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3 years ago

1 point

Physics

1 answer:

stiks02 [169]3 years ago

3 0

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 800 × 35

We have the final answer as

Hope this helps you

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if the volume of a scuba tank filled with air remains constant and its temperature goes down, what happens to its pressure?

Vladimir [108]

Answer:

Decreases

Explanation:

Ideal gas law:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is gas constant,

and T is absolute temperature.

If V is constant and T decreases, then P must decrease.

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3 years ago

What type of Psychologist was B.F. Skinner?

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Answer:

Behaviorist

Explanation:

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2 years ago

Hallar la energía mecánica de un pendulo , en un punto de su oscilación donde la energía cinética es 0,39 J y en ese mismo punto

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Answer:

E = 0,39 + 0,59 = 0,98 J

Explanation:

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3 years ago

An airplane touches down on the runway with a speed of 70 m/s2. Determine the airplane after each second of its deceleration.

ivann1987 [24]

vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
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0 = 55*55 + 2*(-2.5)*d
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3 years ago

A rectangular tank is filled to a depth of 10m with freshwater and open to air at atmospheric pressure.

charle [14.2K]

Answer:

1. 39068.07 N

2. 19534.036 N

Explanation:

depth of water h = 10 m

atmospheric pressure Patm = 101325 Pa

density of water p = 1000 kg/m^3

acceleration due to gravity g = 9.81 m/s^2

pressure due to depth of water = pgh

P = 1000 x 9.81 x 10 = 98100 Pa

total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa

Left plug has diameter = 50 cm = 0.5 m

radius = 0.5/2 = 0.25 m

height = 1 cm = 0.01 m

height below tank surface = 10 - 0.01 = 9.99

pressure at this depth = 1000 x 9.81 x 9.99 = 98001.9 Pa

total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa

surface area of plug = π $r^{2}$ = 3.142 x $0.25^{2}$ = 0.196 $m^{2}$

force required to lift left plug = pressure x area

F = 199326.9 x 0.196 = 39068.07 N

The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug

A = 0.196/2 = 0.098 $m^{2}$

force F required to lift right plug = 199326.9 x 0.098 = 19534.036 N

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3 years ago