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Vera_Pavlovna [14]
3 years ago
14

Why do slippery fluids such as oil reduce sliding friction

Physics
1 answer:
Gwar [14]3 years ago
6 0
   fluid friction<span> occurs when an object moves through a liquid or gas. the force needed to overcome </span>fluid friction <span>is usually less then that needed to overcome </span>sliding friction<span>. the </span>fluid<span> keeps the surface from making direct contact and thus </span>reducing friction<span>.</span>
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How much work in N•m is done when a 10.0-N force moves an object 2.5m?
posledela
10*2.5= 25n.m this is how i got my answer.
5 0
3 years ago
What times what equals 400
Radda [10]

If you're willing to consider fractions or decimals,
then there are an infinite number of answers. 
Like (2.5 x 160), and (15 x 26-2/3).

If you want to stick to only whole numbers,
then these 8 combinations do:

1, 400
2, 200
4, 100
5, 80
8, 50
10, 40
16, 25
20, 20

7 0
4 years ago
How big is the planet earth​
AnnZ [28]

After doing some research on Wiki, Google, and Easyscienceforkids, I have come to a conclusion that:

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Radius: 3958.8 mi

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6 0
3 years ago
An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid
mariarad [96]

Explanation:

It is given that, An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. We need to find the ratio of their masses.

As they are in equilibrium, the force of gravity due to each other is same. So,

\dfrac{Gm_xM}{r^2}=\dfrac{Gm_yM}{r^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{m_y}=(\dfrac{r_x^2}{r_y^2})\\\\\dfrac{m_x}{m_y}=(\dfrac{140^2}{481^2})\\\\\dfrac{m_x}{m_y}=0.0847

So, the ratio of masses X/Y is 0.0847

5 0
3 years ago
Oceanographers use submerged sonar systems, towed by a cable from a ship, to map the ocean floor. In addition to their downward
KATRIN_1 [288]

Answer:

Tension in the cable is T = 16653.32 N

Explanation:

Give data:

Cross section Area A = 1.3 m^2

Drag coefficient CD = 1.2

Velocity V = 4.3 m/s

Angle made by cable with horizontal  =30 degree

Density \rho \ of\  water= 1000 kg/m3

 Drag force FD is given as

F_{D} = \fracP{1}{2} \rho v^{2} C_{D} A

        = 0.5\times 1000\times 4.32\times  1.2\times 1.3

Drag force = 14422.2 N acting opposite to the motion

As cable made angle  of 30 degree with horizontal  thus horizontal component is take into action to calculate drag force

TCos30 = F_D

T = \frac{F_D}{cos30}

T =\frac{ 14422.2}{cos 30}

T = 16653.32 N

7 0
3 years ago
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