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Gelneren [198K]
3 years ago
8

A 10n force is applied to a 25kg mass to slide it across a frictional surface. What is the acceleration of the mass?

Physics
2 answers:
klasskru [66]3 years ago
6 0

Answer: a = 0.4m/s^2 - 9.8*c where c is the coefficient of kinetic friction of the surface

Explanation: We know that, by the second Newton's law, a = F/m

where a is the acceleration, F is the net force and m is the mass of the object.

Then, if the surface is frictionless, the total force applied in the object is 10N, and the mass of the object is 25kg, so the acceleration is:

a =10N/25kg = 0.4m/s^2.

But if the surface is frictional, there will be a force of friction applied in the mass (this depends on the coefficient of friction and the weight of the mass), this means that the acceleration will be reduced.

If = -(9.8*25)*c

where c is a number that is bigger than 0 and smaller than 1, is called the coefficient of kinetic friction.

So the total force is now:

F = (10 - 9.8*25*c)

Then, the acceleration in a frictional surface is equal to:

a = (10 - 9.8*25*c)/25 = 0.4m/s^2 - 9.8*c

kipiarov [429]3 years ago
6 0

Force (in Newtons) equals mass (in kilograms) times acceleration (in meters per second squared).


Plug in the given force and mass to the equation and solve for acceleration:

10=25a

0.4m/s2=a


Hope this helps!!

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A 0.245 kg ball is thrown straight up from 2.07 m above the ground. Its initial vertical speed is 8.00 m/s. A short time later,
iris [78.8K]

Answer:

The work done by gravity is 4.975 \: Joules

Explanation:

The data given in the question is :

Mass is 0.245 kg

Height from ground is 2.07 m

As we know , the work done is state function , it depends on initial and final position not on the path followed.

So, work done by gravity = change in potential energy

Work done = Initial potential energy - final potential energy

Insert values from question

Work done = mass \times gravity \times (change \: in \: height)

Work done = 0.245 kg \times 9.81 m/s^{2} \times 2.07 m

So, work done = 4.975 Joules

Hence the work done by gravity is 4.975 \: Joules

3 0
4 years ago
is dimensionally correct relation necessarily to be a correct physical relation? explain with example.​
Andreas93 [3]

Answer: hope it helps you...❤❤❤❤

Explanation: If your values have dimensions like time, length, temperature, etc, then if the dimensions are not the same then the values are not the same. So a “dimensionally wrong equation” is always false and cannot represent a correct physical relation.

No, not necessarily.

For instance, Newton’s 2nd law is  F=p˙ , or the sum of the applied forces on a body is equal to its time rate of change of its momentum. This is dimensionally correct, and a correct physical relation. It’s fine.

But take a look at this (incorrect) equation for the force of gravity:

F=−G(m+M)Mm√|r|3r  

It has all the nice properties you’d expect: It’s dimensionally correct (assuming the standard traditional value for  G ), it’s attractive, it’s symmetric in the masses, it’s inverse-square, etc. But it doesn’t correspond to a real, physical force.

It’s a counter-example to the claim that a dimensionally correct equation is necessarily a correct physical relation.

A simpler counter example is  1=2 . It is stating the equality of two dimensionless numbers. It is trivially dimensionally correct. But it is false.

4 0
4 years ago
What is the upper temperature range for stars?
padilas [110]
The upper temperature range for stars is : c. 40,000 K

Every object has their own upper and lower temperature
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6 0
3 years ago
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A 30kg bucket of water is lifted vertically 3.0m. How much work was done on the bucket during this movement
Alex

Answer:

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3 0
3 years ago
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A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F
Tanya [424]

Answer: Normal force, N = 141.64 Newton

Explanation:

All the forces acting on the system and described in free body diagram  are:

1) gravitational pull in downward direction  

2) Normal force in upward direction

3) External force of 40 N acting at an angle of 37° with the horizontal can be resolved in two rectangular components:

  i) F Cos 37° along the horizontal plane  in forward direction and

  ii) F Sin 37° along the vertical plane in downward direction

Applying the Newton's second law, net forces in the vertical plane are:

Net force, f = N - (mg + F Sin 37°)

As there is no acceleration in the vertical plane hence, net force f = 0.

So,

N - (mg + F Sin 37°) = 0

Adding (mg + F Sin 37°) both the sides in above equation, we get

N = mg + F Sin 37°

N = 12 \times 9.8 + 40 \times 0.601      because (Sin 37° = 0.601)

N = 117.6 + 24.04

N = 141.64 Newton

3 0
3 years ago
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