Pls help me quick im doing a quiz

A spring with spring constant 13.1 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.

Alenkinab [10]

Answer:

Explanation:

spring constant, K = 13.1 N/m

22 oscillations in 20 seconds

time taken to complete one oscillation is called time period.

T = 20 / 22 second = 0.909 seconds

(a) let m be the mass.

The formula for the time period is

$T = 2\pi \sqrt{\frac{m}{K}}$

$m = \frac{T^{2}K}{4\pi ^{2}}$

$m = \frac{0.909^{2}\times 13.1}{4\pi ^{2}}$

m = 0.275 kg

(b) maximum speed, v = ω A = 2π A / T

v = ( 2 x 31.4 x 0.1) / 0.909

v = 0.691 m/s

6 0

3 years ago

How is solar impulse is different than conventional airplane

Nadusha1986 [10]

Answer:

Solar Impulse is a Swiss long-range experimental solar-powered aircraft project and Conventional Aircraft are the atmosphere-only aircraft that have been around since the Wright Flyer's first takeoff. Though modern craft are much more advanced and capable than that early model, Conventional Aircraft have been superseded in modern times by the AeroSpace Fighter, which can operate in both atmosphere and space.

Explanation:

6 0

3 years ago

What would be your estimate of the age of the universe if you measured a value for Hubble's constant of H0 = 30 km/s/Mly ? You c

Maksim231197 [3]

Answer:

The age of the universe would be 9.9 billion years

Explanation:

We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.

Thus, the time it has taken for the galaxies to reach their current separations is:

$\displaystyle{t=D/v}$

and from Hubble's Law:

$v =H_0D$

Therefore:

$\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}$

With the given value for the Hubble's constant we have:

$H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s$

and thus,

$t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years$

6 0

3 years ago

Listed following are the names and mirror diameters for six of the world’s greatest reflecting telescopes used to gather visible

ziro4ka [17]

Answer:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope

Explanation:

How much light a telescope can collect depends on its diameter, since in a bigger area more photons will be collected.

Remember that in a circle the area is defined as:

$A = \pi r^{2}$ (1)

Where A is the area and r is its radius.

However, the radius can be determined by means of its diameter.

$d = 2r$

$r = \frac{d}{2}$ (1)

Where d is its diameter.

An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

$A = \pi (\frac{d}{2})^{2}$ (3)