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Rus_ich [418]
3 years ago
14

A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra

ight line so that the two balls have the same momentum?
Answer in units of m/s.
Physics
1 answer:
lilavasa [31]3 years ago
6 0

Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg is the mass of the ping-pong ball

V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

Finally:

V_{2}=11369.46 m/s

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Gravity affects weight of an object

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3 0
3 years ago
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Can someone help me with these questions plz
Korolek [52]

Im not 100% on these but i can try

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2.

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8 0
3 years ago
Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 l
Natasha_Volkova [10]

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

          Wₐ = 1500 lb

          W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

             W = mg

             m = W / g

             mₐ = Wₐ / g

             m_b = W_b / g

             mₐ = 1500/32 = 46.875 slug

             m_b = 125/32 = 35,156 slug

Let's reduce to the english system

             v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

           p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

           p_f = (mₐ + m_b) v

the moment is preserved

           p₀ = p_f

           mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

           v = \frac{ m_a \ v_{oa} - m_b \ v_{ob}  }{ m_a +m_b}

we substitute the values

           v = \frac{ 46.875}{82.03} \ v_{oa} -  \frac{35.156}{82.03} \ 61.6

           v = 0.559 v₀ₐ - 26.40                  (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

              M = mₐ + m_b

              M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

              K₀ = ½ M v²

final point. When they stop

             K_f = 0

The work is

             W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

           N-W = 0

           N = W

the friction force has the expression

            fr = μ N

we substitute

            -μ W x = Kf - Ko

             

            -μ W x = 0 - ½ (W / g) v²

            v² = 2 μ g x  

            v = \sqrt{ 2 \ 0.750 \ 32 \ 17.5}Ra (2 0.750 32 17.5  

            v = 28.98 ft / s

3 0
2 years ago
An object moves along the x-axis according to the equation x = 3.00t2 – 2.00t + 3.00,
Ray Of Light [21]

Explanation:

x = 3.00t^{2} – 2.00t + 3.00,

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Average speed = \frac{Total distance}{ Total time}

Average speed = \frac{x (t=2) + x (t=3)}{3}

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Average speed = \frac{35}{3}

Average speed = 11.66 \frac{m}{s}

b) the instantaneous speed at t = 2.00 s and t = 3.00 s,

Instantaneous speed = \frac{dx}{dt}

Instantaneous speed(v) = 6t - 2\left \{ {{t=2} \atop {t=3}} \right.

Instantaneous speed,v(t=2 to t=3) = 18-2-12+2

Instantaneous speed, v = 6 \frac{m}{s}

c) the average acceleration between t = 2.00 s and t = 3.00 s

average acceleration = \frac{average velocity}{time}

average acceleration =  \frac{11.66}{3-2}

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d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s

instantaneous acceleration = \frac{dv}{dt}

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instantaneous acceleration = 6 \frac{m}{s^{2} }

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0 = 3.00t^{2} – 2.00t + 3.00

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7 0
3 years ago
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yulyashka [42]

To solve the problem it is necessary to apply the concepts related to thermal expansion of solids. Thermodynamically the expansion is given by

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7 0
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