.

8 0

3 years ago

An amusement park ride raises people high into the air, suspends them for a moment, and then drops them at a rate of free-fall a

blsea [12.9K]

Answer: apparent weighlessness.

Explanation:

1) Balance of forces on a person falling:

i) To answer this question we will deal with the assumption of non-drag force (abscence of air).

ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).

iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).

2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.

3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.

Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.

3 0

2 years ago

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A dog running at 10 m/s is 30m behind a rabbit moving at 5 m/s. when will the dog catch up with the rabbit assuming both their v

dedylja [7]

The will dog catch up with the rabbit in 6 minutes assuming both their velocities remain constant during the chase.

<h3>What time will the dog catch the rabbit?</h3>

The time that the dog will catch up with the rabbit is given as follows:

Let the distance covered by the rabbit be x.

Distance covered by dog = x + 30