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Lapatulllka [165]
3 years ago
9

Lighting a match changes to :

Physics
2 answers:
Anon25 [30]3 years ago
5 0
I belive it is potential
ale4655 [162]3 years ago
3 0
A, Mechanical and A, Chemical. 
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Convert 3500cm to meter​
qaws [65]
3500/ 100 = 35
the answer is 35m!
5 0
3 years ago
Read 2 more answers
Why can't we say a measurement tool is accurate?
sashaice [31]

There are various reasons why a measurement tool cannot be accurate. One of them is thermal contraction and expansion varies according to seasons.

<h3>What are Accuracy and Precision?</h3>

There are two ways to assess observational error: accuracy and precision. Precision measures how closely two measurements are to one another, whereas accuracy measures how close a group of measurements is to its actual value. In other words, precision is a measure of statistical variability and a description of random errors.

We can say that a tool can be precise, but it cannot be accurate. There are various reasons behind that, some of them are :

  • It may not be calibrated properly. If there are no reliable standards to use for calibration, this may occur.
  • Perhaps it strayed. This is why electronic scales include a tare function—they are terrible in this area.
  • Perhaps the measurements are not linear. Our calipers might have been quite precise at the 2-inch standard, where they were calibrated, but inaccurate at other dimensions.
  • Temperature is one environmental component that the instrument might be sensitive to. These effects might be compensated for, but the compensation might not be ideal. This issue affects both dissolved solids meters and picometers.

These are some of the reasons due to which measurement tool cannot be accurate.

To get more information about Accuracy and Precision :

brainly.com/question/15276983

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7 0
1 year ago
A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

7 0
3 years ago
1. बालू के एक कण की त्रिज्या 1.6 x 10-4मी है। इस कण की त्रिज्या
Over [174]

Answer:

BOIIII english????

Explanation:

5 0
3 years ago
Could an average star, such as our sun, become a neutron star?
algol [13]
No but the sun could be a white dwarf stellar remnant.
3 0
3 years ago
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