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3 years ago

Which of these elements is unlikely to have a reaction with any element or compound?

Physics

1 answer:

Romashka-Z-Leto [24]3 years ago

5 0

Argon (AR) it is a noble gas! ------ Good Luck! ------

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A 1400 kg car is moving at 33.8 m/s when a force is applied the opposite direction of the car's motion. The car slows down to 21

zubka84 [21]

Solve for acceleration:

a = (21.4 m/s - 33.8 m/s) / (4.7 s)

a ≈ -2.6 m/s²

Solve for force:

F = (1400 kg) a ≈ -3700 N

The minus sign tells you the force points in the opposite direction of the car's motion. Its magnitude is always positive, so F = 3700 N.

3 0

3 years ago

A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.

alex41 [277]

The magnitude of the unknown height of the projectile is determined as 16.1 m.

<h3>Magnitude of the height</h3>

The magnitude of the height of the projectile is calculated as follows;

H = u²sin²θ/2g

H = (36.6² x (sin 29)²)/(2 x 9.8)

H = 16.1 m

Thus, the magnitude of the unknown height of the projectile is determined as 16.1 m.

Learn more about height here: brainly.com/question/1739912

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3 0

2 years ago

What practical function is provided by the ionosphere? Produces auroras charges atoms heats oxygen and nitrogen permits worldwid

Karolina [17]

Answer:

Worldwide Radio Communication

Explanation:

The ionosphere is important because it is through the ionosphere that world wide radio communication is possible.

6 0

3 years ago

0.002 written in scientific notation

Oliga [24]

Answer:0,002 = 2 x 10⁻³

Explanation:

0,002 = 2 / 1000 = 2 / 10³ = 2 x 10⁻³

3 0

3 years ago

Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis

solong [7]

Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

$v^{2}=u^2+2as$

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

$v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s$

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

$v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds$

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equals $T=2\times 30.51\approx61seconds$

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals $39.66m/s$

4 0

3 years ago