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maw [93]
3 years ago
6

How much greater is the light-collecting area of a 6-meter telescope than a 3-meter telescope? How much greater is the light-col

lecting area of a 6- telescope than a 3- telescope?
A) two times
B) four times
C) six times
Physics
1 answer:
vovikov84 [41]3 years ago
7 0

Answer:

B) four times

Explanation:

Area of 6 meter telescope

A_1=\pi \frac{D^2}{4}\\\Rightarrow A_1=9\pi

Area of 3 meter telescope

A_2=\pi \frac{D^2}{4}\\\Rightarrow A_1=2.25\pi

Dividing the equations we get

\frac{A_1}{A_2}=\frac{9\pi}{2.25\pi}\\\Rightarrow A_1=4A_2

Hence, the 6-meter telescope has 4 times the light-collecting area of the 3-meter telescope.

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A clarinetist, setting out for a performance, grabs his 3.070 kg clarinet case (including the clarinet) from the top of the pian
Cerrena [4.2K]

Answer:

the vertical acceleration of the case is 1.46 m/s

Explanation:

Given;

mass of the clarinet case, m = 3.07 kg

upward force applied by the man, F = 25.60 N

Apply Newton's second law of motion;

the upward force on the clarinet case = its weight acting downwards + downward force due to its downward accelaration

F = mg + m(-a)

the acceleration is negative due to downward motion from the top of the piano.

F = mg - ma

ma = mg - F

a = \frac{mg - F}{m} \\\\a = \frac{(3.07 \times 9.8) \ - \ 25.6}{3.07} \\\\a = 1.46 \ m/s^2

Therefore, the vertical acceleration of the case is 1.46 m/s²

4 0
3 years ago
Determine the density of a rectangular piece of concrete that measures 3.7 cm by 2.1 cm by 5.8 cm and has a mass of 43.8 grams.
Novay_Z [31]
It is customary to work in SI units.

Calculate the volume of the concrete.
V = 3.7*2.1*5.8 cm³ = 45.066 cm³ = 45.066 x 10 ⁻⁶ m³

The mass is  43.8 g = 43.8 x 10⁻³ kg

The density is mass/volume.
Density = (43.8 x 10⁻³ kg)/(45.066 x 10⁻⁶ m³) = 971.9 kg/m³

Answer: 971.9 kg/m³
5 0
3 years ago
Which statement correctly explains scientific theories?
Elan Coil [88]
<span>Scientific theories are tested and proven over time; they are then considered scientific laws.

Sometimes however, they are proven wrong, and so they do not become laws

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4 0
3 years ago
Read 2 more answers
What is the answer for number 10
Finger [1]

The vertical component is = vsinx m/s

If you know the angle, substitute the value of x.

If you know the velocity at which it is moving, substitute it for v

Hope it helps :)

3 0
3 years ago
PLEASE HELPPPPP
Ganezh [65]

The only graph that accurately depict the given motion is graph D.

The given parameters;

  • initial position of the man = 0
  • direction of the man's first displacement = backward
  • time of first motion, t₁ = 6 seconds
  • velocity of this first displacement = v₁
  • time without any motion (<em>zero movement</em>) = 6 seconds
  • direction of the second displacement = forward
  • velocity of second displacement = 2v₁

Let the acceleration of the first displacement = a

Acceleration of the second displacement = 2a

From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.

The only options with initial motion towards the negative direction are;

  • <em>graph B</em>, and
  • <em>graph D</em>.

The difference between graph B and D;

  • in graph B there is a uniform motion for 6 seconds
  • in graph D there is no motion for 6 seconds (<em>this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds</em>).

Thus, the only graph that accurately depict the given motion is graph D.

Learn more here: brainly.com/question/21095906

5 0
2 years ago
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