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Alecsey [184]
2 years ago
6

Suppose a fireworks shell explodes, breaking into three large pieces for which air resistance is negligible. How is the motion o

f the center of mass affected by the explosion
Physics
1 answer:
denis-greek [22]2 years ago
3 0

The center of mass isn't affected by the explosion.

To find the answer, we need to know about the trajectory of motion at zero external force.

<h3>How is the trajectory of an object changed when the net external force on it is zero?</h3>
  • When there's no net external force acting on an object, its momentum doesn't change with time.
  • As its momentum doesn't change, so it continues with the original trajectory.
<h3>Why doesn't the trajectory of firework change when it's exploded?</h3>
  • When a firework is exploded, its internal forces are changed, but there's no external force.
  • So, although the fragments follow different trajectories, but the trajectory of center of mass remains unchanged.

Thus, we can conclude that the center of mass isn't affected by the explosion.

Learn more about the trajectory of exploded firework here:

brainly.com/question/17151547

#SPJ4

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Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez
Zanzabum

(a)  3.3\cdot 10^{-6} Pa

The radiation pressure exerted by an electromagnetic wave on a surface that totally absorbs the radiation is given by

p=\frac{I}{c}

where

I is the intensity of the wave

c is the speed of light

In this problem,

I=1000 W/m^2

and substituting c=3\cdot 10^8 m/s, we find the radiation pressure

p=\frac{1000 W/m^2}{3\cdot 10^8 m/s}=3.3\cdot 10^{-6}Pa

(b) 4.4\cdot 10^{-8} m/s^2

Since we know the cross-sectional area of the laser beam:

A=6.65\cdot 10^{-29}m^2

starting from the radiation pressure found at point (a), we can calculate the force exerted on a tritium atom:

F=pa=(3.3\cdot 10^{-6}Pa)(6.65\cdot 10^{-29} m^2)=2.2\cdot 10^{-34}N

And then, since we know the mass of the atom

m=5.01\cdot 10^{-27}kg

we can find the acceleration, by using Newton's second law:

a=\frac{F}{m}=\frac{2.2\cdot 10^{-34} N}{5.01\cdot 10^{-27} kg}=4.4\cdot 10^{-8} m/s^2

6 0
3 years ago
Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 5.00 s, it rotates 13.9 rad. Du
alisha [4.7K]

Answer:

(a) Angular acceleration is 1.112 rad/s².

(b) Average angular velocity is 2.78 rad/s .

Explanation:

The equation of motion in Rotational kinematics is:

θ = θ₀ + 0.5αt²

Here θ is angular displacement at time t, θ₀ is angular displacement at time t=0, t is time and α is constant angular acceleration.

(a) According to the problem, θ is 13.9 rad, θ₀ is zero as it is at rest and t is 5 s. Put these values in the above equation:

13.9 = 0 + 0.5α(5)²

α = 1.112 rad/s²

(b) The equation of average angular velocity is:

ω = Δθ/Δt

ω = \frac{13.9}{5}

ω = 2.78 rad/s

3 0
4 years ago
I need to talk to a girl my age
Greeley [361]

Answer:

that is my question

Explanation:

you want a girl to talk your age

<h2>HOW OLD ARE YOU</h2>
6 0
3 years ago
A uniform solid cylindrical log begins rolling without slipping down a ramp that rises 28.0° above the horizontal. After it has
Novay_Z [31]

Answer:

3.07 m/s

Explanation:

6 0
3 years ago
The carbon monoxide molecule (CO) consists of a carbon atom and an oxygen atom separated by a distance of
valkas [14]

Answer:

6.46*10^-11m

Explanation:

We can use the equation of the center of the mass of two atoms

Xcm=\frac{mcxc+moxo}{mc+mo}

If we take the origin at the center of the molecule carbon monoxice, xc will be 0, so

Xcm=\frac{moxo}{mc+mo}=\frac{xo}{mc/mo+1}

Xcm=\frac{1.13.10^-3}{0.750mo/mo+1}=6.46*10^-11m

7 0
3 years ago
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