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3 years ago

11

A 0.010 kg clay ball thrown right at 16 m/s hits a marble moving left at 8.0 m/s, and sticks to it. Both move left at 3.2 m/s af

ter the collision. What is the mass of the marble?

2 answers:

4vir4ik [10]3 years ago

7 0

Answer:

0.040

Explanation:

Anettt [7]3 years ago

6 0

.……… it will be 7.5m/s

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A 500 lines per mm diffraction grating is illuminated by light of wavelength 580 nm . what is the maximum diffraction order seen

vampirchik [111]

The maximum diffraction order seen is 3.

<h3>What is the maximum diffraction order seen?</h3>

We know that the maximum angle of diffraction Q_m of the furthest bright fringe < Q = 90 degrees.

Here we need to compute the nth bright fringe for which is approximated to 90 degrees.

The angle of nth bright fringe is given by;

sin(Q_m) = n(λ)N

Approximating Q_m ≈ 90 degrees.

sin (90) = nλN

n = sin (90) / (λN)

n = 1 / ((580 x 10⁻⁶)500)

n = 3.5 orders

Since, we knew that Q_m < 90 degrees, we will choose n = 3 as the maximum number of orders.

Thus, the maximum diffraction order seen is 3.

Learn more about maximum diffraction here: brainly.com/question/14703089

#SPJ4

6 0

2 years ago

A cue ball of mass m1 = 0.325 kg is shot at another billiard ball, with mass m2 = 0.59 kg, which is at rest. The cue ball has an

Roman55 [17]

Answer:

$v_{2f} = \frac{2vm_1}{m_2 + m_1}$

Explanation:

If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls

Before the collision

$P_i = m_1v$

After the collision

$P_f = m_1v_{1f} + m_2v_{2f}$

So using the law of momentum conservation

$P_i = P_f$

$m_1v = m_1v_{1f} + m_2v_{2f}$

We can solve for the speed of ball 1 post collision in terms of others:

$v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

Their kinetic energy is also conserved before and after collision

$m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2$

$m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2$

From here we can plug in $v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

$m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2$

$m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2$

$m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2$

$v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0$

$v_{2f}(1 + \frac{m_2}{m_1}) = 2v$

$v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}$

8 0

3 years ago

Read 2 more answers

A person on a cruise ship is doing laps on the promenade deck. On one portion of the track the person is moving north with a spe

Aleksandr-060686 [28]

T<u>he direction of motion</u> of the person relative to the water is <u>16.7° north of east.</u>

Why?

We can solve the problem by applying the Pitagorean Theorem, where the first speed (to the north) and the second speed (to the east) corresponds to two legs of the right triangle formed with them. (north and east directions are perpendicular each other)

We can calculate the angle that give the direction using the following formula:

$Tan(\alpha)=\frac{NorthSpeed}{EastSpeed}\\\\Tan(\alpha)^{-1}=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}\\\\\alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}$

Now, substituting the given information we have:

$alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}$

$\alpha =Tan(\frac{3.6\frac{m}{s} }{12\frac{m}{s} })^{-1}\\\\\alpha =Tan(0.3)^{-1}=16.69\°(North-East)=16.7\°(North-East)$

Hence, we have that <u>the direction of motion</u> of the person relative to the water is 16.7° north of east.

Have a nice day!

7 0

3 years ago

6. A ball rolled 125 m from the top of a hill to the bottom of a hill. How long

leva [86]

Time = distance / speed
T = 125/ 5
T = 25 meters per second

7 0

3 years ago

What nuclear weapon is the biggest and who currently owns this bomb

erma4kov [3.2K]

The RDS-220 <span>hydrogen bomb, soviet </span>

5 0

3 years ago