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Reika [66]
4 years ago
8

An object example to decrease pressure.

Physics
2 answers:
svp [43]4 years ago
4 0

Answer:

You can use the following things

Explanation:

Rubber ball - A rubber ball can decrease pressure if you throw it against the wall and catch it.

Rubber squeezy toy - If you squeeze this in your hands also you can decrease pressure.

Hope this helps....

Have a nice day!!!!

Olin [163]4 years ago
3 0

Answer:

To reduce pressure - decrease the force or increase the area the force acts on. If you were standing on a frozen lake and the ice started to crack you could lie down to increase the area in contact with the ice. The same force (your weight) would apply, spread over a larger area, so the pressure would reduce.

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Consider three force vectors F~ 1 with magnitude 43 N and direction 38◦ , F~ 2 with magnitude 26 N and direction −140◦ , and F~
Rama09 [41]

Answer:

34.70 N

Explanation:

Given :

F~ 1 = 43 N in direction 38◦

F~ 2 = 26 N in direction −140◦

F~ 3 = 27 N in direction 110◦

Therefore,

F~x = 43 cos (38) + 26 cos (-140) + 27 cos (110)

      = 43  (0.7) + 26  (-0.7) + 27  (-0.3)

      =  3.8

F~y = 43 sin (38) + 26 sin (-140) + 27 sin (110)

      = 43  (0.6) + 26  (-0.6) + 27  (0.9)

      = 34.5

so, F~ = $ \sqrt{3.8^2 + 34.5^2}$

          = 34.70 N

6 0
3 years ago
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
Vlada [557]

Answer:

the minimum time interval is 0.77 seconds

Explanation:

given data

coefficient of static friction = 0.5

coefficient of static friction shoes= 0.825

travel s = 2.40 m

to find out

what is the minimum time interval

solution

we know newton 2nd law

force = mass × acceleration

and

force acting on person due to friction

force = coefficient of static friction shoes × mg

so we can say

coefficient of static friction shoes × mg = ma

so

a = coefficient of static friction shoes × g

and we know g is 9.8 m/s²

so

distance formula by kinematic relation

distance = ut + 0.5 × at²

here put a value and u is zero because initial speed

2.40 = 0 + 0.5  × coefficient of static friction shoes × g× t²

2.40 = 0.5  × 0.825 × 9.8 × t²

t = 0.77 s

the minimum time interval is 0.77 seconds

5 0
4 years ago
How are inertia and momentum related?
Ipatiy [6.2K]

Explanation:

My sources says "Inertia is an intrinsic characteristic of the object related to its mass. Inertia tells you how much force it will take to cause a particular acceleration on the object. Momentum is a function of an object's mass and velocity. Momentum is a measure of the kinetic energy of the object."

Hopes this helps!

If you feel this answer is correct please mark my answer as the most brainliest, please and thank you!

4 0
3 years ago
Which of the following best describes reverberation?
Liono4ka [1.6K]

Answer:

There are no appropriate descriptions on the list of choices you've provided.

Explanation:

Reverberation, in psychoacoustics and acoustics, is a persistence of sound after the sound is produced.[1] A reverberation, or reverb, is created when a sound or signal is reflected causing a large number of reflections to build up and then decay as the sound is absorbed by the surfaces of objects in the space – which could include furniture, people, and air.[2] This is most noticeable when the sound source stops but the reflections continue, decreasing in amplitude, until they reach zero amplitude.

4 0
3 years ago
Read 2 more answers
An iceskater is turning at a PERIOD of (1/3) second with his arms outstretched. a) What is his ANGULAR VELOCITY w? b) If he pull
Vikentia [17]

Answer:

Explanation:

a )

Time period T = 1/3 s

angular velocity = 2π / T

= 2 x 3.14 x 3

ω = 18.84 radian / s

b )

Applying conservation of angular momentum

I₁ ω₁ = I₂ ω₂

I₁ / I₂ = ω₂ / ω₁

2 = ω₂ / ω

ω₂ = 2 ω

c )

(KE)initial = 1/2 I₁ ω²

(KE)final =  1/2 I₂ ω₂²

= 1/2 (I₁ / 2)  (2ω)²

=  I₁ ω²

c )

Change in rotational kinetic energy

=  I₁ ω² -  1/2 I₁ ω²

=  +  1/2 I₁ ω²

d )

This energy comes from the work done by centripetal force which is increased to increase the speed of rotation.

4 0
3 years ago
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