What is 8 m/s - 3 m/s?

What happens to the energy of gas particles when an elastic collision takes place?

Gnesinka [82]

Answer:

answer 3

Explanation:

7 0

3 years ago

Read 2 more answers

What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 85.0% absorbed, puts 470 J of energy

Advocard [28]

Answer:

26036485.6433 W/m²

Explanation:

E= Energy = 470 J

t = Time = 4 seconds

d = Diameter = 2.6 mm

Power is given by

$P=\dfrac{E}{t}$

Intensity is given by

$I=\dfrac{P}{\pi r^2}\\\Rightarrow 0.85I=\dfrac{E}{t\dfrac{\pi}{4} d^2}\\\Rightarrow I=\dfrac{470}{0.85\times 4\times \dfrac{\pi}{4}\times (2.6\times 10^{-3})^2}\\\Rightarrow I=26036485.6433\ W/m^2$

The intensity of the laser beam is 26036485.6433 W/m²

8 0

4 years ago

A5 kg frisbee is thrown from rest to a final speed of 12 m/s. What is the impulse of the frisbee?

lesya692 [45]

Answer:

60kgm/s

Explanation:

Given parameters:

Mass of frisbee = 5kg

Final speed = 12m/s

Unknown:

Impulse of the frisbee = ?

Solution:

The impulse of the frisbee is the same as the change in momentum.

It is given as:

Impulse = mass (Final velocity - Initial velocity)

Impulse = 5(12 - 0) = 60kgm/s

3 0

3 years ago

30 points!

vlada-n [284]

Answer:

Since strong nuclear forces involve only nuclear particles (not electrons, bonds, etc) items 3 and 4 are eliminated.

Again item 2 refers to bonds between atoms and is eliminated.

This leaves only item 1.

Nuclear forces are very short range forces between components of the nucleus.

Weak nuclear forces are trillions of times smaller than strong forces.

Gravitational forces are much much smaller than the weak nuclear force.

6 0

3 years ago

Galactic Alliance Junior Mission Officer (GAJMO) Bundit Nermalloy is predicting the kinetic energy of a supply spacecraft, which

antiseptic1488 [7]

Answer:

the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

W =∫ F dx = ΔK

Let's replace

∫ (α x³ + β) dx = ΔK

α x⁴ / 4 + β x = ΔK

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

x (α x³ + β) = $K_{f}$ - K₀

$K_{f}$ = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

Let's calculate

$K_{f}$ = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

Kf = 2.7 10¹¹ - 1.1475 10¹¹

Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

W = x F₀ = $K_{f}$ –K₀

$K_{f}$ = K₀ + x F₀

We calculate

$K_{f}$ = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

$K_{f}$ = (2.7 -2.625) 10¹¹

$K_{f}$ = 7.5 10⁹ J

5 0

3 years ago