All categories

3 years ago

Suggest two ways in which the boy could balance the beam?

Physics

2 answers:

Oduvanchick [21]3 years ago

7 0

He can balance it on his hand or look for something that can handle the weight of it

postnew [5]3 years ago

5 0

Answer:

Hope this is what you were looking for!

Explanation:

1) So, if we are talking about balancing a straight stick (beam), he could put a finger from each hand on either side of it and move them together. As a result of friction his fingers will meet at the balance point.

2) Method of trial and error.

You might be interested in

Define malleability?<br>

ad-work [718]

Answer:

Explanation:

the quality or state of being malleable: such as. a : capability of being shaped or extended by hammering, forging, etc. the malleability of tin.

5 0

3 years ago

A 300N box on a 43 degree angle.

Ksenya-84 [330]

Answer:is this a question??? I’m so confused

Explanation:

7 0

3 years ago

Modify how could you charge the electric circuit shown below to allow lightbulb a to stay lit even if lightbulb b is removed fro

shepuryov [24]

When a circuit is complete, or closed, electrons can flow from one end of a battery all the way around, through the wires, to the other end of the battery. Along its way, it will carry electrons to electrical objects that are connected to it – like the light bulb – and make them work!

5 0

3 years ago

How can u create a fair test?<br><br> Short answer only:

Crazy boy [7]

Answer:

making sure that you change one factor at a time while keeping all other conditions the same

Explanation:

5 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

8 0

3 years ago