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PtichkaEL [24]
2 years ago
15

PLEASE HELP ME (stop putting links ) Two objects m1 and m2, each with a mass of 5 kg and 6 kg separated by a distance. A third o

bject m 3= 2 kg is placed between the two. The distance between m1, and m3, is( X) and the distance between m3, from m2, is (2X+1). If the gravitational force experienced by m3, Fg= 0. What is the distance of m3, from m1, and the distance of m1, from m2.​
Physics
1 answer:
BabaBlast [244]2 years ago
7 0

Answer:

Explanation:

Newton's Gravitation Law

\displaystyle \frac{GmM}{d^2}

where G is a constant, M and M the masses e d the distance betwen masses.

\displaystyle G\frac{5\cdot2}{x^2}=G\frac{6\cdot 2}{(2x+1)^2} \quad \sqrt{6}x=(2x+1)\sqrt {5} \quad x=\frac{\sqrt{5}}{\sqrt{6}-2\sqrt{5}}

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A tall cylinder contains 30 cm of water. oil is carefully poured into the cylinder, where it floats on top of the water, until t
Art [367]

The total gauge pressure at the bottom of the cylinder would simply be the sum of the pressure exerted by water and pressure exerted by the oil.

The formula for calculating pressure in a column is:

P = ρ g h

Where,

P = gauge pressure

ρ = density of the liquid

g = gravitational acceleration

h = height of liquid

Adding the two pressures will give the total:

P total = (ρ g h)_water + (ρ g h)_oil

P total = (1000 kg / m^3) (9.8 m / s^2) (0.30 m) + (900 kg / m^3) (9.8 m / s^2) (0.4 - 0.30 m)

P total = 2940 Pa + 882 Pa

P total = 3,822 Pa

 

Answer:

 The total gauge pressure at the bottom is 3,822 Pa.

6 0
2 years ago
Read 2 more answers
Two acrobats flying through the air grab and hold onto each other in midair as part of a circus act.One acrobat has a mass of 60
earnstyle [38]

Answer:

1.36m/s

Explanation:

We are given that

Mass of one acrobat,m_1=60 kg

Mass of another acrobat,m_2=50 kg

v_2=-3 m/s

v_1=5 m/s

We have to find their  velocity immediately after they grab  onto each other.

The collision between two acrobat is inelastic

According to law of conservation of momentum

m_1v_1+m_2v_2=(m_1+m_2)V

Substitute the values

60\times 5+50(-3)=(60+50)V

300-150=110V

V=\frac{300-150}{110}=1.36m/s

4 0
2 years ago
If ice didn't float, but sank instead, this change wouldn't affect aquatic life. true of false
ruslelena [56]

Answer:

No it will not (false)

Explanation:

I say this because if anything the ice will sink and melt from a solid to a liquid and that will lead to i guess more water for the aquatic life. (THIS IS A THEORY SHAWTY)

6 0
3 years ago
Read 2 more answers
a 2000 kg elevator with broken cables in a test rig is falling at 4.00 m/s when it contacts a cushioning spring at the bottom of
Ratling [72]

A. The speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring is 3.65m/s

B. The acceleration when the elevator is 1.00 {\rm m} below point where it first contacts a spring is 4m/s²

In calculating the speed of the elevator and acceleration, first we have to find the force of gravity F on the elevator, which is the force pulling the elevator in downward direction. Using the equation for force of gravity which is:

F = mg

Where:

Mass of the elevator; m= 2000kg

Acceleration due to gravity; g = 9.8m/s

2000kg × 9.8m/s²= 19600N

F = 19600

Force of opposing friction clamp of gravity = 17000N

Net force on the elevator = force of gravity - Force of opposing friction clamp

Net force on the elevator = 19600 - 17000

Net force on the elevator = 2600 N

We will also find the kinetic energy K.E; of the elevator at the point of contact with the spring using:

K.E = 1/2 mv²

Where

Mass of the elevator; m = 2000kg

Velocity of the elevator = 4.00m/s

K.E = (1/2)*2000kg*(4m/s)²

K.E = 16000J

The kinetic energy and energy gained will be absorbed by the spring across the next 2m

Therefore,

Energy; E = K.E + P.E

Where:

Kinetic energy K.E = 16000J

Potential Energy P.E = ?

P.E of spring = net force absorbed × distance at compression

Where:

Net force absorbed = 2600N

Distance at compression = 2.0m

P.E = 2600*2

P.E = 5200J

E = 16000J + 5200J

E = 21200J

Spring constant = k

To find k

Using:

E = (1/2)*k*(x)²

Where:

E = 21200J

k = ?

x = 2m

21200J = (1/2)*k*(2m)²

21200J*2 = (4m)k

K = 42400J/4m

K = 10600N/m

Therefore,

Acceleration at 1m compression = ?

Using:

F = K*X

Where

F is force provided by the spring = 10600N/m,

K = 10600 N/m

X = 1m

F = 10600N/m * 1m

F = 10600N (upward)

A. The speed of the elevator after it has moved downward 1.00 {\rm m} from the point where it first contacts a spring?

Using:

Original Kinetic energy + net force on the elevator = final kinetic energy + spring energy

16000N + 2600N = (1/2)mv² + (1/2)k x²

18600 = (1/2)(2000)(v²) + (1/2)(10600N)(1²)

18600 = 1000(v²) + 5300

18600 - 5300 = 1000(v²)

13300 = 1000(v²)

V² = 13.300

V =3.65m/s

B. The acceleration of the elevator is 1.00m below point where it first contacts a spring

Spring constant = net force on the elevator + resultant force

Where:

Spring constant = 10600N

Net force on the elevator = 2600N

Resultant force = ?

10600N = 2600N + resultant force

Resultant force = 10600N - 2600N

Resultant force = 8000N

Using the equation for Newton's 2nd law where F = ma,

a = F/m

Where:

Resultant force; F =8000N

Mass of the elevator; m =2000kg)

a = 8000 / 2000

a = 4m/s²

Here's the complete question:

In a "worst-case" design scenario, a 2000kg elevator with broken cables is falling at 4.00m/s when it first contacts a cushioning spring at thebottom of the shaft. The spring is supposed to stop the elevator,compressing 2.00m as it does so. During the motion a safety clampapplies a constant 17000N frictional force to the elevator.

1. What is the speed of the elevator after it has moved downward 1.00m from the point where it first contacts aspring?

2. When the elevator is 1.00m below point where it first contacts a spring, what is its acceleration?

Learn more about calculating speed of an elevator from:

brainly.com/question/3850823?referrer=searchResults

#SPJ4

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10 months ago
How long does it take for light to get to earth from the moon?
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About 1.3 SECONDS.............
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