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katovenus [111]
4 years ago
10

Give two examples where friction is a nuisance

Physics
1 answer:
dlinn [17]4 years ago
8 0

Answer:

it can cause wear and tear and reduces efficiency as energy is lost.

Explanation:

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A turtle crawls at 4.32 m/s to cover the short 3.84 m distance to his food bowl. How long does it take?
Lyrx [107]
Speed=Distance/Time. so you need to do distance/speed to get time. 3.84/4.32= .88888888889 seconds to get to his food bowl

6 0
3 years ago
How did the volcano Mt. St. Helens form?
charle [14.2K]

Given what we know about volcanic formations, we can conclude that the Mt. St. Helens volcano was formed by the process described in option D, "Subduction  an oceanic continental convergent boundary caused a mountain  range to form and magma to come to the surface".

Mt. St. Helens is a volcano that falls under the category of a stratovolcano. Stratovolcanoes are volcanoes that commonly form at subduction zones. This process forms chains or clusters of mountains along the boundaries of tectonic plates, where oceanic crust is pulled under another type of crust. This process tends to form mountain ranges which contain volcanoes, such as:

  • Cascade Range (To which Mt. St. Helens belongs)
  • Los Andes
  • Rocky Mountains
  • Campania Mountain range

and so on.

To learn more visit:

brainly.com/question/3080956?referrer=searchResults

6 0
3 years ago
What would be the distance moved if we had a 70 n force and work done is 8j
Lostsunrise [7]

Answer:

0.1143m

Explanation:

W=f×s

8=70s

make s the subject of the formula

s=8/70

=0.1143m

3 0
3 years ago
A 2.44 kg block is pushed 1.55 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle o
svlad2 [7]

Answer:

The work done is L= 49.83 Joules

Explanation:

m= 2.44kg

d= 1.55m

α= 69.6°

g= 9.8 m/s²

μ= 0.691

F= ?

Fy= F*sin(α)

Fx= F*cos(α)

Fr= μ * Fx

Fr= μ * F*cos(α)

W= m*g

W= 23.91 N

Fy - W - Fr = 0

F*sin(α) - W -  μ * F*cos(α) = 0

F* ( sin(α) - μ *cos(α) ) = W

F= W /  ( sin(α) - μ *cos(α) )

F= 34.3 N

L= Fy * d

L= 49.83 J

6 0
4 years ago
when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is remo
Lera25 [3.4K]

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

                              F = k\frac{|q_1|.|q_2|}{r^2}

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             \frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1 ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          \frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2}  = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 =   \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         -\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123

                         

                          q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\

 

6 0
4 years ago
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