Ask question

Ask question

All categories

2 years ago

11

A hockey puck with a mass of 0.17 kg is traveling to the right along the ice at 15 m/s. it strikes a second hockey puck with a m

ass 0.11 kg. the first hockey puck comes to rest after the collision. What is the velocity of the second hockey puck after the collision? ( Round your answer to the nearest integer.)

1 answer:

Igoryamba2 years ago

3 0

Answer:

$v_2 = 23.182m/s$

Explanation:

Given

$m_1 = 0.17kg$

$m_2 = 0.11kg$

$u_1 = 15m/s$ -- Initial Velocity of the first

$u_2 = 0m/s$ -- Initial Velocity of the second

$v_1 = 0m/s$ -- Final Velocity of the first

Required

Determine the final velocity of the second hockey puck ( $v_2$ )

This question illustrates law of conservation of momentum and it'll be solved using the following formula:

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ ---- law of conservation of momentum

Substitute in the right values

$0.17 * 15 + 0.11*0 = 0.17 * 0 + 0.11*v_2$

$0.17 * 15 + 0 = 0 + 0.11*v_2$

$2.55 = 0.11*v_2$

Solve for v2

$v_2 = 2.55/0.11$

$v_2 = 23.182m/s$

<em>Hence, the final velocity of the second hockey puck is 23.182m/s</em>

You might be interested in

Two of the types of infrared light, ir-c and ir-a, are both components of sunlight. their wavelengths range from 3000 to 1,000,0

NikAS [45]

The energy of a light wave is calculated using the formula
E = hc/λ
h is the Planck's constant
c is the speed of light
λ is the wavelength
For the ir-c, the range is
<span>6.63 x 10^-34 (3x10^8) / 3000 = 6.63 x 10 ^-29 J
</span>6.63 x 10^-34 (3x10^8) / 1000000 = 1.99 x 10^-31 J

For the ir-a, the range is
6.63 x 10^-34 (3x10^8) / 700 = 2.84 x 10^-28 J
6.63 x 10^-34 (3x10^8) / 1400 = 1.42 x 10^-28 J

6 0

2 years ago

Read 2 more answers

7. The S.l. unit for force is<br> A. kg<br> C. m/s<br> B. m/s2<br> D. N

netineya [11]

D.N(Newton) this is the S.I unit for force

5 0

2 years ago

A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second

Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

$v_x = 15 m/s$

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

$d= v_x t =(15 m/s)(5 s)=75 m$

5 0

3 years ago

A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp

NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

$f=12GHz=12\cdot 10^9 Hz$ is the frequency

$c=3.0\cdot 10^8 m/s$ is the speed of light

So the wavelength of the beam is

$\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m$

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

$y=\frac{m\lambda D}{a}$

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

$y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m$

and so, the diameter is

$d=2y = 750 m$

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

$r=\frac{750 m}{2}=375 m$

So the area is

$A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2$

And since the power is

$P=100 kW = 1\cdot 10^5 W$

The average intensity is

$I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2$

4 0

2 years ago

Plz help difference between closed system and open system?

frozen [14]

A closed system is a system that is completely isolated from its environment. The physical universe, as we currently understand it, appears to be a closed system. An open system is a system that has flows of information, energy, and/or matter between the system and its environment, and which adapts to the exchange.

D makes the most sense, but you just have to put two and two together and go with your gut feeling, first cross out the answers that don't make sense (A didnt make sense) and go from there! I hope the little bit above me helped you answer or decide :) Good luck!

4 0

3 years ago