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mixer [17]
3 years ago
7

If the standard kilogram bar kept in Paris were subjected to a net force of 1 newton, what acceleration would it have as a resul

t?
Physics
2 answers:
sdas [7]3 years ago
7 0

Answer:

It would have an acceleration of 1\frac{m}{s^{2}}

Explanation:

Let's start defining the newton unit (Newton : N)

The N definition is

N=Kg.(\frac{m}{s^{2}})

Where Kg is kilogram and \frac{m}{s^{2}} is an acceleration unit.

We can think the definition of newton as :

Given an object with a mass of 1 Kg subjected to a force of 1 N the acceleration that it experiments is 1\frac{m}{s^{2}}

As a result, the standard  kilogram bar kept in Paris will experiment an acceleration of 1\frac{m}{s^{2}} If we subject it to a net force of 1 N.

Klio2033 [76]3 years ago
3 0
1 m/s^2
Because of
F=ma
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2 years ago
A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. the coefficient of kinetic friction between the box
forsale [732]
We need to see what forces act on the box:

In the x direction:

Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.

In the y direction:

N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force. 

From N-Gcosα=0 we get: 

N=Gcosα, we will need this for the force of friction.

Now to solve for Fh:

Fh=ma + Ff + Gsinα,

Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²

Fh=ma + μmgcosα+mgsinα

Now we plug in the numbers and get:

Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N

The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
8 0
3 years ago
Force can change the ____________ and _______________ of an object​
slamgirl [31]

Answer:

Force can change the speed and direction of an object.

Explanation:

6 0
3 years ago
An 87.6 g lead ball is dropped from rest from a height of 7.00 m. The collision between the ball and the ground is totally inela
bija089 [108]

Taking specific heat of lead as 0.128 J/gK = c

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When leads gets heated by a temperature ΔT energy needed = mcΔT

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Comparing both the equations

                      87.6*10^{-3}*9.81*7 = 87.6*10^{-3}*0.128*10^3ΔT

                        ΔT = 0.536 K

                        Change in temperature same in degree and kelvin scale

                                      So ΔT = 0.536 ^0C

7 0
3 years ago
Which of the following would result if thermal energy is added to an object?​
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Answer:

is there supposed to be a pic and abcd options?

8 0
3 years ago
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