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rjkz [21]
3 years ago
6

Why is the primary mirror in a telescope curved?

Physics
2 answers:
Crank3 years ago
7 0
The reflector telescope uses a mirror to gather and focus light (which means to reflect lights). All celestial objects are so far way that all of the light rays from them reach the earth as parallel rays

Simple version: it reflects the light so we can see farther away
 <span />
user100 [1]3 years ago
6 0
This collects light, and bends it into focus.
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O prevent tank rupture during deep-space travel, an engineering team is studying the effect of temperature on gases confined to
Ira Lisetskai [31]

To calculate for the pressure of the system, we need an equation that would  relate the number of moles (n), pressure (P), and temperature (T) with volume (V). There are a number of equations that would relate these values however most are very complex equations. For simplification, we assume the gas is an ideal gas. So, we use PV = nRT.<span>

PV = nRT  where R is the universal gas constant
P = nRT / V</span>

<span>P = 3.40 mol ( 0.08205 L-atm / mol-K ) (251 + 273.15 K) / 1.75 L </span>

<span>P = 83.56 atm</span>

<span>
</span>

<span>Therefore, the pressure of the gas at the given conditions of volume and temperature would be 83.56.</span>

7 0
3 years ago
When a rigid body rotates about a fixed axis, all the points in the body have the same A. centripetal acceleration B. tangential
Leto [7]

Answer:

The angular acceleration is same at all the points in the body.

Option (D) is correct.

Explanation:

Given:

When a rigid body rotates about a fixed axis, all the points in the body have the same,

For finding which quantity is same we use pure rotational concept,

 v = \omega r

Where \omega = angular frequency, r = radius of rigid body

When a rigid body rotates about a fixed axis angular velocity of all the points in the body are same.

But the tangential speed, tangential acceleration, linear displacement, and centripetal acceleration depend on the position of the points and hence they are not the same.

Therefore, the angular acceleration is same at all the points in the body.

8 0
3 years ago
The centre of mass of a metre rule is at the 50cm mark. state what is meant by Centre of mass​
Tanya [424]

Answer:

Centre of mass of any body is a point where all mass of a body is supposed to be concentrated

it lies in geometrical centre....

3 0
3 years ago
The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g
Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:  

k=\sqrt{\frac{I}{M} }

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration  k=\sqrt{\frac{I}{M} }

So, moment of rotational inertia (I) of a cylinder about it axis = \frac{MR^2}{2}

k=\sqrt{\frac{\frac{MR^2}{2}}{M} }

k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }

k=\sqrt{{\frac{R^2}{2}}

k={\frac{R}{\sqrt{2}}

k={\frac{1.20m}{\sqrt{2}}

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = \frac{2}{3}MR^2

k = \sqrt{\frac{\frac{2}{3}MR^2}{M}  }

k = \sqrt{\frac{2}{3} R^2}

k = \sqrt{\frac{2}{3} }*R

k = \sqrt{\frac{2}{3}}  *1.20

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of  radius

(I) = \frac{2}{5}MR^2

k = \sqrt{\frac{\frac{2}{5}MR^2}{M}  }

k = \sqrt{\frac{2}{5} R^2}

k = \sqrt{\frac{2}{5} }*R

k = \sqrt{\frac{2}{5}}  *1.20

k = 0.7560

k ≅ 0.76 m

6 0
4 years ago
F.r.e.e points :]]] ​
Colt1911 [192]
Ok? I don’t know what you want me to do though
4 0
3 years ago
Read 2 more answers
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