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Marina CMI [18]
3 years ago
6

A quarterback takes the ball from the line of scrimmage, runs backward for 12.1 yds, then runs sideways parallel to the line of

scrimmage for 19.8 yds. At this point, he throws a 41.5 yd forward pass straight downfield, perpendicular to the line of scrimmage. What is the magnitude of the football's resultant displacement?

Physics
1 answer:
Over [174]3 years ago
3 0

Answer:

The answer is 35.45 yds

Explanation:

You have to picture this to be able to understand it better (see attachment).

Start at the origin which is when the quarterback (QB) takes the ball. He runs backwards 12.1 yards, runs sideways for 19.8 yards (it doesn´t matter if he runs right or left), then he throws the ball forward 41.5 yards. If you look at the attachment, you can see I drew the path that the football followed. And then connected the dots from the origin and finish. The distance between those two points is the magnitude of the resultant displacement.

In order to calculate it, all you need to do is use the Pythagoream theorem, which says that the square of the hypotenuse equals the sum of the squares of the legs a and b of the triangle rectangle.

  1. R^{2} = a^{2} + b^{2} then solve for R
  2. R = \sqrt{a^{2}+b^{2}  }

In this case, you know the length of leg a to be 19.8 yards which how much it moves sideways. And then, to get the length of leg b, all you need to do is substract how much it moved backwards from the 41.5yards forward displacement. This results in b leg being 29.4 yards long.

Now you have a triangle with:

  • a = 19.8 yards
  • b = 29.4 yards

Substituting this numbers in the equation:

  • R = \sqrt{19.8^{2}+29.4^{2}  }
  • R = 35.45 yards

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... then your weight is <em>25.2 lbf</em> on the moon.

6 0
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An athlete completes 1 laps around a track with a radius of 25 meters in 180 seconds. What is the magnitude of the athlete's tan
DanielleElmas [232]

Answer:

0.872<em>m/s</em>

Explanation:

Tangential velocity is given by the formula,

v= 2\pi r/ t

In the question given,

radius= 25meters

time= 180secs

pie= 3.14

number of laps= 1

The magnitude of tangential velocity equals;

\frac{1lap* 2 *3.14 *25m}{180secs}

<em>v </em>= 157<em>m</em>/180<em>secs</em>

Therefore, the magnitude of the tangential velocity

=0.872<em>m/secs</em>

5 0
3 years ago
The sound from a single source can reach point O by two different paths. One path is 20.0 m long and the second path is 21.0 m l
aleksandrvk [35]

Answer:

minimum frequency = 170 Hz

Explanation:

given data

One path long = 20 m

second path long = 21 m

speed of sound = 340 m/s

solution

we get here destructive phase that is path difference of minimum \frac{\lambda}{2}

here  λ is the wavelength of the wave

so path difference will be

21 - 20 = \frac{\lambda}{2}  

λ = 2 m

and

velocity that is express as

velocity = frequency × wavelength    .............1

frequency  = \frac{340}{2}  

minimum frequency = 170 Hz

7 0
4 years ago
It is generally seen that, when a substance absorbs heat energy, its heating curve shows two plateaus. Which phenomenon is indic
Tatiana [17]

Answer:

The correct answer is;

Change of phase.

Explanation:

When the physical state in which a substance exists in nature changes to another physical state such as a change from solid state to a liquid stat or from a liquid state to a gaseous state, the substance is said to have undergone a phase change, or phase transformation.

The state of a substance will change, depending on the surrounding temperature and the applied pressure and the energy required for the substance to change its state is known as latent heat.

The observed plateaus are periods the latent heat is absorbed to completely change the phase of the substance.

5 0
3 years ago
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
4 years ago
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