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RUDIKE [14]
3 years ago
15

According to the conservation of momentum, if a small football player collides with a much larger football player, which of the

following statements will be true? a The larger football player will move, but because of his larger mass, his velocity will be much slower than the smaller player b The smaller football player will be stopped completely, his mass being not enough to move the larger one c The larger football player will have to start moving towards the smaller one as it collides if there is to be any effect d The smaller football player will experience a larger force than the larger football player
Physics
1 answer:
Virty [35]3 years ago
8 0

Answer:

a The larger football player will move, but because of his larger mass, his velocity will be much slower than the smaller player

Explanation:

When the collision takes two equal and opposite forces are created at the point of collision . The force created on the bigger mass will force it to accelerate and the force on moving smaller mass will force it to slow down .

Because of bigger mass , acceleration on bigger mass will be less and hence its velocity will be less .

Hence option a is the right answer.

You might be interested in
The Special Olympics raises money through "plane pull" events in which teams of 25 people compete to see who can pull a 74,000 k
Murrr4er [49]

Answer:

28716.4740661 N

1.2131147541 m/s

51.2474965841%

Explanation:

m = Mass of plane = 74000 kg

s = Displacement = 3.7 m

f = Frictional force = 14000 N

t = Time taken = 6.1 s

u = Initial velocity = 0

v = Final velocity

s=ut+\frac{1}{2}at^2\\\Rightarrow 3.7=0\times 6.1+\frac{1}{2}\times a\times 6.1^2\\\Rightarrow a=\frac{3.7\times 2}{6.1^2}\\\Rightarrow a=0.198871271164\ m/s^2

Force is given by

F=ma+f\\\Rightarrow F=74000\times 0.198871271164+14000\\\Rightarrow F=28716.4740661\ N

The force with which the team pulls the plane is 28716.4740661 N

v=u+at\\\Rightarrow v=0+0.198871271164\times 6.1\\\Rightarrow v=1.2131147541\ m/s

The speed of the plane is 1.2131147541 m/s

Kinetic energy is given by

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}\times 74000\times 1.2131147541^2\\\Rightarrow K=54450.9540448\ J

Work done is given by

W=Fs\\\Rightarrow W=28716.4740661\times 3.7\\\Rightarrow W=106250.954045\ J

The fraction is given by

\dfrac{54450.9540448}{106250.954045}=0.512474965841

The teams 51.2474965841% of the work goes to kinetic energy of the plane.

3 0
3 years ago
A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that incre
Evgen [1.6K]

Complete Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of 1.20  \ \Omega

Answer:

The current is   I =  0.0007 41 \ A

Explanation:

From the question we are told that

   The  area is  A = 8.00 \ cm^2  = 8.0 *10^{-4} \  m^2

   The initial magnetic field at t_o = 0 \ seconds  is B_i = 0.500 \ T

   The magnetic field at t_1 = 0.99 \ seconds is  B_f  = 1.60 \ T

     The resistance is  R = 1.20  \ \Omega

Generally the induced emf is mathematically represented as

      \epsilon  =  A * \frac{B_f - B_i }{ t_f - t_o }

=>   \epsilon  =  8.0 *10^{-4} * \frac{1.60  - 0.500 }{ 0.99- 0  }

=>   \epsilon  = 0.000889 \ V

Generally the current induced is mathematically represented as

     I = \frac{\epsilon}{R }

=>  I = \frac{0.000889}{ 1.20 }  

=>  I =  0.0007 41 \ A  

6 0
2 years ago
What is the work of a 520n student walking a distance of 3 m
Wewaii [24]

Answer:

W=1560 Joules

Explanation:

Use W=fd

so, W=520N(3m)= 1560J

8 0
3 years ago
If 0.035pC of charge is transferred via the movement of Al3+ ions, how's many of these must be transferred in total? Please add
mr Goodwill [35]

Each Al^+^3 ion contains three extra protons. Hence, the extra charge on each  Al^+^3 = 3 \times 1.6 \times 10^-^1^9 C

Total charge = 0.035 pC

Total charge (Q) = 0.035 \times 10^-^1^2 C

Let the number of Al^+^3 ions be n.

According to question:

n \times 3 \times 1.6 \times 10^-^1^9 =0.035 \times 10^-^1^2

n = \frac{0.035 \times 10^-^1^2}{3 \times 1.6 \times 10^-^1^9}

n = 7.29167 \times 10^4

n = 72917

Hence, the total number of ions needed to be transferred is 72917

3 0
3 years ago
A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse
kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

e)

   p_player = (110 kg)(8 m/s) = 880 kg m/s

   p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

a)

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

b)

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

c)

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

d)

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball =  128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

4 0
2 years ago
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