The figure below shows a shooting competition, where air rifles fire soft metal at distant targets

The glass in a window is 35 inches wide and 20 inches tall, and standard atmospheric pressure is 14.7 pounds per square inch. Wh

iogann1982 [59]

Answer:103 pounds

Explanation:

Given

width of window $b=35 in.$

height of window $h=20 in.$

standard atmospheric pressure $P_{outside}=14.7 psi$

Also $P_{inside}=1.01P_{outside}$

Thus Net Force on the window will be the algebraic sum of Force due to outside and inside Pressure .

$F_{net}=(P_{inside}-P_{outside})\cdot A$

$F_{net}=P_{outside}(1.01-1)\times 35\times 20$

$F_{net}=14.7\times 0.01\times 35\times 20$

$F_{net}=102.9\ pounds\approx 103\ pounds$

8 0

2 years ago

A. A child is twirling a 1.52 kg object in a vertical circle with a radius of 67.6

steposvetlana [31]

Answer:

(a) 4.21 m/s

(b) 24.9 N

Explanation:

(a) Draw a free body diagram of the object when it is at the bottom of the circle. There are two forces on the object: tension force T pulling up and weight force mg pulling down.

Sum the forces in the radial (+y) direction:

∑F = ma

T − mg = m v² / r

v = √(r (T − mg) / m)

v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)

v = 4.21 m/s

(b) Draw a free body diagram of the object when it is at the top of the circle. There are two forces on the object: tension force T pulling down and weight force mg pulling down.

Sum the forces in the radial (-y) direction:

∑F = ma

T + mg = m v² / r

T = m v² / r − mg

T = (1.52 kg) (4.21 m/s)² / (0.676 m) − (1.52 kg) (9.8 m/s²)

T = 24.9 N

6 0

3 years ago

A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 562 Hz tone, what is

tester [92]

To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,

Frequency $= f = 562Hz$