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nadezda [96]
3 years ago
14

George is applying a downward force of 50N to and object that has a mass of 50kg. What is the normal force (FN) of the object wh

ile George is applying the pressure?g=10 m/s2
Physics
1 answer:
attashe74 [19]3 years ago
3 0

The normal force acting on the object is 500 N in the upward direction

<u>Explanation:</u>

As George is applying a downward force, the normal force will be in the upward direction. The normal force will be exerted due to the acceleration due to gravity exerted on the object.

So, as per Newton's second law, the normal force acting on the object can be measured by the product of mass of the object and the acceleration due to gravity acting on the object.

But as the acceleration due to gravity is a downward acting acceleration and the normal force is a upward acting force, so the acceleration will be having a negative sign in the formula.

Normal\ force = Mass \times Acceleration\ due\ to\ gravity

Here, acceleration due to gravity g = -10 m/s² and mass is given as 50 kg, then

Normal force = 50 × (-10) = -500 N

So, the normal force acting on the object is 500 N in the upward direction.

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HOPE THIS HELPS YOU MATE!!
I HAVE ALSO GIVEN THE EXPLANATION THINKING THAT IT MIGHT HELP YOU.
THANK YOU.
3 0
2 years ago
Read 2 more answers
A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis
bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

B=\frac{\mu_o I_r}{2\pi r}     (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

I_r=JA_r

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by  A_r=\pi(r^2-R_1^2)

The current density is given by the total area and the total current:

J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current  = 4 A

Then, the current in the wire for a distance r is:

I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}  (2)

You replace the last result of equation (2) into the equation (1):

B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})

Finally. you replace the values of all parameters:

B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

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