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4 years ago

A 0.22-caliber handgun fires a 27-g bullet at a velocity of 755 m/s. is the wave nature of matter significant for bullets?

1 answer:

6 0

<span>All matter can exhibit wave-like behavior, but the question asks if it is significant. A bullet is a very, very large object compared to the light wave-particles where the wave nature is typically observed. So for an object of this magnitude, it is insignificant.</span>

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When drawing a free body diagram, which of the following things would NOT be included

Neko [114]

C. Mass of the object

5 0

3 years ago

Jason launches a model rocket with a mass of 2.0 kg from his spring-powered rocket launcher with a spring constant of 800 N/m. H

Arada [10]

Answer:

121 Joules

6.16717 m

Explanation:

m = Mass of the rocket = 2 kg

k = Spring constant = 800 N/m

x = Compression of spring = 0.55 m

Here, the kinetic energy of the spring and rocket will balance each other

$\frac{1}{2}mu^2=\frac{1}{2}kx^2\\\Rightarrow u=\sqrt{\frac{kx^2}{m}}\\\Rightarrow u=\sqrt{\frac{800\times 0.55^2}{2}}\\\Rightarrow u=11\ m/s$

The initial velocity of the rocket is 11 m/s = u.

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² = g

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11^2}{2\times -9.81}\\\Rightarrow s=6.16717\ m$

The maximum height of the rocket will be 6.16717 m

Potential energy is given by

$P=mgh\\\Rightarrow P=2\times 9.81\times \frac{0^2-11^2}{2\times -9.81}\\\Rightarrow P=121\ J$

The potential energy of the rocket at the maximum height will be 121 Joules

5 0

3 years ago

A force of 1.150×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 795 N. If he starts

nalin [4]

Answer:

He has a speed of 16.60m/s after 35.0 meters.

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

$v_{f}^{2} = v_{i}^{2} + 2ad$

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$ (1)

The acceleration can be found by means of Newton's second law:

$\sum F_{net} = ma$

Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$ (2)

All the forces can be easily represented in a free body diagram, as it is shown below.

Forces in the x axis:

$F_{x} = F - F_{air}$ (3)

Forces in the y axis:

$F_{y} = 0$ (4)

Solving for the forces in the x axis:

$F_{x} = F - F_{air}$

Where $F = 1.150x10^{3} N$ and $F_{air} = 795 N$ :

$F_{x} = 1.150x10^{3} N - 795 N$

$F_{x} = 355 N$

Replacing in equation (2) it is gotten:

$Fx + Fy = ma$

$355 N + 0 N = (90.0 Kg)a$

$355 N = (90.0 Kg)a$

$a = \frac{355 N}{90.0Kg}$

$a = \frac{355 Kg.m/s^{2}}{90.0Kg}$

$a = 3.94 m/s^{2}$

So the acceleration for the cyclist is $3.94 m/s^{2}$ , now that the acceleration is known, equation (1) can be used:

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$

However, since he was originally at rest its initial velocity will be zero ( $v_{i} = 0$ ).

$v_{f} = \sqrt{2ad}$

$v_{f} = \sqrt{2(3.94m/s^{2})(35.0m)}$

$v_{f} = 16.60m/s$

He has a speed of 16.60m/s after 35.0 meters

8 0

3 years ago

What is the circle of least confusion?

Artist 52 [7]

Answer and Explanation:

In optics, a CoC(Circle of Confusion) is defined the minimum cross section of a paraxial bundle of rays made by a lens which is sphero-cylindrical type and can be viewed as an optical spot, which do not converge perfectly at the focus while a point source is being imaged due to spherical aberration.

The Circle of Confusion is also referred to as circle of indistinctness or a blur spot

5 0

3 years ago

1 Calculate the size of the quantum involved in the excitation of (a) an electronic motion of frequency 1.0 × 1015 Hz, (b) a mol

torisob [31]

Answer: a) E= 6.63x10^-19J

E= 3.97×10^2KJ/mol

b) E = 3.31×10^-19J

E= 18.8×10^4 KJ/mol

C) E = 1.32×10^-33J

E= 8.01×10^-10KJ/mol

Explanation:

a) E = h ×f

h= planks constant= 6.626×10^-34

E=(6.626×10^-34)×(1.0×10^15)

E=6.63×10^-19J

1mole =6.02×10^23

E=( 6.63×10^-19)×(6.02×10^23)

E=3.97×10^2KJ/mol

b) E =(6.626×10^-34)/(1.0×10^15)

E=3.13×10^-19J

E= 3.13×10^-19) ×(6.02×10^23)

E= 18.8×10^3KJ/MOL

c) E= (6.626×10^-34) /0.5

E= 1.33×10^-33J

E= (1.33×10^-33) ×(6.02×10^23)

E= 8.01×10^-10KJ/mol

8 0

3 years ago