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3 years ago

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Two pool balls, each moving at 2 m/s, roll toward each other and collide. Suppose after bouncing apart, each moves at 2 m/s. Thi

s collision violates conservation of
A) Momentum.
B) Energy.
C) Both momentum and energy.
D) None of the above choices.

1 answer:

stira [4]3 years ago

4 0

Answer:D

Explanation:according to the law of conservation of energy/momentum, when two bodies collides, their total momentum and energy before and after collision are equal. Given that the two bodies move with the same velocities after collision, means that the law has not been violated since momentum = mass x velocity (where mass is constant)

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A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 37.6 vibrations

tangare [24]

Answer:

$\lambda = 25.79\ cm$

Explanation:

given,

Wave vibrates = 37.6

time = 27.9 s

maximum distance travel = 450 cm

time = 11.3 s

wavelength = ?

frequency of wave

$f=\dfrac{37.6}{27.9}$

f = 1.35 Hz

Speed of wave

$v = \dfrac{450}{11.3}$

v = 39.82 cm/s

wavelength of wave

v = fλ

$\lambda =\dfrac{v}{f}$

$\lambda =\dfrac{34.82}{1.35}$

$\lambda = 25.79\ cm$

Hence, wavelength of the wave is equal to 25.79 cm.

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3 years ago

PLEASE HELP Scientists have observed that the average sunspots have increased in the last 50 years. Which of these statements be

Serga [27]

Answer:

option c is correct, earth temp increase

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3 years ago

A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0

3 years ago

You are traveling 70 mph (31.3 m/s) and slam on the brakes to avoid hitting another car. How far do you travel if it takes you 8

Arlecino [84]

You traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.

<u>Given the following data:</u>

Initial velocity, U = 31.3 m/s

Time, t = 8.5 seconds.

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find the distance traveled, we would use the second equation of motion:

Mathematically, the second equation of motion is given by the formula;

$S = ut + \frac{1}{2} at^2$

Where:

S is the distance travelled.

u is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting the parameters into the formula, we have;

$S = 31.3(8.5) + \frac{x}{y} (9.8)(8.5^2)\\\\S = 266.05 + 4.9(72.25)\\\\S = 266.05 + 354.025$

<em>Distance, S</em><em> = </em><em>620.075 meters.</em>

Therefore, you traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.

Read more: brainly.com/question/8898885

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2 years ago

A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po

aleksandrvk [35]

Answer:

Explanation:

Let the equilibrium position of third charge be x distance from q₁.

Force on third charge due to q₁

= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²

Force on third charge due to q₂

= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

Both the force will act in opposite direction and for balancing , they should be equal.

9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

5 / x² = 2 / ( .4 - x )²

Taking square root on both sides

2.236 / x = 1.414 / .4 - x

2.236 ( .4 - x ) = 1.414 x

.8944 - 2.236 x = 1.414 x

.8944 = 3.65 x

x = .245 m

24.5 cm

So the third charge should be at a distance of 24.5 cm from q₁ .

4 0

2 years ago