Choose the correct statement about the acceleration of the car.

List the largest-aperture single telescope currently in use in each of the following bands of the electromagnetic spectrum: radi

tankabanditka [31]

Explanation:

Radio Telescope:

1. FAST (Five Hundred Meter Aperture Spherical Telescope), China is the largest single dish radio telescope in the world with a dish of diameter 500 meters. Although its construction is over but its not yet functional.

2. Arecibo Observatory, Puerto Rico is the largest single dish operational radio telescope having a single dish of diameter 305 meters..

X - ray Telescope: Chandra X ray observatory also known as Advanced X-ray Astrophysics Facility (AXAF) is the largest x-ray telescope with aperture of 1.2 meters. It can detect X-ray sources which are 100 times fainter in comparison to any previous telescope.

Gamma ray telescope: Fermi Gamma ray telescope is the largest gamma ray telescope.

4 0

3 years ago

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. a

USPshnik [31]

Refer to the diagram shown below.

Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force

The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m

By definition,
T = I*α

Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²

Answer: 105.93 kg-m²