An object has an acceleration of 12.0 m/s/s. If the net force

cupoosta [38]

Answer:

Yes, it's correct

Explanation:

Newton's second Law states that the acceleration of an object is proportional to the net force applied on it, according to the equation:

$F=ma$

where

F is the net force on the object

m is the mass of the object

a is the acceleration of the object

We can re-arrange the previous equation in order to solve explicitely for a, the acceleration, and we find:

$F=ma\\\frac{F}{m}=\frac{ma}{m}\\\frac{F}{m}=a\\a=\frac{F}{m}$

So, we see that the acceleration is proportional to the net force and inversely proportional to the mass of the object.

4 0

3 years ago

The figure below shows two forces A=15 N and B=5 N acting on an object. What is the

alexandr1967 [171]

Answer:

18.03 N

Explanation:

From the fiqure below,

Using parallelogram law of vector

R² = 15²+5²-2×5×15cos(180-60)

R² = 225+25-150cos120°

R² = 250-150(-0.5)

R² = 250+75

R² = 325

R = √325

R = 18.03 N

Hence the resultant force of the object is 18.03 N

7 0

2 years ago

A table exerts a 4.0 Newton force on a book which lies at rest on its top. The force exerted by the book on the table is

SOVA2 [1]

I believe the correct answer from the choices listed above is the third option. <span>The force exerted by the book on the table is equal to the force exerted by the table which is 4.0 N. The book does not move so it must be that the forces are balanced. Hope this answers the question.</span>

4 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:

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