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White raven [17]
3 years ago
14

If the magnets are removed from the skateboard, the height of the skateboard should [increase, decrease, stay the same].

Physics
2 answers:
galina1969 [7]3 years ago
8 0
Decrease Is what I think
vagabundo [1.1K]3 years ago
6 0

Maybe its decrease I'm sceptical though

Explanation:

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What information is necessary to describe motion scientifically?
tamaranim1 [39]
Answer is A
to determine an objects true motion, you need velocity, which requires direction, time and distance!
8 0
3 years ago
Read 2 more answers
A worker pushes horizontally on a 35.0 kg crate with a force of magnitude 112 N. The coefficient of static friction between the
OLga [1]

Answer: a) -127 N b) No. c) -112 N  d) 40 N e) 15 N

Explanation:

a) Friction force always oppose to the relative movement between two surfaces, and, provided that be less than the fs max, adopt any value to counteract the applied force.

The fs max, is the horizontal component of the contact force, and can be written as follows:

Fs max = us . Fn  

As the block is at rest in the vertical direction, this means that Fn must be numerically equal to the weight of the object:

Fn = m g = 35 kg. 9.8 m/s2 = 343 N → Fs max = 0.37. 343 N = 127 N

b) Now, as the applied force is smaller than Fs max, this means that the friction force, is equal and opposite to the applied forcé, i.e., -112 N, so the crate doesn´t move.

c) Please see above.

d) As explained above, the maximum friction force, is proportional to the normal force, which adopts any value needed to satisfy the Newton´s 2nd Law.

So, if we diminish the normal force, we can lower the máximum friction force, helping to the worker to move the crate.

The mínimum needed normal force, will be the one that satisfies the following:

Fs max = F applied = us Fn = 112 N = 0.37. Fn.

Solving for Fn, we get Fn= 303 N

So, the difference between the original normal force and the new one, will be the mínimum upward force needed to make the crate to move, as follows:

Fup = 343 N -303 N = 40 N  

e) If we keep the normal force unchanged, but add an horizontal force to help the worker, we will need that the sum of both forces, will be equal to the Fs max, as follows:

Fh + Fapp = 127 N → Fh = 127 N – 112 N = 15 N

5 0
3 years ago
How is work affected when an object is lifted straight up instead of using a ramp? Work will increase.
kifflom [539]
Without friction, the amount of work only depends on the final height,
and is not affected by the route used to get there. 

If the ramp has no friction, then it has no effect on the total amount
of work done.  The work to lift the load straight up is the same.

If the ramp has some friction, then it takes more work to use the ramp
than to lift the load straight up.  Then the work to lift the load straight up
would be less than when the ramp is used.


3 0
3 years ago
A freight car with a mass of 10 metric tons is rolling at 108 km/h along a level track when it collides with another freight car
Rufina [12.5K]

Answer:

20 metric tons

Explanation:

Using the Principle of Conservation of Linear Momentum which states that the total momentum of a system is constant in any direction in which no external force acts.

momentum before collision= momentum after collision

mathematically expressed as

m1u1 + m2u2 = m1v1+m2v2 =0

where:

m1= mass of rolling freight car = 10 metric tons = 10Mg

1 ton = 1000kg= 1Mg

u1 =initial velocity of freight car = 108km/h,

m2 = mass of stationary freight car = ?, this the unknown we are finding

u2 = initial velocity of

m2 = 0, because it is at rest,

v1= final velocity of m1,

v2= final velocity of m2

since they move together in the same direction they share a common final velocity as such

v1=v2= 36km/h

substituting valves in expression give

m1u1 + m2u2 = m1v1 + m2v2= 0

10Mgx 108km/hr + m2 x 0 = 10Mg x 36km/h + m2x36km/h

1080( Mg x km/hr) + 0 = 360 (Mg x Km/h) + 36m2 (km/h)

1080 (Mgx km/hr) - 360(Mg x km/h) = 36m2 (km/h)

720(Mg x km/h) = 36m2 (km/h)

divide both side by  36 (km/h) gives

m2 = 20Mg =20 metric tons

3 0
4 years ago
A spring is compressed by 15 cm when 400 N of force is applied. What is the spring constant of the spring?
Helga [31]

Answer:

2666.7 N/m

Explanation:

Spring constant = force / displacement

k = F / x

k = 400 N / 0.15 m

k = 2666.7 N/m

8 0
4 years ago
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