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11111nata11111 [884]
1 year ago
14

If a scientist gets unexpected results during the first trial of an experiment,

Physics
2 answers:
vladimir2022 [97]1 year ago
8 0

Answer:

the correct answer is A to repeat the expirement to be sure the results are valid

My name is Ann [436]1 year ago
5 0

Answer:

A. Repeat the experiment to be sure the results are valid.

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what times are the moon phases visible? My science teacher said that the phases of the moon are only visible at sertant times of
klio [65]
Your teacher is right. The moon can be seen early in the morning sometimes and late at night. Different phases are only visible on certain days as one day might be full quarter, the next full moon, the next first quarter, etc.
6 0
3 years ago
When we look at a star
stiv31 [10]

I'm not really sure what specific answer they're looking for, but if it's an open-ended question, then let's think about it this way...

A light year, is the distance it takes for light to travel in a year. If an object is 50,000 light years away, then by the time the light travels to us, 50,000 years has passed. We are looking at a 50,000 year old image of that object. (ignoring gravity and spatial expansion fun stuffs)

4 0
3 years ago
The slope of the the line is the _______ of the object
Nat2105 [25]

Answer:

rate of change

Explanation:

8 0
3 years ago
The horizontal bar rises at a constant rate of three hundred mm/s causing peg P to ride in the quarter circular slot. When coord
SVETLANKA909090 [29]

Answer:

Explanation:

Given

Horizontal bar rises with 300 mm/s

Let us take the horizontal component of P be

P_x=rcos\theta

P_y=rsin\theta

where \thetais angle made by horizontal bar with x axis

Velocity at y=150 mm

150=300sin\theta

thus \theta =30^{\circ}

position ofP_x=rcos\theta =300\cdot cos30=300\times \frac{\sqrt{3}}{2}

P_x=259.80 mm

P=259.80\hat{i}+150\hat{j}

Velocity at this instant

u_x=-rsin\theta =300\times sin30=-150 mm/s

u_y=rcos\theta =300\times cos30=259.80 mm/s

4 0
3 years ago
julia throws a ball vertically upward from the ground with a speed of 5.89m/s. Andrew catches it when it is on its way down at a
aliya0001 [1]
Vo = 5.89 m/s Y = 1.27 m g = 9.81 m/s^2 
Time to height 
Tr = Vo / g Tr = (5.89 m/s) / (9.81 m/s^2) Tr = 0.60 s 
Max height achieved is:
H = Vo^2 / [2g] H = (5.89 )^2 / [ 2 * (9.81) ] H = (34.69) / [19.62] H = 1.77 m 
It falls that distance, minus Andrew's catch distance:
h = H - Y h = (1.77 m) - (1.27 m) h = 0.5 m 
Time to descend is therefore:
Tf = √ { [2h] / g ] Tf = √ { [ 2 * (0.5 m) ] / (9.81 m/s^2) } Tf = √ { [ 1.0 m ] / (9.81 m/s^2) } Tf = √ { 0.102 s^2 } Tf = 0.32 s 
Total time is rise plus fall therefore:
Tt = Tr + Tf Tt = (0.60 s) + (0.32 s) Tt = 0.92 s           (ANSWER)
8 0
3 years ago
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