What is the velocity of the dart just before the collision?

Answers

Answer 1

Answer:

Answer: 1.38 m/s to the right.

Explanation: We can solve this problem using the principle of conservation of momentum. The momentum of an object is defined as the product of its mass and velocity, and the total momentum of a closed system is conserved, meaning that the total momentum before a collision is equal to the total momentum after the collision.Let the initial velocity of the dart be v, and let the final velocity of the dart and the block be vf. The momentum of the dart before the collision is given by p = mv, where m is the mass of the dart. The momentum of the dart and block after the collision is given by (m + M)vf, where M is the mass of the block.Using the principle of conservation of momentum, we have:p = (m + M)vfSubstituting the given values, we get:0.012 kg v = (0.012 kg + 0.2 kg) 0.78 m/sSimplifying, we get:v = (0.212 kg) (0.78 m/s) / 0.012 kgv ≈ 1.38 m/sTherefore, the velocity of the dart just before it hits the block is approximately 1.38 m/s to the right.

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HELP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Some people consider the earth to be _____ because of its dynamic processes or cycles.

stagnant
alive
changing

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Alive
The many processes and cycles on Earth make people think of it as being alive.

Total discharge head includes?

Answers

The term “total” indicates that it includes the static pressure head (h), the velocity head (hv), and the elevation head (Z).

Help with number 32
Apparently the answer is 4 but I have no idea why

Answers

All you have to remember is that when the source is APPROACHING the observer,
the frequency seems HIGHER than what's actually leaving the source.

The siren is wailing at 1,000 Hz.

(1). 30.0 Hz . . . that's way lower

(2). 9.19 x 10² Hz = 919 Hz . . . that's lower

(3). 1.00 x 10³ Hz = 1,000 Hz . . . that's exactly the same

(4). 1.10 x 10³ Hz = 1,100 Hz . . . that's the only choice that's higher than 1,000 Hz.

A ball is thrown horizontally from the top ofa building 140 m high. The ball strikes the
ground 54 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.

Answers

Answer:

v = d / t = 54 / 5.34 = 10.11 m/s (rounded to two decimal places)

Explanation:

A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point at coordinates (2.50 m, 2.50 m) with a velocity of -5.00i hat m/s and an acceleration of -10.0j m/s2. What are the coordinates of the center of the circular path?

Answers

Answer:

(2.5,0)

Explanation:

The particle can be described by the following equations:

$x=Rsin(-\omega t)+2.5\ny=Rcos(-\omega t)\n(dx)/(dt)=-\omega Rcos(-\omega t)\n(dy)/(dt)=\omega Rsin(-\omega t)\n(d^2x)/(dt^2)=-\omega^2Rsin(-\omega t)\n(d^2y)/(dt^2)=-\omega^2Rcos(-\omega t)$

For R = 2.5, ω = 2 and t = 0:

$x=2.5\ny=2.5\n\n(dx)/(dt)=-5\n (d^2y)/(dt^2)=-10$

The center of the circle would be at point (2.5,0)

A 2.75 kg sample of a substance occupies a volume of 250 cm³ . Find the density in g/cm³ and kg/m³