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3 years ago

In which direction does the sun appear to move across the sky?

Physics

2 answers:

sergiy2304 [10]3 years ago

3 0

Answer:

If we consider a system where the y-axis as the South-North line, and the x-axis as the West-East line (where North and East are the positive sides)

We know that the sun goes from East to West, so in our system, the sun goes from the positive side of the x-axis to the negative side of the x-axis.

Where we would see this if we were standing right in the equator line.

If we where in other point of the planet, the Sun will stil move from East to West, but it will have a little tilt along the path, so we will have a little displacement in the y-axis. This displacement will depend on where we are, if we are at the North of the equator, we will se that the sun seems to go a little towards South as it goes to the West side.

jasenka [17]3 years ago

3 0

Answer:

the sun appears from east to west were it sets

Explanation:

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While following the directions on a treasure map a pirate walks 37.0 m north and then turns and walks 8.5 m east what is the mag

kogti [31]

Answer: $38\ m$

Explanation:

For this exercise you can use the Pythagorean theorem to find the magnitude of the resultant displacement.

Then:

$d^2=\triangle x^2+\triangle y^2$

You can observe that the square of the displacement is equal to the sum of the square of the horizontal displacement and the square of the vertical displacement.

Since the pirate walks 37.0 meters north and then turns and walks 8.5 meters east:

$\triangle x=37.0\ m\\\triangle y=8.5\ m$

Substituting values and solving for "d", you get:

$d=\sqrt{(37.0\ m)^2+(8.5\ m)^2}\\\\d=38\ m$

8 0

3 years ago

What is the speed of a car that travels 150 km in 3.00 hrs?

zhuklara [117]

You would do distance divided by speed. So 150÷3, which would equal 5km per hour.

4 0

3 years ago

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a 5.0 kg tool box is raised from the ground by a rope. if the upward acceleration of the bucket is 2.5 m/s^2 , find the force ex

denpristay [2]

Answer:

62 N

Explanation:

Sum of the forces on the toolbox:

∑F = ma

T − mg = ma

T = mg + ma

T = m (g + a)

T = (5.0 kg) (9.8 m/s² + 2.5 m/s²)

T = 61.5 N

Rounded to two significant figures, the force exerted by the rope is 62 N.

3 0

4 years ago

Compare these two collisions of a PE student with a wall.

Stolb23 [73]

1) The variable that is different in the two cases is $\Delta t$ , the duration of the collision

2) The change in momentum is the same in the two cases

3) The impulse is the same in the two cases

4) Case B will experience a greater force

Explanation:

The variable that is different in the two cases is $\Delta t$ , the duration of the collision.

In fact, in the first case the wall is padded: this means that the collision will be "softer" and therefore will last longer, so the duration of the collision, $\Delta t$ , will be larger.

In the second case instead, the wall is unpadded: this means that the collision is "harder" and so it will last less time, therefore the duration of the collision $\Delta t$ will be smaller.

The change in momentum in the two cases is the same.

In fact, the change in momentum is given by:

$\Delta p = m(v-u)$

where:

m is the mass of the student

u is the initial velocity

v is the final velocity

In both cases, we have:

m = 75 kg

u = 8 m/s

v = 0 (they both comes to rest)

Therefore, the change in momentum is

$\Delta p = (75)(0-8)=-600 kg m/s$

The impulse in the two cases is the same.

In fact, impulse is defined as the product of force applied, F, and duration of the collision, $\Delta t$ :

$J=F \Delta t$

However, the force can be rewritten as product of mass (m) and acceleration (a), according to Newton's second law:

$F=ma$

So the impulse is

$J=ma\Delta t$

The acceleration can be rewritten as rate of change of velocity:

$a=\frac{\Delta v}{\Delta t}$

So the impulse becomes

$J=m\frac{\Delta v}{\Delta t}\Delta t = m\Delta v$

So, the impulse is equal to the change in momentum: and since in the two cases the change in momentum is the same, the impulse is the same as well.

The force in the collision is related to the impulse by

$J=F\Delta t$

where

J is the impulse

F is the force

$\Delta t$ is the duration of the collision

The equation can be rewritten as

$F=\frac{J}{\Delta t}$

In the two situations described in the problem (A and B), we already said that the impulse is the same (because the change in momentum is the same). However, in case A (padded wall) the time $\Delta t$ is longer, while in case B (unpadded wall) the time $\Delta t$ is shorter: since the force F is inversely proportional to the duration of the collision, this means that in case B the student will experience a greatest force compared to case A.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

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4 years ago

Kelli drew a diagram to compare cast and imprint fossils. Which label belongs in the area marked Y? Involves minerals replacing

STatiana [176]

Answer:

the answer is c

Explanation:

an imprint gives evidence of an organisms activity they don't normally fill with sediments

8 0

3 years ago

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