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Grace [21]
3 years ago
13

A 6.0 kg bowling ball moving at 3.5m/s to the right makes a collision, head-on, with a stationary 0.70 kg bowling pin. If the ba

ll is moving 2.77 m/s to the right after the collision, what will be the velocity ( magnitude and direction) of the pin?
Physics
1 answer:
devlian [24]3 years ago
4 0

Answer:

The velocity of the pin will be 6.26 m/s in the right direction.

Explanation:

Let's use the momentum conservation equation.

p_{i}=p_{f}

Initially, we have:

p_{i}=m_{b}*v_{ib}

Where:

  • m(b) is the ball mass
  • v(ib) is the initial velocity of the ball

Now, the final momentum will be:

p_{f}=m_{b}*v_{fb}+m_{p}*v_{fp}

Where:

  • m(p) is the pin mass
  • v(fb) is the final velocity of the ball
  • v(fp) is the final velocity of the pin

Then, using the equation of the conservation we have:

m_{b}*v_{ib}=m_{b}*v_{fb}+m_{p}*v_{fp}

6*3.5=6*2.77+0.7*v_{fp}

6*3.5=6*2.77+0.7*v_{fp}

v_{fp}=6.26 m/s

Therefore the velocity of the pin will be 6.26 m/s in the right direction.

I hope it helps you!

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When an object is falling and reaches a constant velocity, the net force on the object is ____ and the weight of the object is e
Maslowich

When an object is falling and reaches a constant velocity, the net force on the object is <em>zero</em> (it's not accelerating), and the weight of the object is equal to <em>the force of air resistance against the object</em>.  (choice-D)

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An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?
Serjik [45]

The image of the object is 8cm to the left of the lens (D)

<h3></h3>

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

<h3>How to calculate the image of an object</h3>

From the formula

1÷v = 1÷f - 1÷u

<h3>Where </h3>

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

Therefore, the image of the object is 8cm to the left of the lens (D)

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6 0
2 years ago
The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive p
aliina [53]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

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and, we know that, F = ma

So,

m_{p}a = q_{p}E

a = \frac{q_{p}.E }{m_{p} }      Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2at^{2}

Here, u = 0.

S = 1/2at^{2}

Put equation 1 into the above equation:

S = 1/2 x (\frac{q_{p}.E }{m_{p} }  )t^{2}      Equation 2

So,  

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x (\frac{q_{e}.E }{m_{e} }  )t^{2}    Equation 3

We know that the charge of electron is equal to the charge of proton so,

q_{p} = q_{e} = q

By dividing the equation 2 by equation 3, we get:

\frac{S}{D-S} = \frac{m_{e} }{m_{p} }

Solve the above equation for S,

Sm_{p} = m_{e}D - m_{e}S

So,

S = \frac{m_{e}.D }{(m_{e} + m_{p})  }

Plugging in the values,

As we know the mass of electron is 9.1 x 10^{-31} and the mass of proton is 1.67 x 10^{-27}

S = \frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27}  }

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = \frac{m_{Cl}.D }{(m_{Cl} + m_{Na})  }

As we know the mass of chlorine is 35.5 and of sodium is 23

S = \frac{35.5 . 4.22}{(35.5 + 23)}

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Where does pressure come from?
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Answer:

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Explanation:

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3 years ago
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A standing wave pattern is created on a string with mass density μ = 3.4 × 10-4 kg/m. A wave generator with frequency f = 61 Hz
uranmaximum [27]

Answer:

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Explanation:

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3) let's use the equation

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     T = v² μ

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4) the rope tension is proportional to the hanging weight

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    m = W / g

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5) n = 2

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7) T = v² μ

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8) m = W / g

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9) n = 1

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    m = 262.4 10⁻³ kg

5 0
3 years ago
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