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2 years ago

Decreasing a telescope's eyepiece focal length will _____.

Physics

2 answers:

krek1111 [17]2 years ago

8 0

Answer:

increase magnification

Explanation:

Novay_Z [31]2 years ago

7 0

I believe the correct answer from the choices listed above is the first option. Decreasing a telescope's eyepiece focal length will increase magnification. The magnification of the telescope image is (focal length of the objective) divided by (focal length of the eyepiece). Hope this answers the question.

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Astronauts are trained for take-off in a high-speed centrifuge of 4.7 m radius that spins in the horizontal plane.

Leokris [45]

Answer:

a) $1.94 \frac{rad}{s}$

b) $9.12\frac{m}{s}$

c) Towards the center of the centrifuge

Explanation:

Becuse the centrifuge rotates in circular motion, there's an angular acceleration tha simulates high gravity accelerations

$a_{rad}=\omega r^{2}$

with r the radius and ω the amgular velocity, in or case $a_rad=3.5g$ so:

$3.5g=\omega r^{2}$ and g=9.8 $\frac{m}{s^{2}}$

solving for ω:

$\omega=\frac{3.5g}{r^2}=\frac{3.5*9.8}{4.2^2}$

$\omega = 1.94 \frac{rad}{s}$

b) Linear speed (v) and angular speed are related by:

$v=\omega r =(1.94)(4.7)$

$v= 9.12\frac{m}{s}$

c) The apparent weigth is pointing towards the center of the circle, becuse angular acceleration is pointing in that direction.

8 0

2 years ago

How tightly does mass need to be compacted in order to become a black hole??? (2 words)

olga55 [171]

is it infinite density?

8 0

3 years ago

Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of

max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

$Q =\frac{KA(\delta T)}{L}$

$Q =\frac{25.87*1*20}{1}$

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0

2 years ago

Read 2 more answers

Assume that the radius ????r of a sphere is expanding at a rate of 70 cm/min.70 cm/min. The volume of a sphere is ????=43???????

rodikova [14]

Answer:

the rate of change in volume with time is 280πr² cm³/min

Explanation:

Data provided in the question:

Radius of the sphere as 'r'

$\frac{d\textup{r}}{\textup{dt}}$ = 70 cm/min

Volume of the sphere, V = $\frac{\textup{4}}{\textup{3}}\pi r^3$

Surface area of the sphere as 4πr²

Now,

Rate of change in volume with time, $\frac{d\textup{V}}{\textup{dt}}$

= $\frac{d(\frac{\textup{4}}{\textup{3}}\pi r^3)}{dt}$

= $3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times\frac{dr}{dt}$

Substituting the value of $\frac{dr}{dt}$

= $3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times70$

= 280πr² cm³/min

Hence, the rate of change in volume with time is 280πr² cm³/min

4 0

2 years ago

Net force of 200 newtons is applied to a wagon for 3 seconds

AveGali [126]

Momentum of the wagon increases by (200 x 3)

= 600 newton-sec

= 600 kg-m/sec

6 0

3 years ago