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aleksley [76]
3 years ago
7

Select the statement that is not true about the Hubble Telescope:

Physics
1 answer:
xxMikexx [17]3 years ago
4 0

Answer:

The option is B is not true for Hubble telescope.

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Help!!! I need it today <br> Thank you in advance
svlad2 [7]

Answer:

 F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.

Explanation:

The student wants to prove hooke's law which has the form

          F = - k (x-xo)

To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.

Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,

we must be careful when hanging the weights so as not to create oscillations in the spring

we look for the mass of each weight

         W = mg

          m = W / g

and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.

The fact of obtaining a line already proves Hooke's law.

5 0
3 years ago
How was the gravitational constant G first determined
Oliga [24]
<span>In equation form, this is often expressed as follows: The constant of proportionality in this equation is G - the universal gravitation constant. The value of G was not experimentally determined until nearly a century later (1798) by Lord Henry Cavendish using a torsion balance.</span>
6 0
3 years ago
An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m lon
telo118 [61]

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N  = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

L = \frac{\lambda}{2} \\\\\lambda = 2L

λ  = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

6 0
3 years ago
A graph of x vs. t2 is linear, and intercepts the vertical axis at 12 m and the horizontal axis at 4 s2. What is the function?
gogolik [260]

Answer:

x = -3t² + 12

Explanation:

x vs t² is a line.

x = at² + b

The y intercept is 12.

x = at² + 12

At t² = 4s², x = 0.

0 = a(4) + 12

a = -3

Therefore, the function is:

x = -3t² + 12

4 0
4 years ago
A truck is traveling due north at 75 km/hr approaching a crossroad. On a perpendicular road a police car is traveling west towar
Gre4nikov [31]

Answer:

109.66km

Explanation:

Velocity is defined as the change in displacement of a body with respect to time.

Velocity = displacement/Time

Displacement = velocity * time

A truck is traveling due north at 75 km/hr approaching a crossroad and perpendicular road a police car is traveling west toward the intersection at 80 km/hr, then the resultant velocity is gotten using the pythagoras theorem since they are both perpendicular to each other.

v² = 75²+80²

v² = 5625+6400

v² = 12,025

v = √12,025

v = 109.66km/hr

If both vechicles reaches cross road in exactly one hour, the displacement of the truck with respect to the police car =  109.66km/hr * 1 hour

Displacement of the truck with respect to the police car = 109.66km

7 0
3 years ago
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