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Zigmanuir [339]
2 years ago
13

In making the determination that bullets, shells, or cartridge cases were fired from a specific weapon, the criminalist would us

e which piece of laboratory equipment?
Physics
1 answer:
ss7ja [257]2 years ago
7 0

Answer:

Comparison Microscope

Explanation:

The Comparison Microscope allows for comparison between two objects or samples by placing them side by side.

It is primarily used in criminology for ballistics which makes it ideal to find out if bullets, shells, or cartridge cases were fired from a specific weapon.

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Number of valence electrons for calcium?
zimovet [89]
Calcium has 2 valence electrons
7 0
3 years ago
Read 2 more answers
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
3 years ago
Tres móviles, A, B y C, con las mismas características tienen los siguientes estados: A se encuentra inicialmente en reposo y ca
Morgarella [4.7K]

Answer:

La respuesta es sí, hay una fuerza que actúa sobre el móvil A y es la única fuerza ya que A cae libremente bajo la influencia de la fuerza.

Explanation:

Según la primera ley de movimiento de Newton, un cuerpo continuará en su estado de reposo o en un movimiento uniforme en línea recta a menos que actúen sobre él fuerzas impresas.

Dado que el móvil A cae libremente, desde su estado de reposo inicial, según la primera ley de movimiento de Newton, experimenta una fuerza que actúa sobre él para hacer que caiga y continúe en caída libre.

El móvil B se mueve con una velocidad constante, por lo tanto, de acuerdo con la primera ley de movimiento de Newton, no hay fuerzas impresas que actúen sobre él.

El móvil C está completamente en reposo en el suelo, por lo tanto, tampoco hay fuerzas que actúen sobre él.

La respuesta es sí, hay una fuerza actuando sobre el móvil A y es la única fuerza cuando A cae libremente bajo la influencia de la fuerza.

6 0
3 years ago
A hoop and a disk with uniform mass distribution have the same radius but the total masses are not known. Can they both roll dow
ser-zykov [4K]

Answer:

Explanation:

radius of hoop and the radius of disk is same = R

Let the mass of hoop is M and the mass of disk is M'.

As they reach the bottom of teh surface in same time so they travel equal distance thus, they have same acceleration.

The acceleration is given by

a=\frac{gSin\theta }{1+\frac{I}{MR^{2}}}

As the acceleration is same so that the moment of inertia is also same.

Moment of inertia of disk = moment of inertia of hoop

1/2 x mass of disk x R² =  mass of hoop x R²

So, mass of disk = 2 x mass of hoop

Option (c) is correct.

5 0
3 years ago
A copper wire of length 16 cm is in a magnetic field of 0.19 T. If it has a mass of 9.0 g, what is the minimum current through t
Sergeeva-Olga [200]

Answer:

2.9 A

Explanation:

L = 16 cm = 0.16 m

B = 0.19 T

m = 9 g = 0.009 kg

Let the minimum current be i.

Magnetic force is balanced by the gravitational force

B x i x L = m x g

0.19 x i x 0.16 = 0.009 x 9.8

i = 2.9 A

7 0
3 years ago
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