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3 years ago

7

Do all engines operate at less than 100 percent efficiency because they conduct heat

2 answers:

Alika [10]3 years ago

4 0

all engines operate at less than 100 percent efficiency because they EMIT heat. ;)

Julli [10]3 years ago

3 0

They EMIT heat is the correct answer

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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

3 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

3 years ago

How fast would 40 Newtons of force accelerate a 2 kg object?

Digiron [165]

Answer:

$20 m/s^2$

Explanation:

We can solve this problem by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between its mass and its acceleration:

$F=ma$

where

F is the net force on the object

m is its mass

a is its acceleration

In this problem:

F = 40 N is the force on the object

m = 2 kg is its mass

Therefore, the acceleration of the object is

$a=\frac{F}{m}=\frac{40}{2}=20 m/s^2$

8 0

3 years ago

Please help me! Science is not my best subject

Paul [167]

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8 0

3 years ago

A 4000 kg rocket is launched, shooting 50 kg of burned fuel from its exhaust at a velocity of -625 m/s. What is the velocity of

AnnZ [28]

Hopefully this will help you.

8 0

4 years ago