The reactivity of an atom arises primarily from __

The pressure of a sample of gas is measured as 49 torrent. Convert this to atmosphere

spin [16.1K]

Answer:

P = 0.0644 atm

Explanation:

Given that,

The pressure of a sample of gas is measured as 49 torr.

We need to convert this temperature to atmosphere.

The relation between torr and atmosphere is as follow :

1 atm = 760 torr

1 torr = (1/760) atm

49 torr = (49/760) atm

= 0.0644 atm

Hence, the presssure of the sample of gas is equal to 0.0644 atm.

5 0

2 years ago

The graph below shows the position of an ant as it crawls over a flat picnic blanket. The total time for the ant to go from the

Mice21 [21]

The average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

The correct answer is option D.

In the given graph, we can deduce the following;

the total time of the motion, = 1 mins + 45 s = 60 s + 45 s = 105 s

The average speed of the ant is calculated as;

$average \ speed = \frac{total \ distance }{total \ time }$

The total distance from the graph is calculated as follows;

first horizontal distance from 2 cm to 8 cm = 8 - 2 = 6 cm

first upward distance from 3 cm to 5 cm = 5 - 3 = 2 cm

second horizontal distance from 8 cm to 6 cm = 8 - 6 = 2 cm

second upward distance from 5 cm to 12 cm = 12 - 5 = 7 cm

third horizontal distance from 6 cm to 13 cm = 13 - 6 = 7 cm

fourth downward distance from 12 cm to 9 cm = 3 cm

final horizontal distance from 13 cm to 15 cm = 2cm

The total distance = (6 + 2 + 2 + 7 + 7 + 3 + 2) cm = 29 cm

$average \ speed = \frac{total \ distance }{total \ time } = \frac{29 \ cm}{105 \ s} = 0.276 \ cm/s$

The average velocity is calculated as the change in displacement per change in time.

The displacement is the shortest distance between the start and end positions.

This shortest distance is the straight line connecting the start and end position. Call this line P

From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm

Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm

Notice, you have a right triangle, now calculate the length of line P.

↓end

↓

↓ 6cm

↓

start -------------13 cm------------

Use Pythagoras theorem to solve for P.

$P^2 = 6^2 + 13^2\\\\P^2 = 36 + 169\\\\P^2 = 205\\\\P= \sqrt{205} \\\\P = 14.318 \ cm$

The average velocity of the ant is calculated as;

$average \ velocity= \frac{\Delta displacemnt }{total\ time }= \frac{14.318 \ cm}{105 \ s} = 0.136 \ cm/s \\\\$

Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

Learn more here: brainly.com/question/589950

5 0

2 years ago

A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal

Setler [38]

Answer

given,

mass of the ball = 3 kg

swing in vertical circle with radius = 2 m

work done by the gravity = ?

work done by the tension = ?

Work done by the gravity = - m g Δh

Δ h = 2 + 2 = 4 m

Work done by the gravity = $- 3 \times 9.8 \times 4$

= -117.6 J

work done by gravity is equal to -117.6 J

Work done by tension will be equal to zero.

Zero because tension is always perpendicular to velocity

work done by tension is equal to 0 J

7 0

3 years ago

Pls help im begging you

Lostsunrise [7]

Answer:

I think it's TRUE because forces change an object's motion but dont quote me on it ok? Cause I'm not a 100 percent sure

7 0

2 years ago

The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if i

Sergio [31]

Answer with Explanation:

We are given that

Diameter of fighter plane=2.3 m

Radius= $r=\frac{d}{2}=\frac{2.3}{2}=1.15 m$

a.We have to find the angular velocity in radians per second if it spins=1200 rev/min

Frequency= $\frac{1200}{60}=20 Hz$

1 minute=60 seconds

Angular velocity= $\omega=2\pi f$

Angular velocity= $2\times \frac{22}{7}\times 20=125.7 rad/s$

b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.

$v=r\omega=1.15\times 125.7=144.56 m/s$

c.Centripetal acceleration= $\omega^2 r=(125.7)^2(1.15)=18170.56 m/s^2$

Centripetal acceleration== $\frac{18170.56\times g}{9.81}=1852.25 g m/s^2$

7 0

3 years ago

The reactivity of an atom arises primarily from ___.