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1 year ago

Two hockey pucks with mass 0.1 kg slide across the ice and collide. Before

Physics

1 answer:

RideAnS [48]1 year ago

7 0

The velocity of pluck 1 is 12 m/s west.

<h3>What is the conservation of momentum?</h3>

The principle of the conservation of the linear momentum states that momentum before collision is equal to momentum after collision.

Now given that;

m1u1 + m2u2 = m1v1 + m2v2

(0.1 * 15) - (0.1 * 12) = 0.1* v + (0.1 * 15)

1.5 - 1.2 = 0.1v + 1.5

0.3 - 1.5 = 0.1v

v = -1.2/0.1

v = - 12 m/s

Hence, the velocity of pluck 1 is 12 m/s west.

Learn more about linear momentum:brainly.com/question/27988315

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8. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott

nasty-shy [4]

<u>Answer:</u> The ball is travelling with a speed of 5.5 m/s after hitting the <u>bottle.</u>

<u>Explanation:</u>

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

$m_1,u_1\text{ and }v_1$ are the mass, initial velocity and final velocity of ball.

$m_2,u_2\text{ and }v_2$ are the mass, initial velocity and final velocity of bottle.

We are given:

$m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s$

Putting values in above equation, we get:

$(0.4\times 18)+(0.2\times 0)=(0.4\times v_1)+(0.2\times 25)\\\\v_1=5.5m/s$

Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

5 0

3 years ago

The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would

natima [27]

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a) $v_{es} =v_{rms}$ = 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

$v_{rms} = \sqrt{ \frac{3RT}{M}}$

So, $v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}$

multiply each side by M_{o2}, so we have

$v_{rms,o2}^2 *M_{o2} =3RT_{o2}$

solving for temperature T_{o2}

$T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}$

In the question given, $v_{rms} =v_{es}$

$T_{o2} = \frac{(11.1*10^3)^2 *32.0*10^{-3}}{3*8.31}$

T_{o2} =1.58*10^5 K

7 0

3 years ago

Consider a thin, spherical shell of radius 14.5 cm with a total charge of 29.5 µC distributed uniformly on its surface. (a) Find

ra1l [238]

Answer:

10.0 zero, by Gauss' Law the symmetrical distribution will produce no internal electric fields

21.5 E = k Q / R^2 behaves as if all charge were at center

E = 9 E9 * 29.5 E-6 / .215^2 = 5.74E6 N/C

8 0

3 years ago

A piece of clay sits 0.10 m from the center of a potter’s wheel. If the potter spins the wheel at an angular speed of 15.5 rad/s

finlep [7]

Answer:

$a=24.025\ m/s^2$

Explanation:

Given that

Distance from the center ,r= 0.1 m

The angular speed ,ω = 15.5 rad/s

We know that centripetal acceleration is given as

a=ω² r

a=Acceleration

r=Radius

ω=angular speed

a=ω² r

Now by putting the values in the above equation we get

$a=15.5^2\times 0.1\ m/s^2$

$a=24.025\ m/s^2$

Therefore the acceleration of the clay will be $a=24.025\ m/s^2$ .

4 0

3 years ago

Which of the following equations defines the law of conservation of energy?

loris [4]

NONE of those choices defines the law of conservation of energy,
and none even describes it. It's as if these choices belong to
a whole different question.

In fact, none of the choices is even a true statement by itself !

A statement that defines the law of conservation of energy
would have to say something like this:

The total amount of energy in any closed system is constant.
The total amount of energy after any event or process is the
same as the total amount was before it.

7 0

3 years ago