On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s meets the runner in a head-on collision. If the two players stick together, a) what is their velocity immediately after collision? b) What is the kinetic energy of the system just before the collision and a moment after the collision?

Answers

Answer 1
Answer:

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

                                               = 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2                

                                                = 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J


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Answers

I need a good phisics teacher for one day online please123

Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 2.10 kg) moves is frictionless. The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s. a)What is the tension force that the rope exerts on block B? b)What is the tension force that the rope exerts on block A? c)What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?

Answers

The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is \rm 0.430 \; kg\;m^2.

Given :

  • Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope.
  • The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle.
  • The horizontal surface on which block A (mass 2.10 kg) moves is frictionless.
  • The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s.

a) First, determine the acceleration of the B block.

\rm s = ut + (1)/(2)at^2

\rm 1.8 = (1)/(2)* a* (2)^2

\rm a = 0.9\; m/sec^2

Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.

\rm \sum F=ma

\rm mg-T_b=ma

\rm T_b = m(g-a)

\rm T_b = 7* (9.8-0.9)

\rm T_b = 62.3\;N

b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.

\rm \sum F=ma

\rm T_a=ma

\rm T_a = 2.1* 0.9

\rm T_a = 1.89\;N

c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.

\rm \sum \tau = I\alpha

\rm T_br-T_ar = I\alpha

\rm I = ((T_b-T_a)r^2)/(a)

Now, substitute the values of the known terms in the above expression.

\rm I = ((62.3-1.89)(0.080)^2)/(0.90)

\rm I = 0.430 \; kg\;m^2

For more information, refer to the link given below:

brainly.com/question/2287912

Answer:

(a) 62.3 N

(b) 1.89 N

(c) 0.430 kg m²

Explanation:

(a) Find the acceleration of block B.

Δy = v₀ t + ½ at²

1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²

a = 0.90 m/s²

Draw a free body diagram of block B.  There are two forces:

Weight force mg pulling down,

and tension force Tb pulling up.

Sum of forces in the -y direction:

∑F = ma

mg − Tb = ma

Tb = m (g − a)

Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)

Tb = 62.3 N

(b) Draw a free body diagram of block A.  There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and tension force Ta pulling right.

Sum of forces in the +x direction:

∑F = ma

Ta = ma

Ta = (2.10 kg) (0.90 m/s²)

Ta = 1.89 N

(c) Draw a free body diagram of the pulley.  There are two forces:

Tension force Tb pulling down,

and tension force Ta pulling left.

Sum of torques in the clockwise direction:

∑τ = Iα

Tb r − Ta r = Iα

(Tb − Ta) r = I (a/r)

I = (Tb − Ta) r² / a

I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)

I = 0.430 kg m²

The cycle is a process that returns to its beginning, but it does not repeatitself.
True
False

Answers

False. It does repeat itself

The radius of Earth is 6370 km in the Earth reference frame. The cosmic ray is moving at 0.880Co relative to Earth.a. In the reference frame of a cosmic ray how wide does Earth seem along the flight direction?
b. In the reference frame of a cosmic ray how wide does Earth seem perpendicular to the flight direction?
Express your answer with the appropriate units.

Answers

Answer:

6052114.67492 m

12.742* 10^(6)\ m

Explanation:

v = Velocity of cosmic ray = 0.88c

c = Speed of light = 3* 10^8\ m/s

d = Width of Earth = Diameter of Earth = 12.742* 10^(6)\ m

When the cosmic ray is moving towards Earth then in the frame of the cosmic ray the width of the Earth appears smaller than the original

This happens due to length contraction

Length contraction is given by

d_e=d\sqrt{1-(v^2)/(c^2)}\n\Rightarrow d_e=12.742* 10^(6)\sqrt{1-(0.88^2c^2)/(c^2)}\n\Rightarrow d_e=6052114.67492\ m

The Earth's width is 6052114.67492 m

Contraction only occurs in the cosmic ray's frame of reference in the direction of the ray. But in perpendicular direction the width remains unchanged.

Hence, the width is12.742* 10^(6)\ m

*Which of the following cannot be an example of projectile motion
A. A football flying through the air
B. An apple falling from a tree
C. A pencil rolling on the ground
D.A rocket dropping from its maximum height

Answers

A. football flying through the air
A cause it flying through the also a projectile is a object flying in the air like a arrow for example/also can I get Brainly

Early in the morning, when the temperature is 5.5 °C, gasoline is pumped into a car’s 53-L steel gas tank until it is filled to the top. Later in the day the temperature rises to 27 °C. Since the volume of gasoline increases more for a given temperature increase than the volume of the steel tank, gasoline will spill out of the tank. How much gasoline spills out in this case?

Answers

Answer:

Volume of gasoline spills out is 0.943 L.

Explanation:

Volumetric expansion of both gasoline and steel tank is :

\beta_(gas)=9.5 *10^(-4)/K\n\beta_(steel \ gas)=3.6 * 10^(-5)/K.  { source Internet}

We know expansion due to temperature change is :

\Delta V=\beta*\Delta T* V

For gasoline:

\Delta V_g=0.98 \ L.\n

Similarly for Steel tank:

\Delta V_(steel \ gas)=0.037\ L.

Now, volume of gasoline spills out is equal to difference between expansion in volume.

\Delta V_(gas)-\Delta V_(Steel \ gas)=0.98-0.037\ L=0.943\ L.