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inysia [295]
3 years ago
7

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

e goalie and the puck have an elastic collision, and the puck is reflected back in the direction from which it came. What would the final velocities of the goalie and the puck be in this case? Assume that the collision is completely elastic.v goalie = _____ m / sv puck = ______ m / s
Physics
1 answer:
NISA [10]3 years ago
4 0

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

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No force is required to lift that balloon. In fact, force is required to hold it down, and if you let go, it's up, up, and away.

Since the balloon's density is less than the density of the air around it, it's lighter than the air it displaces, there is a net upward buoyant force acting on it, and it floats up !

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4 years ago
Professor Sanchez believes that children acquire language through reinforcement from adults, especially their parents. He believ
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The answer is “behaviorist”
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3 years ago
A student throws a baseball upwards at an angle of 60 degrees to the horizontal. The initial horizontal and vertical components
frez [133]

Given data

ball throws upwards at an angle 60°

Horizontal component (Vh) = 12.5 m/s,

Vertical component (Vv) = 21.7 m/s ,

The magnitude of throw/resultant velocity (V) = ?

The resultant velocity /the velocity with which ball is throws is determined by the following equation

                                   V = √[(Vh)² + (Vv)²]

                                       = √[(12.5)² + (21.7)²]

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4 0
3 years ago
The current theory of the structure of the
irina1246 [14]

Answer:

pt 1: m=1.66698*10^{21} kg

Pt 2: KE=1212.23531 J

Explanation:

Information Given: (p = density)

l = 5200km  d = 35km p = 2700kg/m^{2}

Part 1: Mass

  • Find volume
  1. V=(l)^2(d)
  2. V=(4.2*10^6)^2(35*10^3)
  3. V=61.74*10^{16}
  • Find Mass
  1. m=Vp
  2. m=(61.74*10^{16})(2700)
  3. m=1.66698*10^{21}

Part 2: Kinetic Energy

  1. v=\frac{3.8cm}{yr}*\frac{m}{100cm}*\frac{yr}{365d}*\frac{d}{24hr}*\frac{hr}{3600s}
  2. v=1.20497*10^{-9}

KE=\frac{1}{2}mv^2

KE=\frac{1}{2} (1.66698*10^{21})(1.20497*10^{-9})^2

KE=1212.23531 J

Part 3: Jogger Speed

set up, because I don't have the mass :(

Information given:

KE_{jogger}

  1. KE=\frac{1}{2}mv^2
  2. v_{jogger} =\sqrt{\frac{2KE}{m_{jogger} } }
  • Input the values

Hope it helps :)

6 0
3 years ago
Calculate the number of cells in a hummingbird assuming the mass of an average cell is ten times the mass of a bacterium
Mazyrski [523]

Answer:

4\times 10^{12}

Explanation:

Given that, the mass of an average cell is 10 times the mass of a bacterium.

And we know that the mass of the bacterium is,

m_{B} =10^{-15}kg

Now, the mass of an average cell can be calculate by the mass of bacterium,

m_{c}=10\times m_{B}\\m_{c}=10\times 10^{-15}kg\\m_{c}=10^{-14}kg

And the average mass of the humming bird is,

m_{H} =4\times 10^{-2}kg

And also we know that,

The no cell is defined as the ratio of mass of an object to the mass of the average cell.

Mathematically,

N=\frac{m_{H}}{m_{c}}

Therefore,

N=\frac{4\times 10^{-2}kg}{10^{-14}kg}\\ N=4\times 10^{12}

Therefore, the number of cell in the humming bird is 4\times 10^{12}

4 0
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