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shepuryov [24]
3 years ago
12

A 2 kg ball going 10 m/s follows a path that is perpendicular to the surface of a wall. It impacts the wall and looses 20 % of i

ts momentum in the collision. What is the balls impulse?
Physics
1 answer:
Ostrovityanka [42]3 years ago
4 0

Answer:

If two objects make a head on collision, they can bounce and move along the same direction they approached from (i.e. only a single dimension). However, if two objects make a glancing collision, they'll move off in two dimensions after the collision (like a glancing collision between two billiard balls).

For a collision where objects will be moving in 2 dimensions (e.g. x and y), the momentum will be conserved in each direction independently (as long as there's no external impulse in that direction).

In other words, the total momentum in the x direction will be the same before and after the collision.

\Large \Sigma p_{xi}=\Sigma p_{xf}Σp

xi

​

=Σp

xf

​

\Sigma, p, start subscript, x, i, end subscript, equals, \Sigma, p, start subscript, x, f, end subscript

Also, the total momentum in the y direction will be the same before and after the collision.

\Large \Sigma p_{yi}=\Sigma p_{yf}Σp

yi

​

=Σp

yf

​

\Sigma, p, start subscript, y, i, end subscript, equals, \Sigma, p, start subscript, y, f, end subscript

In solving 2 dimensional collision problems, a good approach usually follows a general procedure:

Identify all the bodies in the system. Assign clear symbols to each and draw a simple diagram if necessary.

Write down all the values you know and decide exactly what you need to find out to solve the problem.

Select a coordinate system. If many of the forces and velocities fall along a particular direction, it is advisable to use this direction as your x or y axis to simplify calculation; even if it makes your axes not parallel to the page in your diagram.

Explanation:

i think this what your aswer is let me know if you got  it right if not i fix it or i will look in my answer book

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Answer:

3.605551275463989

Explanation:

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A thin spherical shell has a radius of 0.70 m. An applied torque of 860 N m gives the shell an angular acceleration of 4.70 rad/
Artyom0805 [142]

Answer:

I=182.97\ kg-m^2

Explanation:

Given that,

Radius of a spherical shell, r = 0.7 m

Torque acting on the shell, \tau=860\ N

Angular acceleration of the shell, \alpha =4.7\ m/s^2

We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

\tau=I\alpha

I is the rotational inertia of the shell

I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{860}{4.7}\\\\I=182.97\ kg-m^2

So, the rotational inertia of the shell is 182.97\ kg-m^2.

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3 years ago
a pool ball leaves a 0.60-meter high table with an initial high table with an initial horizontal velocity of 2.4m/s. what is the
inysia [295]

Answer:

0.84 m

Explanation:

Given in the y direction:

Δy = 0.60 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

0.60 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.35 s

Given in the x direction:

v₀ = 2.4 m/s

a = 0 m/s²

t = 0.35 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (2.4 m/s) (0.35 s) + ½ (0 m/s²) (0.35 s)²

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3 years ago
Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat
Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e \frac{dv}{dt} =10 cm^3/s

Consider r be the radius of the balloon.

The volume of of a sphere is

v=\frac{4}{3} \pi r^3

Differentiate with respect to t

\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}

\Rightarrow 10 =4\pi r^2\frac{dr}{dt}

\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}

The surface of area of the balloon is(S) = 4\pi r^2

S=4\pi r^2

Differentiate with respect to t

\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}

Putting the value of \frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}

\Rightarrow \frac{dS}{dt} =\frac{20}{ r}

Given that r = 5 cm

[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}  =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

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answer is chemical process

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