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3 years ago

The blank is the sun's apparent annual path through the

Physics

1 answer:

zaharov [31]3 years ago

6 0

<span>The ecliptic is the sun's apparent annual path through the stars.</span>

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A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.

alisha [4.7K]

A)

It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

$S=S_{o}+v_{o}t+ \frac{at^2}{2} \\ 20= \frac{10t^2}{2} \\ t=2s$

Through Definition of Velocity, comes:

$\Delta v= \frac{\Delta S}{\Delta t} \\ v_x= \frac{36}{2} \\ \boxed {v_{x}=18m/s}$

B)

Using the Velocity Hourly Equation in vertical direction, we have:

$v_{y}=v_{y_{o}}+gt \\ v_{y}=10\times2 \\ \boxed {v_{y}=20m/s}$

The angle of impact is given by:

$cos(\theta) =\frac{v_{x}}{v_{y}} \\ cos(\theta) = \frac{18}{20} \\ cos(\theta) =0.9 \\ arccos(0.9)=\theta \\ \boxed {\theta \approx 25.84}$

If you notice any mistake in my english, please let me know, because i am not native.

7 0

3 years ago

Read 2 more answers

A bumblebee darts past at 3 m/s. The frequency of the hum made by its wings is 152 Hz. Assume the speed of sound to be 342 m/s.

Veronika [31]

It'll be 152 Hz at the exact instant the bumblebee
is right at the tip of your nose, on his way past you.

Before he gets there, while he's coming at you,
he sounds like a frequency higher than 152 Hz.

After he passes by, and is going away from you,
he sounds like a frequency lower than 152 Hz.

8 0

3 years ago

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The doppler effect is used to determine?

Korolek [52]

Uhh it is used to detirmine heat

7 0

3 years ago

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How much work is done if 10 N is applied to a 5kg object for 10 meters if there is an opposing force of 5 N

BlackZzzverrR [31]

Answer:

50 J

Explanation:

The net force acting on the box is given by the algebraic sum of the two forces, so:

$F=10 N -5 N = 5 N$

The net work done on the box is equal to (assuming the net force is parallel to the displacement of the object)

$W=Fd$

where

F = 5 N is the net force on the object

d = 10 m is the displacement of the object

Substituting,

$W=(5 N)(10 m)=50 J$

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3 years ago

What happens to electric charges within a conductor when a charged object is brought close to it

Agata [3.3K]

Well, if a charger conductor is touched to another object or close enough to touching the object then the conductor can transfer its charge to that object. Conductors allow for electrons to be transported from particle to particle, so a charged object will always distribute its charge until the repulsive forces are minimized.

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3 years ago