Ground-based radio telescopes can collect data from distant objects in space

Physics

1 answer:

Butoxors [25]3 years ago

6 0

Answer:

A. during the day or night and in any weather conditions.

Explanation:

Ground-based radio telescopes can be used to collect data from distant objects in space during the day or night in any weather condition.

They do not depend or are they affected by weather and they pass well through them.

Telescopes are devices used to obtain information about distant bodies usually astronomical in nature.
Optical telescopes use the visible range of light and they are overwhelmed by the sun during the day.
Bad weather conditions can also diminish the reception of light.
They work best at night.
Radio telescopes uses electromagnetic radiations and can work at any time and during any weather.

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A 35.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 4.00 cm on a frictionles

slamgirl [31]

Answer:

(a) The total energy of the spring system is 0.032 J

(b) The speed of the object when its position is 1.20 cm is approximately 1.28996 m/s

Explanation:

The given parameters are;

The mass of the object connected to the spring, m = 35.0 g = 0.00

The force constant, k = 40.0 N/m

The amplitude of the oscillation, a = 4.00 cm = 0.04 m

Therefore, we have

(a) The total energy of the spring system, E given as follows;

E = PE + KE = 1/2·m·v² + 1/2·k·x²

Where;

v = The velocity of the spring

x = The extension of the spring

When the spring is completely extended, x = a, and v = 0, therefore;

The total energy of the spring system, E = 1/2 × k × a² = 1/2 × 40.0 N/m × (0.04 m)² = 0.032 J

(b) At x = 1.20 cm = 0.012 m, we have;

E = 1/2·m·v² + 1/2·k·x²

0.032 = 1/2 × 0.035 × v² + 1/2 × 40 × 0.012²

0.032 - 1/2 × 40 × 0.012² = 1/2 × 0.035 × v²

0.02912 = 1/2 × 0.035 × v²

1/2 × 0.035 × v² = 0.02912

v² = 0.02912/(1/2 × 0.035) = 1.664

v = √1.664 ≈ 1.28996

The speed of the object when its position is 1.20 cm, v ≈ 1.28996 m/s

E = PE + KE

0.032 = 1/2 × 40 × 0.025² + KE

KE = 0.032 - 1/2 × 40 × 0.025² = 0.0195

The kinetic energy when its position is 2.50 cm = 0.0195 J.

4 0

3 years ago

If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what wo

Elena L [17]

Answer: The question is incomplete or some details are missing. Here is the complete question ; (a) The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.55 m/s2 for 4.05 s, making straight skid marks 63.0 m long, all the way to the tree. With what speed (in m/s) does the car then strike the tree? m/s

(b) What If? If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2

a ) With what speed (in m/s) does the car then strike the tree? m/s = 4.3125m/s

b) then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2 = -5.696m/s2

Explanation:

The detailed steps and calculation is as shown in the attached file.