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3 years ago

Why is a football firm when it is inflated to its proper pressure

Physics

1 answer:

Molodets [167]3 years ago

8 0

Answer:

Proper Inflation and Feel When inflating your ball, you can use either a hand pump or an air pump equipped with a gauge that gives readings in pounds per square inch, also called psi. Footballs used in the NFL are inflated to 13 psi, but a proper range can fall between 12.5 and 13.5 psi, according to Wilson Sporting Goods.

Explanation:

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A paper-filled capacitor is charged to a potential difference of V0=2.5 V. The dielectric constant of paper is k=3.7 . The capac

Pavlova-9 [17]

Answer:

k = 6.72

Explanation:

K of paper = 3.7

k of air = 1

Given that charge Q on the capacitor is constant because cell is disconnected from the circuit. So

V = Q / C = 2.5

Capacity becomes C / 3.7 in air .

capacity becomes C/3.7 when paper is replaced by air .

V₁ = Q / (C/3.7)

= 3.7 Q/C

3.7 x 2.5

= 9.25 V

In the second case ,

capacitance due to new unknown dielectric k

= C/3.7 x k

= kC / 3.7 ( Capacitance in air is C/3.7 )

V ( new ) = Q / ( kC/3.7 )

= 3.7 Q/kC

.55 x 2.5 = 3.7 x( 2.5 / k )

k = 3.7 / .55

= 6.72

7 0

3 years ago

On March 21, a stick casts the following shadow. What is the most likely time of day?

jenyasd209 [6]

Answer:

9:00 AM

Explanation:

I took the test and that was the answer

5 0

4 years ago

Read 2 more answers

Kinetic energy of an object is equal to

Gnom [1K]

See
K.E=1/2(mass*velocity²)
so option B is the correct answer.
Brainliest pls :-)

8 0

3 years ago

PLEASE HELP!!!! I’ll give brainliest!

Jet001 [13]

Ok well I know measure of long leg is 30 degrees and short leg is 60 degrees

3 0

3 years ago

A typical running track is an oval with 74-mm-diameter half circles at each end. A runner going once around the track covers a d

lisabon 2012 [21]

The centripetal acceleration a is 4.32 $\times$ 10^-4 m/s^2.

<u>Explanation:</u>

The speed is constant and computing the speed from the distance and time for one full lap.

Given, distance = 400 mm = 0.4 m, Time = 100 s.

Computing the v = 0.4 m / 100 s

v = 4 $\times$ 10^-3 m/s.

radius of the circular end r = 37 mm = 0.037 m.

centripetal acceleration a = v^2 / r

= (4 $\times$ 10^-3)^2 / 0.037

a = 4.32 $\times$ 10^-4 m/s^2.

6 0

3 years ago