6. Which of the following statements is true for displacement ?

a moving billiard ball collides with an identical stationary billiard ball in an elastic collision. after the collision, the sec

MArishka [77]

A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.

<h3>Why does the first ball comes to rest after collision ?</h3>

Let m be the mass of the two identical balls.

u1 = velocity before the collision of ball 1

u2 = 0 = velocity of second ball that is at rest

v1 and v2 are the velocities of the balls after the collision.

From the conservation of momentum,

∴ mu1 + mu2 = mv1 + mv2

∴ mu1 = mv1 + mv2

∴ u1 = v1 + v2

In an elastic collision, the kinetic energy of the system before and after collision remains same.

$\frac{1}{2} mu_1^2+0=\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2$

∴ $\frac{1}{2} m(v_1+v_2 )^2=\frac{1}{2} mv_1^2+\frac{1}{2}mv_2^2$

∴ $\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2+mv_1 v_2=\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2$

∴ $mv$ ₁ $v$ ₂ = 0

It is impossible for the mass to be zero.
Because the second ball moves, velocity v2 cannot be zero.
As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.

<h3>What is collision ?</h3>

An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

Can learn more about elastic collision from brainly.com/question/12644900

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3 0

1 year ago

What is the volume V of a sample of 4.00 mol of copper? The atomic mass of copper (Cu) is 63.5 g/mol, and the density of copper

rusak2 [61]

Answer : The volume of a sample of 4.00 mol of copper is $28.5cm^3$

Explanation :

First we have to calculate the mass of copper.

$\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}$

$\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g$

Now we have to calculate the volume of copper.

Formula used :

$Density=\frac{Mass}{Volume}$

Now put all the given values in this formula, we get:

$8.92\times 10^3kg/m^3=\frac{254g}{Volume}$

$Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3$

Conversion used :

$1kg/m^3=1g/L\\\\1L=10^3cm^3$

Therefore, the volume of a sample of 4.00 mol of copper is $28.5cm^3$

7 0

3 years ago

Read 2 more answers

A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th

Andre45 [30]

The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

A uniform motion along the horizontal (x) direction
A uniformly accelerated motion along the vertical (y) direction

We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}at^2$

where:

s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)

$u_y$ is the initial vertical velocity of the ball, which is given by

$u_y = u sin \theta$

where

u = 23.5 m/s is the initial velocity

$\theta=33.5^{\circ}$ is the angle of projection

Substituting,

$u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity, downward

Substituting everything into the equation we get:

$-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0$

Solving the equation for t, we find the time of flight of the ball:

t = -0.94 s

t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

$v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s$

Therefore, the horizontal distance covered in a time t is

$d=v_x t$

And substituting t = 3.59 s, we find

$d=(19.6)(3.59)=70.4 m$

So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

brainly.com/question/8751410

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4 0

3 years ago

You are given a vector in the xy plane that has a magnitude of 84.0 units and a y-component of -67.0 units.

melomori [17]

Answer:

Explanation:

a)Magnitude = $\sqrt{(x1-y2)^{2} + (x1-x2)^{2} }$

84= $\sqrt{(0- (-67))^{2} + (x-0)^{2} }$

x= +50.67 or -50.67 units

b) We are given that the resultant is entirely in the -ve x direction which means that the y-component of the resultant is 0; It means that the y-component of the next vector = -ve of the y component of the initial vector i.e 67.

To make the magnitude 80 units in the negative x direction where the y component is 0, the x component must be -130.67(-50.67 - 80) as the x component is + 50.67units.

Magnitude = $\sqrt{(0- (67))^{2} + (-130.67)^{2} }$ = 146.85 units

c) The direction vector = 67/146.85 i - 130.67/146.85 j where i corresponds to the vector in y direction and j corresponds to the vector in x direction. Or this vector is at an angle of 180 - $Tan^{-1}(67/130.67)degrees$ i.e 152.85 degrees from the +ve x-axis.

5 0

3 years ago

5

mina [271]

Answer:

B West

Explanation:

5 0

3 years ago