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3 years ago

What are some of the challenges that you might face if you wanted to observe and study stars from Earth?

2 answers:

7 0

<span>The earth moves, which is unstoppable and if the earth moves the telescope won't be able to see it clearly because the telescope needs to be able to move at the same pace as earth to keep up to the objects</span>

AlexFokin [52]3 years ago

4 0

Another main issue with astronomical sighting is light pollution. The lights from our cities and such drown out the light of the stars. You could also say clouds, humidity, fog, or anything of the kind.

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If a Cu 2+ ion that is initially at rest accelerates through a potential difference of 12 V without friction, how much kinetic e

IgorC [24]

Answer:

23,8eV

Explanation:

A=Uq=12 (V)* 2* 1,6*10^-19 (C)=3,84*10^-18 (J)= 23,8 (eV)

8 0

4 years ago

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

$x=ut$

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

$t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s$

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

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

Substitute 9.8 m/s² for g and 0.461 s for t.

$y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m$

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by, $(2.0 m)-(1.04 m)=0.96 m$

The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>

8 0

3 years ago

in california the pacific plate slides past the north american plate. If the pcific plate is moving at a speed of 5 centimeters

frozen [14]

Answer:

6.34×10⁹ seconds

Explanation:

Applying,

V = d/t............. Equation 1

Where V = speed plate, d = distance covered by the plate, t = time it takes the plate to travel

make t the subject of the equation

t = d/V............ Equation 2

From the question,

Given: d = 5 cm/yr = (5×3.154×10⁻⁹) = 1.577×10⁻⁸ m/s, d = 100 meter

Substitute these values into equation 2

t = 100/1.577×10⁻⁸

t = 6.34×10⁹ seconds

Hence the time it takes the the plate to travel is 6.34×10⁹ seconds

3 0

3 years ago

What is the acceleration of a ball rolling down a ramp that starts from rest and travels 0.9 m in 3 s?

cupoosta [38]

Given:
u = 0, initial velocity
s 0.9 m, distance traveled.
t = 3 s, the time taken.

Let a = the acceleration. Then
s = ut + (1/2)*a*t²
(0.9 m) = 0.5*(a m/s²)*(3 s)²
0.9 = 4.5a
a = 0.2 m/s²

Answer: 0.2 m/s²

3 0

3 years ago

Lexy says that the diffraction angle is the one labeled R, and Jin says that the diffraction angle is the one labeled S. Which b

marusya05 [52]

Answer: A

Explanation: Lexy is correct because angle R is between the incident wave, n = 0, and a diffracted wave.

9 0

3 years ago