All categories

2 years ago

What is first and second conditions of equilibrium?

1 answer:

7 0

I think the answer is the first condition sum of forces acting on a body is zero ( ∑ F =0 ) and the second condition sum of torque acting on a body is ( ∑ τ = 0 )

Could you put me as brainliest?

You might be interested in

What is important when it comes to winning

Dmitry_Shevchenko [17]

Answer:

satisfaction, enjoyment and fair play

3 0

2 years ago

Read 2 more answers

A 400 gram sample of alcohol at 10o C [Ca=2.64 J/g*oC] is mixed with 400 grams of warm water [Cw=4.186 J/g*Co] at 88o C. Assume

alukav5142 [94]

The heat lost by the water will be equivalent to the energy gained by the alcohol. Thus:
maCaΔT = -mwCwΔT
400 x 2.64 x (T - 10) = 400 x 4.186 x (88 - T)
T = 57.8 °C

8 0

2 years ago

What must be part of a quantitative observation?

inna [77]

a number hope this helps

8 0

2 years ago

Read 2 more answers

The Moon is made up mostly of _______, similar to minerals on Earth.

yawa3891 [41]

The moon is made up mostly of Silicates, hope this answer helps.

3 0

2 years ago

Read 2 more answers

What are the units, if any, of the particle in a box wavefunction. What does this mean?

SIZIF [17.4K]

Answer:

$[\psi]= [Length^{-3/2}]$
This means that the integral of the square modulus over the space is dimensionless.

Explanation:

We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral

$\int\limits^{x_f}_{x_0} \int\limits^{yf}_{y_0} \int\limits^{z_f}_{z_0} |\psi|^2 \, dz \, dy \, dx$

must be dimensionless, as represents a probability.

As the differentials has units of length

$[dx]=[dy]=[dz]=[Length]$

for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:

$[\psi]^2 = [Length^{-3}]$

taking the square root this gives us :

$[\psi] = [Length^{-3/2}]$

5 0

2 years ago