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ehidna [41]
2 years ago
6

A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k

ilograms. What is the kinetic energy of the system after the collision if the collision is elastic?
Physics
1 answer:
rewona [7]2 years ago
3 0

The kinetic energy of the small ball before the collision is

                             KE  =  (1/2) (mass) (speed)²

                                     = (1/2) (2 kg) (1.5 m/s)

                                     =    (1 kg)  (2.25 m²/s²)

                                     =        2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

                     Energy is never created or destroyed. 
                     If it seems that some energy disappeared,
                     it actually had to go somewhere.
                     And if it seems like some energy magically appeared,
                     it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision.  The large ball
and the small ball will just have to share the same 2.25 joules.

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Ice fishermen sit on top of frozen lakes in the winter and catch fish in the liquid water below through holes cut in the ice she
melomori [17]

Answer:

This is because below 4°c, water unlike other materials becomes less dense when it's temperature is further lowered.

Explanation:

Due to the unusual nature of water; at about 4°c, the behavior of the density of water in relation to its temperature reverses. This means that water becomes less dense as it becomes colder below 4°c. The colder parts therefore floats to the top of the water body while the warmer part sinks allowing the top to freeze and the remaining body below to remain in its liquid state.

The freezing of the top of the lake alone protects the remaining depth of water from freezing by acting as an insulator and preventing further heat loss from the water to the ambient space. If this had not been the case, and water froze all through, marine lives will freeze to death and it will be more difficult to melt the ice come the next summer.

This behavior is due to the hydrogen bonding of the water molecules.

8 0
2 years ago
the distance between the sun and earth is about 1.5X10^11 m. express this distance with an SI prefix and kilometers
Angelina_Jolie [31]
First, we write the SI prefixed. The SI unit for distance is meters.

Kilo = 10³
Mega = 10⁶
Giga = 10⁹
Terra = 10¹²

Because our value has ten to the power of 11, we will use the closest and lowest power prefix, which is giga. 

1.5 x 10¹¹ /  10⁹
= 1.5 x 10² Gm or 150 Gm

Writing in kilometers, we simply repeat the procedure except we divide by 10³ this time.

1.5 x 10¹¹ / 10³
= 1.5 x 10⁸ km
5 0
2 years ago
A steel guitar string with a diameter of 0.300 mm and a length of 70.0 cm is stretched by 0.500 mm while being tuned. How much f
Ganezh [65]

Answer:

10.1 N

Explanation:

Your answer is 10.1 N, I don't actually know how to do it but I hope it helps.

8 0
2 years ago
A 57 kg paratrooper falls through the air. How much force is pulling him down?
just olya [345]

Answer:

Explanation:

Force = Mass * acceleration due to gravity.

Given

Mass of the paratrooper = 57kg

Acceleration due to gravity = 9.81m/s²

Required

Force pulling him down

Substitute unto formula;

F = 57 * 9.81

Force = 559.17 N

Hence the force pulling him down is 559.17N

7 0
2 years ago
A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold
LUCKY_DIMON [66]

Answer:

(a). The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

Q=\dfrac{V}{t}

Put the value into the formula

Q=\dfrac{405}{90}

Q=4.5\ cm^3/s

(a). We need to calculate the coefficient of permeability

Using formula of coefficient of permeability

Q=kiA

k=\dfrac{Q}{iA}

k=\dfrac{Ql}{Ah}

Where, Q=discharge

l = length

A = cross section area

h=constant head causing flow

Put the value into the formula

k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}

k=8.6\times10^{-3}\ cm/s

The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(c). We need to calculate the discharge velocity during the test

Using formula of discharge velocity

v=ki

v=\dfrac{kh}{l}

Put the value into the formula

v=\dfrac{8.6\times10^{-3}\times38}{17.5}

v=0.0187\ cm/s

The discharge velocity during the test is 0.0187 cm/s.

(b). We need to calculate the volume of solid in the ample

Using formula of volume

V_{s}=\dfrac{M_{s}}{V_{s}}

Put the value into the formula

V_{s}=\dfrac{4950\times10^{-3}}{2710}

V_{s}=1826.56\ cm^3

We need to calculate the volume of the soil specimen

Using formula of volume

V=A\times L

Put the value into the formula

V=\dfrac{\pi(17.5)^2}{4}\times17.5

V=4209.24\ cm^3

We need to calculate the volume of the voids

V_{v}=V-V_{s}

Put the value into the formula

V_{v}=4209.24-1826.56

V_{v}=2382.68\ cm^3

We need to calculate the seepage velocity

Using formula of velocity

Av=A_{v}v_{s}

v_{s}=\dfrac{Av}{A_{v}}

v_{s}=\dfrac{V}{V_{v}}\times v

Put the value into the formula

v_{s}=\dfrac{4209.24}{2382.68}\times0.0187

v_{s}=0.0330\ cm/s

The seepage velocity is 0.0330 cm/s.

Hence, (a). The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

8 0
2 years ago
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