Ask question

Ask question

All categories

3 years ago

11

Object a and object b are both in motion when they collide with each other. They then continue in a new direction unaffected by

any other force this is a

1 answer:

Harrizon [31]3 years ago

8 0

This is an elastic collision

bcuz i think they move apart after the collision

sorry if im wrong

You might be interested in

Which of the following illustrates 2 resistors in a series circuit?

umka2103 [35]

Explanation :

An electrical circuit can be either connected in series or parallel.

Series combination: The current flowing in each resistor is the same and the equivalent resistance is given by :

$R{eq}=R_1+R_2+R_3+.................+R_n$

Parallel combination: The voltage for all components in the parallel combination is the same. The equivalent resistance is given by :

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}......................+\dfrac{1}{R_{n}}$

So, our of given options the correct one is (b) because the current flowing in both the resistors is the same.

5 0

4 years ago

Read 2 more answers

Un reloj de péndulo de largo L y período T, aumenta su largo en ΔL (ΔL << L). Demuestre que su período aumenta en: ΔT = π

Kruka [31]

Answer:

ΔT = $\pi \ \frac{\Delta L}{\sqrt{Lg} }$

Explanation:

In a simple harmonic motion, specifically in the simple pendulum, the angular velocity

w = $\sqrt{\frac{g}{L} }$

angular velocity and period are related

w = 2π / T

we substitute

2π / T = \sqrt{\frac{g}{L} }

T = $2\pi \ \sqrt{\frac{L}{g} }$

In this exercise indicate that for a long Lo the period is To, then and increase the long

L = L₀ + ΔL

we substitute

T = $2\pi \ \sqrt{\frac{L + \Delta L}{g} }$

T = $2\pi \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }$

in general the length increments are small ΔL/L «1, let's use a series expansion

$\sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...$

we keep the linear term, let's substitute

T = $2\pi \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L} )$

if we do

T = T₀ + ΔT

T₀ + ΔT = $2\pi \sqrt{\frac{\Delta L}{g} } + \pi \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}$

T₀ + ΔT = T₀ + $\pi \sqrt{\frac{1}{Lg} } \ \Delta L$

ΔT = $\pi \ \frac{\Delta L}{\sqrt{Lg} }$

4 0

3 years ago

calculate the percentage contraction of a rod moving with a velocity of 0.8c in a direction inclined at 60° to its own length

hram777 [196]

Answer:

Let lo be the length of the rod in the frame in which it is at rest and s' is the frame which is moving with a speed 0.8c in a direction making an angle 60° with x-axis. The components of lo along and perpendicular to the direction of motion are lo cos 60° and lo sin 60° respectively.

Now length of the rod along the direction of motion

= lo cos 60°_/1-(0.8) 2/c2

= lo/2×0.6

= 0.3 lo.

Length of the rod perpendicular to the direction of motion.

= lo sin 60°

=_/3/2 lo

Length of moving rod

l = [(0.3lo)2+{lo_/3/2} 2] 1/2

= 0.916 lo.

Percentage contraction

= lo-0.916lo/lo×100

= 8.4%.

Explanation:

<h2><u><em>Brainliest?</em></u></h2>

4 0

3 years ago

A spring scale hung from the ceiling stretches by 6.1 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl

Hunter-Best [27]

The ideal spring equation is

   Stretch = K times Force .

This says that the stretch is directly proportional to the force.
In simple English, that means that if you double the force, then
you double the stretch, and if you multiply the force by π or
any other number, you multiply the stretch by the same number.

So you can always write a proportion for a spring:

   Stretch₁ / Force₁ = Stretch₂ / Force₂ .

Part A:

In Part-A of this question, the force is increased to (2.5 / 2.0) = 1.25 times .

So the stretch is also increased to 1.25 times .

   (1.25) x (6.1 cm) = 7.625 cm .

6 0

4 years ago

An lc circuit oscillates at a frequency of 2000 Hz. What will the frequency be if the inductance is quadrupled?

Irina-Kira [14]

<h2>Answer:</h2>

An LC circuits if formed by an inductor and a capacitor. The charge on the capacitor and the current through the inductor both vary sinusoidally with time. Also, energy is transferred between magnetic energy in the inductor and electrical energy in the capacitor. But <em>what happens with the frequency if the inductance is quadrupled? </em>that is, if initially the inductance is $L$ and the frecuency $f=2000Hz$ if now $L_{New}=4L$ What will the frequency be? Well, we know that the frequency, inductance and capacitance are related as:

$f=\frac{1}{2\pi}\sqrt{\frac{1}{LC}}$

and this equals 2000Hz. If now L is quadrupled:

$f_{1}=\frac{1}{2\pi}\sqrt{\frac{1}{4LC}}=\frac{1}{2\pi}\sqrt{\frac{1}{4}}\sqrt{\frac{1}{LC}} \\ \\ \therefore f_{1}=(\frac{1}{2})\frac{1}{2\pi}\sqrt{\frac{1}{LC}} \\ \\ \\ Since \ f=\frac{1}{2\pi}\sqrt{\frac{1}{LC}} \ then: \\ \\ f_{1}=\frac{1}{2}f \therefore f_{1}=\frac{2000}{2} \\ \\ \therefore \boxed{f_{1}=1000Hz}$

<em>Finally, if L is quadrupled the frequency is half the original frequency and equals 1000Hz</em>

3 0

3 years ago