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xenn [34]
3 years ago
8

A certain comet has a parallax 1/40 as large as the moon. How far away is it (compared to the Moons distance)? Show Work

Physics
1 answer:
Ad libitum [116K]3 years ago
3 0
The distance and parallax are inversely related. We can find the distance using the following equation:

d= \frac{1}{p}

where d is distance and p is parallax.

We are given the parallax of the comet relative to the moon, and we are looking for the distance to the comet relative to the moon's distance, so wee can plug in the following value:

d= \frac{1}{ \frac{1}{40} }

The distance is 40 times as far away as the moon.
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An input force is applied to a 2 cm diameter piston. The piston compresses oil and is connected to a 4 cm diameter output piston
____ [38]

Answer:

4 times

Explanation:

Given:

diameter of the input piston, d_i=2\ cm=0.02\ m

diameter of the output piston, d_o=4\ cm=0.04\ m

In such mechanical arrangements the fluid acts as an incompressible link and the pressure remains uniform throughout the fluid bulk.

Then, by the Pascal's law:

\frac{F_i}{A_i} =\frac{F_o}{A_o}

where:

F_i\ \&\ F_o are the input and the output forces respectively.

A_i\ \&\ A_o are the input and the output area of pistons respectively.

\frac{F_i\times 4}{\pi.d_i^2} =\frac{F_o\times 4}{\pi.d_o^2}

\frac{F_i}{d_i^2} =\frac{F_o}{d_o^2}

\frac{F_o}{0.04^2}=\frac{F_i}{0.02^2}

F_o=4\times F_i

8 0
3 years ago
A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help.
uranmaximum [27]

Answer:

532.0725 m

102.17270893 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = g

H = Height of cliff

Distance traveled in 3 seconds

s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 3^2\\\Rightarrow s=44.145\ m

Distance traveled by sound = 2H-44.145 m

2H-44.145=ut+\dfrac{1}{2}at^2\\\Rightarrow 2H-44.145=340\times 3\\\Rightarrow H=\dfrac{340\times 3+44.145}{2}\\\Rightarrow H=532.0725\ m

The height of the cliff is 532.0725 m

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 532.0725+0^2}\\\Rightarrow v=102.17270893\ m/s

Her speed just before she hits the ground is 102.17270893 m/s

5 0
3 years ago
What is science explain​
vladimir2022 [97]

Answer:

Science is the pursuit and application of knowledge and understanding of the natural and social world following a systematic methodology based on evidence. Scientific methodology includes the following: ... Evidence. Experiment and/or observation as benchmarks for testing hypotheses.

3 0
3 years ago
Read 2 more answers
A car collides into a concrete wall going 25.0 m/s . It stops in 0.141 seconds and has a change in momentum of 39,400. What is t
vodka [1.7K]

Answer:

Mass of the car is 1576 kg.

Explanation:

Let the mass of the car be m kg.

Given:

Initial velocity of the car is, u=25\ m/s

As the car stops, final velocity of the car is, v=0\ m/s

Change in momentum is, \Delta p=39400

Now, we know that, momentum is given as the product of mass and velocity.

So, change in momentum is given as:

\Delta p=m(u-v)\\39400=m(25-0)\\39400=25m\\m=\frac{39400}{25}\\m=1576\ kg

Therefore, the mass of the car is 1576 kg.

4 0
3 years ago
A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
bagirrra123 [75]

Answer:

a) 520m

b) 10.30 s

c) 100,95 m/s

Explanation:

a) According the given information, the rocket suddenly stops when it reach the height of 520m, because the engines fail, and then it begins the free fall.

This means the maximum height this rocket reached before falling  was 520 m.

b) As we are dealing with constant acceleration (due gravity) g=9.8 \frac{m}{s^{2}} we can use the following formula:

y=y_{o}+V_{o} t-\frac{gt^{2}}{2}   (1)

Where:

y_{o}=520 m  is the initial height of the rocket (at the exact moment in which it stops due engines fail)

y=0  is the final height of the rocket (when it finally hits the launch pad)

V_{o}=0 is the initial velocity of the rocket (at the exact moment in which it stops the velocity is zero and then it begins to fall)

g=9.8m/s^{2}  is the acceleration due gravity

t is the time it takes to the rocket to hit the launch pad

Clearing t:

0=520 m+0-\frac{9.8m/s^{2} t^{2}}{2}   (2)

t^{2}=\frac{-520 m}{-4.9 m/s^{2}}   (3)

t=\sqrt{106.12 s^{2}   (4)

t=10.30 s   (5)  This is the time

c) Now we need to find the final velocity V_{f} for this rocket, and the following equation will be perfect to find it:

V_{f}=V_{o}-gt  (6)

V_{f}=0-(9.8 m/s^{2})(10.30 s)  (7)

V_{f}=-100.95 m/s  (8) This is the final velocity of the rocket. Note the negative sign indicates its direction is downwards (to the launch pad)

7 0
3 years ago
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