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just olya [345]
2 years ago
15

A particle released during the fission of uranium-235 is a(n) ________

Physics
1 answer:
Kaylis [27]2 years ago
6 0
<span>A particle released during the fission of uranium-235 is a "Neutron"</span>
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The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c
eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

X_f = X_i +\frac{1}{2}(V_i+V_f)t

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

X_f = \frac{1}{2} (V_i+V_f)t

Clearing for time,

t = \frac{2X_f}{(V_i+V_f)}

t = \frac{2*1.2}{24+0}

t = 0.1s

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

a= \frac{\Delta V}{t}

Then

F = m\frac{\Delta V}{t}

F = m\frac{V_f-V_i}{t}

F = m\frac{-V_i}{t}

F = \frac{(1600kg)(-24m/s)}{(0.1s)}

F = -384000N

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

V_f^2=V_i^2-2ax

0 = (24m/s)^2-2*a(1.2m)

a = \frac{(24m/s)^2}{1.2m}

a=480m/s^2

The gravity is equal to 0.8, then the acceleration is

a = 480*\frac{g}{9.8}

a = 53.3g

3 0
2 years ago
What does the speedometer of a car tell you?
Gelneren [198K]

Answer:

Speedometer, instrument that indicates the speed of a vehicle, usually combined with a device known as an odometer that records the distance traveled.

Explanation:

5 0
2 years ago
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.
SSSSS [86.1K]

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

\rho(r) = ar^2 - br^3

r is the radius of the spherical shell

dr is the thickness

volume of shell

dV = 4 \pi r^2 dr

mass of shell

dM = \rho(r)dV

\rho = \rho_0 - br

now,

dM = (\rho_0 - br)(4 \pi r^2)dr

integrating both side

M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr

M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})

M = \pi R^3(\dfrac{\rho_0}{3}+\rho)

we know,

a = \dfrac{GM}{R^2}

a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}

a =\pi RG(\dfrac{\rho_0}{3}+\rho)

a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)

a = 9.94 m/s²

7 0
2 years ago
Please help me
Setler79 [48]

Answer:

This motion is known as Brownian motion.

Explanation:

This motion is known as Brownian motion.

5 0
2 years ago
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