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zalisa [80]
3 years ago
7

A table tennis ball with a mass of 0.003kg and soccer with a mass of 0.43 kg. are both set in motion at 16 m"s. Calculate and co

mpare the momenta of both balls.
Physics
2 answers:
Aliun [14]3 years ago
7 0

Let us first know the given: Tennis ball has a mass of 0.003 kg, Soccer ball has a mass of 0.43 kg. Having the same velocity at 16 m/s. First the equation for momentum is P=MV P=Momentum M=Mass V=Velocity. Now let us have the solution for the momentum of tennis ball. Pt=0.003 x 16 m/s= (    kg-m/s ) I use the subscript "t" for tennis.  Momentum of Soccer ball Ps= 0.43 x 13m/s = (      km-m/s). If we going to compare the momentum of both balls, the heavier object will surely have a greater momentum because it has a larger mass, unless otherwise  the tennis ball with a lesser mass will have a greater velocity to be equal or greater than the momentum of a soccer ball.

Mashcka [7]3 years ago
5 0

Answer:

\texttt{Ration of moment of soccer ball to table tennis ball}=\frac{6.88}{0.048}=143.33

Explanation:

Linear momentum is product of mass and velocity.

Mass of table tennis ball = 0.003 kg

Mass of table soccer ball = 0.43 kg

Velocity of both balls = 16 m/s

Momentum of tennis ball = 0.003 x 16 = 0.048 kgm/s

Momentum of soccer ball = 0.43 x 16 = 6.88 kgm/s

\texttt{Ration of moment of soccer ball to table tennis ball}=\frac{6.88}{0.048}=143.33

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The nebular theory describes the formation of the solar system and states that the system began as a gigantic cloud of gas and dust called a nebula which eventually condensed to form the sun, planets and other objects in the solar system. The first fact speaks to the formation of the planets, where gravity pulled larger clumps of material closer to form solid rocky planets closer to the sun and gas giants further out. The second requirement is that a nearby explosion or super nova would have to disturb our nebula to trigger rotation and the eventual formation of the sun. The third requirement/fact is that the planets go around the sun in the same direction. the last fact is that the planets go around the sun within 6 degrees of a common plane. This indicates that the solar system formed from a spinning disk of materials.
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3 years ago
A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.
Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

1)

The motion of the shell is a projectile motion, so we  can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

y=(u sin \theta)t-\frac{1}{2}gt^2 (1)

where:

u sin \theta is the initial vertical velocity of the shell, with u=156 ft/s and \theta=49.0^{\circ}

g=32 ft/s^2 is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

x=(ucos \theta)t

where u cos \theta is the initial horizontal velocity of the shell.

We can re-write this last equation as

t=\frac{x}{u cos \theta} (1b)

And substituting into (1),

y=xtan\theta -\frac{1}{2}gt^2 (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of \alpha=-41.0^{\circ} from the horizontal, so we can write the position (x,y) of the hill as

y=x tan \alpha (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

xtan\alpha = xtan \theta - \frac{1}{2}gt^2

Substituting (1b) into this equation,

xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0

Which has 2 solutions:

x = 0 (origin)

and

tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft

So, the distance d down the hill at which the shell strikes the hill is

d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m

2)

In order to find how long the mortar shell remain in the air, we can use the equation:

t=\frac{x}{u cos \theta}

where:

x = 1322 ft is the final position of the shell when it strikes the hill

u=156 ft/s is the initial velocity of the shell

\theta=49.0^{\circ} is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s

3)

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s

Instead, the vertical component of the velocity is given by

v_y=usin \theta -gt

And substituting at t = 12.9 s (time at which the shell strikes the hill),

v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s

Therefore, the  final speed of the shell is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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There are two types of stress, axial and tangential. Since we are only given with the dimension of the radius (and not the length), the possible stress is axial. So, the area is,

A = πr² = π(0.75 in)² = 1.767 in²

So,
σ = F/A = 500 lb/1.767 in² = <em>282.94 psi</em>
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In a catapult, energy is transferred into useful kinetic energy stores. What store is the energy in when it is the catapult?
yawa3891 [41]

Answer:

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Where do plants obtain their energy
Ne4ueva [31]

Answer:

Sunlight

Explanation:

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