# If a barometer reads 772 mm hg, what is the atmospheric pressure expressed in pounds per square inch?

Answer: I think it would be 15.23 not so sure but
hope this helps! (:

## Related Questions

As a science project, you drop a watermelon off the top of the Empire State Building. 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a constant speed of 30 m/s. A) How much time passes before the watermelon has the same velocity? B) How fast is the watermelon going when it passes Superman?C) How fast is the watermelon traveling when it hits the ground?

3.06 seconds time passes before the watermelon has the same velocity

watermelon going at speed 59.9 m/s

watermelon traveling when it hits the ground at speed is 79.19 m/s

Explanation:

given data

height = 320 m

speed = 30 m/s

to find out

How much time passes before the watermelon has the same velocity and How fast is the watermelon going and How fast is the watermelon traveling

solution

we will use here equation of motion that is

v = u + at    ....................1

here v is velocity 30 m/s and u is initial speed i.e zero and a is acceleration i.e 9.8 m/s²

put the value and find time t

30 = 0 + 9.8 (t)

t = 3.06 s

so 3.06 seconds time passes before the watermelon has the same velocity

and

we know superman cover distance is = velocity × time

so distance = 30 × t

and distance formula for watermelon is

distance = ut + 0.5×a×t²    .............2

here u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and h is 30 × t

30 × t = 0 + 0.5×9.8×t²

t = 6.12 s

so  by equation 1

v = u + at

v = 0 + 9.8 ( 6.12)

v = 59.9 m/s

so watermelon going at speed 59.9 m/s

and

watermelon traveling speed formula is by equation of motion

v² - u² = 2as      ......................3

here v is speed and u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and s is distance i.e 320 m

v² - 0 = 2(9.8) 320

v = 79.19 m/s

so watermelon traveling when it hits the ground at speed is 79.19 m/s

25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression in terms of the variables d, vcmax and vg for the time, tc, it takes the cheetah to catch the gazelle.

maximum speed of cheetah is

speed of gazelle is given as

Now the relative speed of Cheetah with respect to Gazelle

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

so by rearranging the terms we can say

so above is the relation between all given variable

Which of the following are electromagnetic waves?a. Water wavesb. Radio wavesc. Sound wavesd. Seismic waves

Explanation:

Hope this helped!

Which describes one feature of the image formed by a convex mirror?????

The image formed by a convex mirror will always have its smaller than the size of the object no matter what the position of the object.

Explanation:

The image formed by a convex mirror will always have its smaller than the size of the object no matter what the position of the object.

Also notice that convex mirror always makes virtual images.

Another feature of the convex mirror is that an upright image is always formed by the convex mirror.

An important mirror formula to remember which is applicable for both convex and mirrors

• 1/f= 1/u + 1/v

Here:

'u' is an object which gets placed in front of a spherical mirror of focal

length 'f' and image 'u' is formed by the mirror.

right side up

Explanation:

When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to make this force twice as strong, the distance would have to be changed toA) √2d
B) d/√2
C) d/4
D) 2d
E) d/2

Explanation:

Given

Force of attraction is F when charges are d distance apart.

Electrostatic force is given by

where k=constant

and are charges

d=distance between them

In order to double the force i.e. 2F

divide 1 and 2 we get

A cars mass is 950kg and it travels at a speed of 35 m/s when it rounds a flat curve of radius 215 m.a. Determine the value of the frictional force exerted on the car.

b. Determine the value of the coefficient of friction between the tires and the road.

(a) It's the force of (static) friction that keeps the car on the road and prevents it from skidding, and this friction is directed toward the center of the curve.

Recall that centripetal acceleration has a magnitude a of

a = v ² / R

where

v = tangential speed

R = radius of the curve

so that

a = (35 m/s)² / (215 m) ≈ 5.69767 m/s² ≈ 5.7 m/s²

Parallel to the road, the only force acting on the car is friction. So by Newton's second law, we have

F = Fs = ma

where

Fs = magnitude of static friction

m = mass of the car

Then

Fs = (950 kg) (5.7 m/s²) ≈ 5412.79 N ≈ 5400 N

(b) Perpendicular to the road, the car is in equilbrium, so its weight and the normal force of the road on the car are equal in magnitude. By Newton's second law,

N - W = 0

where

N = magnitude of normal force

W = weight

so that

N = W = m g = (950 kg) (9.8 m/s²) = 9310 N

Friction is proportional to the normal force by a factor of µ, the coefficient of static friction:

Fs = µN

Assuming 35 m/s is the maximum speed the car can travel without skidding, we find

µ = Fs / N = (5400 N) / (9310 N) ≈ 0.581395 ≈ 0.58