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3 years ago

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Ball A (7kg) and ball B (3kg) hit each other in a perfectly elastic collision. If ball A has an initial velocity of 12 m/s and b

all B has an initial velocity of -1 m/s. What are the final velocities of each ball?

1 answer:

dusya [7]3 years ago

4 0

Explanation:

Mass of ball A, $m_A=7\ kg$

Mass of ball B, $m_B=3\ kg$

Initial velocity of ball A, $u_A=12\ m/s$

Initial velocity of ball B, $u_B=-1\ m/s$

We need to find the final velocity of each ball. For a perfectly elastic collision, the coefficient of restitution is equal to 1. It is given by :

$e=\dfrac{v_B-v_A}{u_A-u_B}$

$v_A\ and\ v_B$ are final velocities of ball A and B

$1=\dfrac{v_B-v_A}{13}$

$v_B-v_A=13$ ...........(1)

Using the conservation of linear momentum as :

$m_Au_A+m_Bu_B=m_Av_A+m_Bu_B$

$7(12)+3(-1)=7v_A+3u_B$

$7v_A+3u_B=81$ ..............(2)

On solving equation (1) and (2) using calculator we get :

$v_A=4.2\ m/s$

$v_B=17.2\ m/s$

So, the final velocities of ball A and B are 4.2 m/s and 17.2 m/s. Hence, this is the required solution.

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The period of the wave is 4.35 ms. The sound waves are called longitudinal waves

Explanation:

The period of a wave is related to its frequency by the equation:

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For the bee in this problem, the frequency of the sound wave emitted by it is

$f=2.3 \cdot 10^2 Hz = 230 Hz$

Therefore, the period of the sound wave is

$T=\frac{1}{230}=4.35\cdot 10^{-3}s = 4.35 ms$

The sound wave is a type of wave called longitudinal wave. In longitudinal waves, the oscillation of the medium occurs in a direction parallel to the direction of motion of the wave: therefore in a sound wave, the particle of the medium (air, in this case) oscillate back and forth along the direction of propagation of the wave, forming alternating areas of higher density of particles (called compressions) and of lower density of particle (called rarefactions).

The other type of wave, instead, is called transverse wave. In a transverse wave, the oscillation of the wave occurs in a direction perpendicular to the direction of motion of the wave. An example of transverse waves are the electromagnetic waves, which consists of electric field and magnetic fields that vibrate in a plane perpendicular to the direction of motion of the wave itself.

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3 years ago

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

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Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

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Answer:

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Explanation:

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W =∫ F dx = ΔK

Let's replace

∫ (α x³ + β) dx = ΔK

α x⁴ / 4 + β x = ΔK

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

x (α x³ + β) = $K_{f}$ - K₀

$K_{f}$ = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

Let's calculate

$K_{f}$ = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

Kf = 2.7 10¹¹ - 1.1475 10¹¹

Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

W = x F₀ = $K_{f}$ –K₀

$K_{f}$ = K₀ + x F₀

We calculate

$K_{f}$ = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

$K_{f}$ = (2.7 -2.625) 10¹¹

$K_{f}$ = 7.5 10⁹ J

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