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mojhsa [17]
3 years ago
10

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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A shopping cart given an initial velocity of 2.0 m/s undergoes a constant acceleration to a velocity of 13 m/s. What is the magn
olga55 [171]

Answer:

The acceleration is a = 2.75 [m/s^2]

Explanation:

In order to solve this problem we must use kinematics equations.

v_{f} = v_{i} + a*t\\

where:

Vf = final velocity = 13 [m/s]

Vi = initial velocity = 2 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Now replacing:

13 = 2 + (4*a)

(13 - 2) = 4*a

a = 2.75 [m/s^2]

5 0
3 years ago
1 2 3 4 5 6 7 8 9 10
7nadin3 [17]

Answer:

<h2>C. <u>0.55 m/s towards the right</u></h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

<u>x is the final velocity of the 0.25kg ball</u>

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>

<u></u>

4 0
3 years ago
A rock is thrown straight down, not dropped, from the roof of a building that is 61 m above the ground. If it takes 3.1 s to rea
PtichkaEL [24]
Hiii !!!
I am sending the soluction !!!

If you have any question, let me now =)

Jorge:)

4 0
3 years ago
A block is attached to a horizontal spring. On top of this block rests another block. The two-block system slides back and forth
Minchanka [31]

Answer:

Angular frequency will increase

No change in the amplitude

Explanation:

At extreme end of the SHM the energy of the SHM is given by

E = \frac{1}{2} (m_1 + m_2)\omega^2 A^2

here we know that

\omega^2 = \frac{k}{m_1 + m_2}

now at the extreme end when one of the mass is removed from it

then in that case the angular frequency will change

\omega'^2 = \frac{k}{m_1}

So angular frequency will increase

but the position of extreme end will not change as it is given here that the top block is removed without disturbing the lower block

so here no change in the amplitude

6 0
3 years ago
One kind of slingshot consists of a pocket that holds a pebble and is whirled on a circle of radius r. The pebble is released fr
Debora [2.8K]

Answer:

θ=142.9°

Explanation:

d=1 *r

angle ϕ= 37.1°

the line connecting pebble and target should be tangent to a circle so

cos(180-ϕ-θ)=\frac{r}{d}=\frac{1}{1}

∴ θ=180-ϕ-cos^{-1} (\frac{1}{1} )

  θ= 180-37.1-0

  θ=142.9°

 

5 0
3 years ago
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