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2 years ago

What is the pressure 100m below the surface of the sea , if the density of sea water is 1150kg/m3

Physics

1 answer:

Olin [163]2 years ago

4 0

The pressure at 100 meters below the surface of sea water with a density of 1150kg is 145.96 psi.

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My name is scooter hey fat boy you waa fight

zavuch27 [327]

Answer:

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YES

Explanation:

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2 years ago

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Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between m₁ and m₂ has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between m₁ and m₃ has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where x is measured in m.

The mass m₁ is attracted to m₂ in one direction, and attracted to m₃ in the opposite direction such that m₁ in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for x :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

5 0

2 years ago

A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca

igomit [66]

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

$f=mg\ sin\theta$

$f=1100\ kg\times 9.8\ m/s^2\ sin(4)$

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

3 0

2 years ago

1. What are 3 main categories of waste and what is the largest contribution of municipal solid waste?

Tasya [4]

Organic materials continue to be the largest component of MSW. Paper and paperboard account for 27 percent and yard trimmings and food account for another 28 percent. Plastics comprise about 13 percent; metals make up 9 percent; and rubber, leather, and textiles account for 9 percent.

5 0

2 years ago

A car of mass 2200 kg collides with a truck of mass 4500 kg, and just after the collision the car and truck slide along, stuck t

katovenus [111]

Answer:

37.7m/s: principle of conservation of momentum

Explanation:

The principle to make use of is the principle of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of momentum of bodies after collision. This bodies will move with the same velocity after collision.

Momentum = Mass × velocity

For car of mass 2200kg moving with velocity 33m/s:

Momentum of car before collision = 2200×33

= 72,600kgm/s

For the truck of mass 4500kg;

Momentum = 4500 ×(22-(-18)

= 4500×40

= 180000kgm/s

After collision, their momentum is:

Momentum after collision = (2200+4500)v

= 6700v

Using the principle above to get the common velocity v we have

72600+180000 = 6700v

252600 = 6700v

v = 252600/6700

v = 37.7m/s

7 0

3 years ago