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juin [17]
3 years ago
7

A 102 kg football player runs at a speed of 8 m/s to sack the quarterback. What is

Physics
1 answer:
leva [86]3 years ago
3 0

The mass of the quarterback is 61.2 kg.

Explanation:

mass of the football player = m1 = 102 kg

mass of the quarterback = m2 = ?

velocity of the football player = v1 = 8 m/s

According to the law of conservation of momentum:

The total momentum of a system before and after the collision remains constant. Assuming the situation as an isolated system which is not affected by any external factors, we have:

m₁v₁ + m₂v₂ = (m₁+m₂)V

Here, we need to find m₂.

We assume that the quarterback is standing still when he is attacked by the football player so v₂ = 0 m/s

After the collision both of them fall to the ground with a velocity of 5 m/s so V = 5 m/s

102(8) + m2(0) = (102 + m2)(5)\\816 + 0 = (102 + m2)(5)\\816/5 = 102 + m2\\163.2 - 102 = m2\\m2 = 61.2 kg

Keywords: momentum, velocity, law of conservation of momentum

Learn more about Law of Conservation of Momentum from brainly.com/question/7538238

#learnwithBrainly

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True or false? If two components are connected in series, the current through one component will
Molodets [167]

the answer is ( True ) .

the current is the same in series circuits .

8 0
3 years ago
The entropy of an isolated system must be conserved, so it never changes.a. Trueb. Fasle
Snowcat [4.5K]

Answer:

B: False

Explanation:

The second law of thermodynamics states that: the entropy of an isolated system will never decrease because isolated systems always tend to evolve towards thermodynamic equilibrium which is a state with maximum entropy.

Thus, it means that the entropy change will always be positive.

Therefore, the given statement in the question is false.

6 0
3 years ago
PLEASE HELPPPPP
Ganezh [65]

The only graph that accurately depict the given motion is graph D.

The given parameters;

  • initial position of the man = 0
  • direction of the man's first displacement = backward
  • time of first motion, t₁ = 6 seconds
  • velocity of this first displacement = v₁
  • time without any motion (<em>zero movement</em>) = 6 seconds
  • direction of the second displacement = forward
  • velocity of second displacement = 2v₁

Let the acceleration of the first displacement = a

Acceleration of the second displacement = 2a

From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.

The only options with initial motion towards the negative direction are;

  • <em>graph B</em>, and
  • <em>graph D</em>.

The difference between graph B and D;

  • in graph B there is a uniform motion for 6 seconds
  • in graph D there is no motion for 6 seconds (<em>this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds</em>).

Thus, the only graph that accurately depict the given motion is graph D.

Learn more here: brainly.com/question/21095906

5 0
2 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
3. The expression 0.62 x10^3 is equivalent to...
Korolek [52]

\\ \sf\longmapsto 0.62\times 10^3

\\ \sf\longmapsto 62\times 10^{-2}\times 10^3

\\ \sf\longmapsto 62\times 10^{-2+3}

\\ \sf\longmapsto 62\times 10^1

\\ \sf\longmapsto 62\times 10

\\ \sf\longmapsto 620

5 0
3 years ago
Read 2 more answers
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