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andrew11 [14]
4 years ago
12

A table tennis ball with a mass of 0.003 kg and a soccer ball with a mass of 0.43 kg or both Serta name motion at 16 M/S calcula

te and compare the moments of both balls
Physics
2 answers:
poizon [28]4 years ago
8 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.
Zepler [3.9K]4 years ago
7 0

Answer:

Momentum of tennis ball = 0.048 \frac{kg m}{s}

Momentum of soccer ball = 6.88 \frac{kg m}{s}

momentum of soccer ball is greater than that of tennis ball.

Explanation:

For table tennis ball,

Momentum of tennis the ball is given by,

Momentum = mass × volume

Momentum =  0.003 × 16

Momentum = 0.048 \frac{kg m}{s}

For table soccer ball,

Momentum of the soccer ball is given by,

Momentum = mass × volume

Momentum =  0.43 × 16

Momentum = 6.88 \frac{kg m}{s}

Thus, momentum of soccer ball is greater than that of tennis ball.

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A body of mass 400 kg is suspended at a lower end of a light vertical chain and is being pulled up vertically. Initially the bod
yKpoI14uk [10]

Answer:31.62 m/s

Explanation:

Given

mass of body m=400 kg

Pull on chain is F_1=6000g N=60 kN

Pull get smaller at the rate of F_2=360g N/m

Net Upward Force F=6000 g-360 g\times 10=24 kN

net acceleration a=\frac{F}{m}

a=\frac{24\times 1000}{m}

a=\frac{24000}{400}

a=60 m/s^2

but g is acting downward

a_{net}=a-g=60-10=50 m/s^2

using v^2-u^2=2 as

here initial velocity is zero

v^2=2\times 50\times 10

v=31.62 m/s

7 0
4 years ago
An automobile tire having a temperature of −1.6 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 22 lb/in2 . What
r-ruslan [8.4K]

Answer:

25.8 lb/in²

Explanation:

Gay-Lussac's law tells us that given an ideal gas of a certain mass has a constant volume, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.

\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} } \\\\\frac{22}{-1.6+273.15} =\frac{P_{2} }{45+273.15} \\\\P_{2} = \frac{22*318.15}{271.55} = 25.8lb/in^{2}

4 0
3 years ago
a police officer parked on the side of the rode clockes a driver at a constant 130km/h (passing the stopped car). if the officer
nignag [31]

Answer

given,

speed of the vehicle = 130 Km/h

officers accelerates at = 12 km/h/s

time taken by the driver to catch the vehicle = ?

at time t second

130 t = \dfrac{1}{2}at^2

260 t - 12 t^2 = 0

t (12 t - 260 )= 0

12 t =  260

t = 21.67 s

v = u + at

v = 12 × 21.67

v = 260 Km/h

8 0
4 years ago
Carbon is most abundant in the and . <br><br> Fill the blanks
earnstyle [38]

Answer: yes it is

Explanation:

6 0
3 years ago
A hoop, a solid cylinder, a spherical shell, and a solid sphere are placed at rest at the top of an inclined plane. All the obje
iogann1982 [59]

Answer:

Solid sphere will reach first

Explanation:

When an object is released from the top of inclined plane

Then in that case we can use energy conservation to find the final speed at the bottom of the inclined plane

initial gravitational potential energy = final total kinetic energy

mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

now we have

I = mk^2

here k = radius of gyration of object

also for pure rolling we have

v = R\omega

so now we will have

mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(\frac{v^2}{R^2})

mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})

v^2 = \frac{2gh}{1 + \frac{k^2}{R^2}}

so we will say that more the value of radius of gyration then less velocity of the object at the bottom

So it has less acceleration while moving on inclined plane for object which has more value of k

So it will take more time for the object to reach the bottom which will have more radius of gyration

Now we know that for hoop

mk^2 = mR^2

k = R

For spherical shell

mk^2 = \frac{2}{3}mR^2

k = \sqrt{\frac{2}{3}} R

For solid sphere

mk^2 = \frac{2}{5}mR^2

k = \sqrt{\frac{2}{5}} R

So maximum value of radius of gyration is for hoop and minimum value is for solid sphere

so solid sphere will reach the bottom at first

7 0
3 years ago
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