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2 years ago

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A student stays at her initial position for a bit of time, then walks slowly in a straight line for a while, then stops to rest

awhile and finally runs quickly back to her initial position along a straight line.
Required:
Draw displacement versus time plots best represents the student’s trip?

1 answer:

True [87]2 years ago

6 0

Answer:

The first interval is walked slowly, this is a straight line with a small slope

Second interval stops, which gives a horizontal line, indicating the same position

Third interval, walk back, straight downhill

Explanation:

In this problem we have a uniform movement, this means that the acceleration in each intervals

x = v t

The first interval is walked slowly, this is a straight line with a small slope

Second interval stops, which gives a horizontal line, indicating the same position

Third interval, walk back, straight downhill

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A girl delivering newspapers covers her route by traveling 3.00 blocks west, 4.00 blocks north, and then 6.00 blocks east. What

Harrizon [31]

The concepts used to solve this problem are those related to the Pythagorean theorem for which we will calculate the distance and the pitch.

According to the attached diagram we have that the expression of the resulting displacement is

$R = \sqrt{(3b)^2+(4b)^2}$

Therefore the resultant displacement of the girl is

$R = \sqrt{(9b^2+16b^2)}$

$R = \sqrt{25b^2}$

$R = 5b$

Therefore the girl has displaced around of 5 blocks

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2 years ago

Are light waves longitudinal or transverse

Marysya12 [62]

Answer:

Transverse

Explanation:

There are two types of waves, according to the direction of their oscillation:

- Transverse waves: in a transverse wave, the direction of the oscillation is perpendicular to the direction of motion of the wave. Examples of transverse waves are electromagnetic waves

- Longitudinal waves: in a longitudinal wave, the direction of the oscillation is parallel to the direction of motion of the wave. Examples of longitudinal waves are sound waves.

Light waves corresponds to the visible part of the electromagnetic spectrum, which includes all the different types of electromagnetic waves (which consist of oscillations of electric and magnetic fields that are perpendicular to the direction of propagation of the wave): therefore, they are transverse waves.

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2 years ago

Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

Gennadij [26K]

Answer:

a. The station is rotating at $1.496 \frac{rev}{min}$

b. the rotation needed is $2.8502 \frac{rev}{min}$

Explanation:

We know that the centripetal acceleration is

$a_{c}= \omega ^2 r$

where $\omega$ is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

<h3>a. </h3>

The rotational speed is :

$2.7 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.02454 \frac{rad^2}{s^2} }$

$\omega = 0.1567 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.1567 \frac{rad}{s} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 1.496 \frac{rev}{min}$

<h3>b. </h3>

The rotational speed needed is :

$9.8 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.08909 \frac{rad^2}{s^2} }$

$\omega = 0.2985 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.2985 \frac{rev}{min} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 2.8502 \frac{rev}{min}$

3 0

2 years ago

Read 2 more answers

stephen buys a new moped . he travels 4km south and then 6km east. how far does he need to go to get back where he started??

stiks02 [169]

Answer:

My answer is 7.2 km

Explanation:

When Stephen goes to the south and then to the east, he is drawing a right triangle, where the 4 km and 6 km sides are the cathetus of a right triangle.

Then we use the Pithagorean theorem to solve this problem. We need to find the hypotenuse.

c² = a² + b²

c² = 4² + 6²

c² = 16 + 36

c² = 52

c = 7.2 km

7 0

3 years ago

Pls answer quick I need to get the answer rn

lora16 [44]

I think it is False because as the Gad relajases fuel it doesn’t move as much anymore

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2 years ago