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Vitek1552 [10]
3 years ago
11

Can you pls give a question about science. like why does the sky blue? PLS HELPPP ASAP ​

Physics
1 answer:
Aleonysh [2.5K]3 years ago
6 0
I can why do plants grow fast?
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If a train is going 60 m/s hits the brakes, and it takes the train 1 minute 25 seconds to stop, what is the train’s acceleration
Cloud [144]
Hey there, the answer is ..............................  About 0.7 m/sec^2
<span>
Acceleration is the change in speed / time </span>

<span>Change in speed is 60 m/sec </span>

<span>Time is 1 minute 25 second. Convert that to seconds. </span>

<span>Divide the change in speed by the time in seconds.

About 0.7 m/sec^2

 </span><span>So the acceleration is - 60 / 85 = - 0.71 m/s^2 

HOPE I HELPED!!!!!!!!!!
</span>
8 0
3 years ago
A straight fin is made from copper (k = 388 W/m-K) and is 0.5 cm in diameter and 30 cm long. The temperature at the base of the
erastovalidia [21]

Answer:

The rate of transfer of heat is 0.119 W

Solution:

As per the question:

Diameter of the fin, D = 0.5 cm = 0.005 m

Length of the fin, l =30 cm = 0.3 m

Base temperature, T_{b} = 75^{\circ}C

Air temperature, T_{infty} = 20^{\circ}

k = 388 W/mK

h = 20\ W/m^{2}K

Now,

Perimeter of the fin, p = \pi D = 0.005\pi \ m

Cross-sectional area of the fin, A = \frac{\pi}{4}D^{2}

A = \frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}

To calculate the heat transfer rate:

Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})

where

m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237

Now,

Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W

5 0
3 years ago
Draw velocity-ime graph for uniform moion of an object , when initially body is at rest.
Klio2033 [76]
When the body is at rest, its speed is zero, and the graph lies on the x-axis.

When the body is in uniform motion, the speed is constant, and the graph is a horizontal line, parallel to the x-axis and some distance above it.

It's impossible to tell, based on the given information, how these two parts of the
graph are connected.  There must be some sloping (accelerated) portion of the graph
that joins the two sections, but it cannot be accounted for in either the statement
that the body is at rest or that it is in uniform motion, since acceleration ... that is,
any change of speed or direction ... is not 'uniform' motion'.
8 0
3 years ago
8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h
pantera1 [17]

(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

<em>d</em> = - <em>v</em>² / (2<em>a</em>)

Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

0² - (13.9 m/s)² = 2 <em>a</em> (15 m) → <em>a</em> ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) <em>d</em> → <em>d</em> ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that

60 J = <em>m g h</em>

where <em>g</em> = 10 m/s² and <em>h</em> is the height it is lifted, 1.2 m. Solving for <em>m</em> gives

<em>m</em> = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

4 0
3 years ago
Determine the angular velocity ω of the telescope as it orbits around the Sun.
lara31 [8.8K]
The JWST is postioned about 1.5 million kilometers from the earth on the side facing away from the sun
5 0
2 years ago
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