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3 years ago

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Two balls of mass m1 and m2, with velocities v1 and v2 collide head on. Is there any way for both balls to have zero velocity af

ter the collision? If so, find the conditions under which this can occur.

1 answer:

nordsb [41]3 years ago

8 0

Answer:

Explanation:

As the final Kinetic energy is zero or less than initial kinetic energy, the collision must be inelastic.

In Inelastic collision both the bodies must stick together as final velocity is zero for both the bodies.

To conserve the momentum, momentum associated before the collision of first must be equal and opposite to the momentum associated with the second ball.

i.e.

$m_1v_1=m_2v_2$

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If a battery causes a wire to carry a current of 4 Amps how many coulombs of charge flow past any point in the wire in 3 seconds

BabaBlast [244]

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According to above question ~

Current (I) = 4 Amperes

Time (t) = 3 seconds

Charge (q) = ?

Let's find the charge (q) by using formula ~

$I = \dfrac{q}{t}$

$4 = \dfrac{q}{3}$

$q = 4 \times 3$

$q = 12 \: \: coulombs$

Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds

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3 years ago

Which part of a river would have animals with very muscular bodies and adaptations that let survive in turbulent water? Source z

lidiya [134]

The part of a river that would have animals with muscular bodies and adaptations that let survive in turbulent water is in the transition zone, the mid-transition zone to be precise. Water at the source zone possesses a lot of potential energy and as it flows from the upper reaches the potential energy is turned into kinetic energy when the course of the river begins to gradually level out and this translates into increase in velocity. By the time river water reaches the middle of the transition zone, most of the potential energy would have been turned into kinetic energy and thus water velocity would be quite high here. Animals living here would develop muscles because of constantly fighting against the strong current to avoid being swept downstream.

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3 years ago

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swat32

Answer:

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A crocodile cannot stick its tongue out

It is physically impossible for pigs to look up into the sky

Grapes light on fire in the microwave

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2 years ago

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Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates

IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

Sphere A : ma = 80 kg , ( 0 , 0 )

Sphere B : ma = 60 kg , ( 0.25 , 0 )

Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

Determine the angle (α) between vectors rac and rab using cosine rule:

$cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}$

Determine the angle (β) between vectors rbc and rab using cosine rule:

$cos ( \beta ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta ) = 0.6\\\\\beta = 53.13^{\circ \:}$

- Now determine the scalar gravitational forces due to sphere A and B on C:

Between sphere A and C:

Fac = G*ma*mc / rac^2

Fac = (6.674×10−11)*80*0.2 / 0.2^2

Fac = 2.67*10^-8 N

vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

Between sphere B and C:

Fbc = G*mb*mc / rbc^2

Fbc = (6.674×10−11)*60*0.2 / 0.15^2

Fbc = 3.56*10^-8 N

vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

Fc = vector Fac + vector Fbc

Fc = [ - 2.136 i - 1.602 j ]*10^-8 + [ 2.136 i - 2.848 j ]*10^-8

Fc = [ - 4.45 * 10^-8 j ] N

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3 years ago

The radioactive 60co isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequenc

Ivanshal [37]

1. Frequency: $3.23\cdot 10^{20} Hz$

The energy given is the energy per mole of particles:

$E=1.29\cdot 10^{11} J/mol$

1 mole contains a number of Avogadro of particles, $N_A$ , equal to

$N_A=6.022\cdot 10^{23}$ particles

So, by setting the following proportion, we can calculate the energy of a single photon:

$1.29 \cdot 10^{11} J/mol : 6.022 \cdot 10^{23} ph/mol = E_1 : 1 ph\\E_1 = \frac{(1.29\cdot 10^{11} J/mol)(1 ph)}{6.022\cdot 10^{23} ph/mol}=2.14\cdot 10^{-13} J$

This is the energy of a single photon; now we can calculate its frequency by using the formula:

$E_1 = hf$

where

$h=6.63\cdot 10^{-34} Js$ is the Planck's constant

f is the photon frequency

Solving for f, we find

$f=\frac{E_1}{h}=\frac{2.14\cdot 10^{-13} J}{6.63\cdot 10^{-34} Js}=3.23\cdot 10^{20} Hz$

2. Wavelength: $9.29\cdot 10^{-13} m$

The wavelength of the photon is given by the equation:

$\lambda=\frac{c}{f}$

where

$c=3\cdot 10^8 m/s$

is the speed of the photon (the speed of light). Substituting,

$\lambda=\frac{3 \cdot 10^8 m/s}{3.23\cdot 10^{20} Hz}=9.29\cdot 10^{-13} m$

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3 years ago