Pe cine trebuia să îi fac de la școală cu sfârșit de<br> groază

The cue ball is deflected so that it makes an angle of 30.0° with its original direction of travel.

Alexus [3.1K]

Answer:

(a) θ= 43.89°

(b) $v_{1} = 1.88\sqrt{3} \frac{m}{s}$

$v_{2} = 6.79 \frac{m}{s}$

Explanation:

Ball 1:

$u_{1} = 7.52\frac{m}{s}$

Ball 2:

$u_{2} = 0\frac{m}{s}$

As the collision is elastic, it means that kinetic energy and momentum are conserved. Following that, we apply the law of conservation of energy and momentum:

$\frac{1}{2} m_{1}u_{1}^{2} = \frac{1}{2} m_{1}v_{1}^{2} + \frac{1}{2} m_{2}v_{2}^{2}$

and

$m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2}$

Where u is the velocity before the collision, and v are the velocities after the collision. Both previous equations can be simplified as:

$u_{1}^{2} = v_{1}^{2} + v_{2}^{2}$

and

$u_{1} = v_{1} + v_{2}$

This because the two balls have the same mass. We know that the cue ball is deflected and makes an angle of 30°. From conservation in x-direction, we get::

$56.55 = v_{1} ^{2} + v_{2} ^{2}$

and

$u_{1} = v_{1}cos(30) + v_{2}cos(\theta)$

Solving we get:

$(\frac{2u_{1}-\sqrt{3}v_{1} }{2cos(\theta)})^{2} = 56.55-v_{1} ^{2}$

From conservation in y-direction, we get:

$0 = v_{1}sin(30) - v_{2}sin(\theta)$

From this, we solve this equation system and get the answers. Remember to add 30° to the angle obtained.

8 0

3 years ago

A trapeze artist swings in simple harmonic motion with a period of 3.8 s.

zavuch27 [327]

As we know that time period of simple pendulum is given as

T = 2π √L/g

3 0

3 years ago

Calculate the efficiency for each machine, then answer the question. Machine A has an input work of 1,000 J and an output work o

Usimov [2.4K]

Machine C. Machine A has an efficiency of 250/1000 = 0.25 = 25%, Machine B's efficiency is 350/500 = 0.7 = 70% and Machine C has an efficiency of 150/200 = 0.75 = 75%.

7 0

3 years ago

Read 2 more answers

Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 9 m into 32 cm. of water. Assu

baherus [9]

Answer:

823.46 kgm/s

Explanation:

At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.

So, mgh = 1/2mv²

From here, his velocity just as he reaches the surface of the water is

v = √2gh

h = 9 m and g = 9.8 m/s²

v = √(2 × 9 × 9.8) m/s

v = √176.4 m/s

v₁ = 13.28 m/s

So his velocity just as he reaches the surface of the water is 13.28 m/s.

Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.

So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.

His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,

J = (62 × 0 + 62 × 13.28) kgm/s = 0 + 823.46 kgm/s = 823.46 kgm/s

So the magnitude of the impulse J of the water on him is 823.46 kgm/s

6 0

3 years ago

A 2 kg object is a distance of 10,000,000 m away from the center of Earth, which has a mass of nearly 6×1024 kg. What is the app

Gelneren [198K]

Answer:

4 (N/kg) or B

Explanation:

An application of the equation for Newton’s law of universal gravitation can be used to determine the gravitational field strength at the 2 kg object’s location.

7 0

3 years ago

Pe cine trebuia să îi fac de la școală cu sfârșit de groază​

Pe cine trebuia să îi fac de la școală cu sfârșit de groază