4 years ago

What is

125x12} " align="absmiddle" class="latex-formula">

Physics

1 answer:

Whitepunk [10]4 years ago

5 0

Answer:

What are you asking me? Tex{t} x1|2?3 squ{R] oo|t?

Explanation:

You might be interested in

Consider an oil droplet of mass m and charge q. We want to determine the charge on the droplet in a Millikan-type experiment. We

Katen [24]

Answer:

$q=\frac{mg}{E_o}$

Explanation:

Given:

Charge = <em>q</em>

Electric field strength = $E_o$

weight of the droplet = <em>mg</em>

The charge is suspended motionless. This is because the electric force on the charge is balanced by the weight of the droplet.

electric force on charged droplet, $F=qE_o$

This is balanced by the weight, $mg$

Equating the two:

$qE_o=mg\\\Rightarrow q=\frac{mg}{E_o}$

4 0

4 years ago

Acceleration changes and objects...

brilliants [131]

Answer:

speed and direction

Explanation:

Acceleration is the rate of change of speed and direction.

3 0

3 years ago

If two 100 ohms resistors are placed in series, their total resistance is what?

Sindrei [870]

Answer:

add the Resistance

2(100)= 200

2) 200 ohms

3 0

3 years ago

Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface tempe

stiks02 [169]

Answer:

6.0 × $10^{11}$ W/ $m^{2}$

Explanation:

From Wien's displacement formula;

Q = e A $T^{4}$

Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = $\frac{Q}{A}$ = e $T^{4}$

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × $(1000)^{4}$

= 0.6 × 1.0 × $10^{12}$

= 6.0 × $10^{11}$ $\frac{W}{m^{2} }$

Therefore, the emissive intensity coming out of the surface is 6.0 × $10^{11}$ W/ $m^{2}$ .

3 0

3 years ago

PLEASE HELP!!!!

Alex_Xolod [135]

I would say A Because it weighs more than the water

5 0

3 years ago