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ehidna [41]
3 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]3 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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calculate the mass of displaced water when a piece of 30cm Iceberg with surface area 1000 cm^2 floats on water density of ice is
sweet [91]

Vi = As * h = 1000 * 30 = 30,000 cm^3 = Vol. of the ice.

Vb = (Di/Dw) * Vi = (0.9/1.0) * 30,000 = 27,000 cm^3 = Vol. below surface - Vol. of water displaced.

27,000cm^3 * 1g/cm^3 = 27,000 grams = 27 kg = Mass of water displaced.

3 0
3 years ago
Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to
jok3333 [9.3K]

a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) ​K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of  Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1​

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

​=K 2 +GmM (r 11− r 21)=6.9×10 7 J

Learn more about Kinetic energy on:

brainly.com/question/12337396

#SPJ4

7 0
2 years ago
The standard metric unit of density for a solid is
prisoha [69]
There is no SI "base unit" of density.
(Any unit of mass) divided by (any unit of volume) is
a valid unit of density.

The units of density that are seen most often are

(gram per cm³) and (kgm per meter³) .
5 0
3 years ago
A 4.0 cm × 4.2 cm rectangle lies in the xy-plane. You may want to review (Pages 664 - 668) . Part A What is the electric flux th
padilas [110]

Answer:

(A). The flux is 0.336 N.m²/C

(B). The flux is zero.

Explanation:

Given that,

Length = 4.2 cm

Width = 4.0 cm

Electric field E=(150 i-200 k)\ N/C

Area vector is perpendicular to xy plane

(A). We need to calculate the flux

Using formula of flux

\phi=E\cdot A

Where, E = electric field

A = area

Put the value into the formula

\phi=(150 i-200 k)\times(4.2\times10^{-2}\times4.0\times10^{-2})k

\phi=-200\times4.2\times10^{-2}\times4.0\times10^{-2}

\phi=-0.336\ N.m^2/C

(B). Given electric field

E=(150i-200j)\ N/C

We need to calculate the flux

Using formula of flux

\phi=E\cdot A

Put the value into the formula

\phi=(150 i-200 j)\times(4.2\times10^{-2}\times4.0\times10^{-2})k

Here, The component of k is not given

So, the flux is

\phi=0

Hence, (A). The flux is -0.336 N.m²/C

(B). The flux is zero.

7 0
3 years ago
Read 2 more answers
Voidddddddddddddddddddddddd<br>​
bogdanovich [222]

Answer: understandable have a nice day

Explanation:

5 0
3 years ago
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