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serg [7]
2 years ago
5

A football player kicks a 0.41-kg football initially at rest; and the ball flies through the air. If the kicker's foot was in co

ntact with the ball for 0.051 s and the ball's initial speed after the collision is 21 m/s, what was the magnitude of the average force on the football?
Physics
1 answer:
LUCKY_DIMON [66]2 years ago
4 0

Answer:

Average force on the football = 168.82 N

Explanation:

Force = Mass x Acceleration

F = ma

Mass, m = 0.41 kg

We have equation of motion, v = u + at

Initial velocity, u = 0 m/s

Final velocity, v = 21 m/s

Time, t = 0.051 s

Substituting

                  21 = 0 + a x 0.051

                     a = 411.76 m/s²

Substituting in force equation,

                   F = ma = 0.41 x 411.76 = 168.82 N

Average force on the football = 168.82 N

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luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 34.0 ∘ above the horizontal by a force F⃗ of magnitude 165 N that
Andreas93 [3]

Answer:

a)  W = 643.5 J, b) W = -427.4 J  

Explanation:

a) Work is defined by

       W = F. x = F x cos θ

in this case they ask us for the work done by the external force F = 165 N parallel to the ramp, therefore the angle between this force and the displacement is zero

       W = F x

let's calculate

       W = 165  3.9

        W = 643.5 J

b) the work of the gravitational force, which is the weight of the body, in ramp problems the coordinate system is one axis parallel to the plane and the other perpendicular, let's use trigonometry to decompose the weight in these two axes

       sin θ = Wₓ / W

       cos θ = Wy / W

        Wₓ = W sinθ = mg sin θ

        Wy = W cos θ

the work carried out by each of these components is even Wₓ, it has to be antiparallel to the displacement, so the angle is zero

      W = Wₓ x cos 180

      W = - mg sin 34  x

     

let's calculate

       W = -20 9.8 sin 34 3.9

        W = -427.4 J

The work done by the component perpendicular to the plane is ero because the angle between the displacement and the weight component is 90º, so the cosine is zero.

3 0
2 years ago
A student decides to spend spring break by driving 50 miles due east, then 50 miles 30 degrees south of east, then 50 miles 30 d
stira [4]

Answer:600 miles, 12

Explanation: The movement described in the question exhibits that of a polygon. Exhibiting a constant distance and angle with only varying direction until the starting point is reached.

The sum of exterior angles of a polygon = 360 degrees.

Exterior angle of a polygon = (360 ÷ number of sides)

Therefore,

Number of sides = 360 ÷ exterior angle

Exterior angle = 30 degrees

Hence,

Number of sides = 360 ÷ 30 = 12 sides

Since distance traveled of 50 miles is the same for each displacement ;

Total displacement = distance traveled * number of sides

Total displacement = 50 * 12 = 600 miles.

5 0
2 years ago
A slender uniform rod 100.00 cm long is used as a meter stick. Two parallel axes that are perpendicular to the rod are considere
Nataliya [291]

Answer:

The correct answer is D    I_{30} /I_{50} =   1.5

Explanation:

In this exercise the moment of inertia equation should be used

    I = ∫ r² dm

In addition to the parallel axis theorem

    I = I_{cm} + M D²

Where  I_{cm} is the moment of the center of mass, M is the total mass of the body and D the distance from this point to the axis of interest

Let's apply these relationships to our problem, the center of mass of a uniform rod coincides with its geometric center, in this case the rod is 1 m long, so the center of mass is in

    L = 100.00 cm (1m / 100 cm) = 1.0000 m

     x_{cm} = 50 cm = 0.50 m

Let's calculate the moment of inertia for this point, suppose the rod is on the x-axis and use the concept of linear density

    λ = M / L = dm / dx

    dm =  λ dx

Let's replace in the moment of inertia equation

    I = ∫ x² ( λ dx)

We integrate

    I =  λ x³ / 3

We evaluate between the lower limits x = -L/2 to the upper limit x = L/2

    I =  λ/3 [(L/2)³ - (-L/2)³] = lam/3  [L³/8 - (-L³ / 8)]

    I =  λ/3  L³/4

    I = 1/12  λ L³

Let's replace the linear density with its value

    I = 1/12  (M/L)  L³

    I = 1/12  M L²

Let's calculate with the given values

   I = 1/12  M 1²

   I = 1/12 M

This point is the center of mass of the rod

    Icm = I = 1/12 M  = 8.333 10-2 M

Now let's use the parallel axis theorem to calculate the moment of resection of the new axis, which is 0.30 m from one end, in this case the distance is

    D = x_{cm} - x

    D = 0.50 - 0.30

    D = 0.20  m

Let's calculate

   I_{30} =I_{cm} + M D²

   I_{30} = 1/12 M + M 0.202

   I_{30} = M (1/12 + 0.04)

   I_{30} = M 0.123

To find the relationship between the two moments of inertia, divide the quantities

  I_{30} / I_{50} = M 0.123 / (M 8.3 10-2)

   I_{30} /I_{50} = 1.48

The correct answer is d 1.5

6 0
2 years ago
A steel guitar string with a diameter of 0.300 mm and a length of 70.0 cm is stretched by 0.500 mm while being tuned. How much f
Ganezh [65]

Answer:

10.1 N

Explanation:

Your answer is 10.1 N, I don't actually know how to do it but I hope it helps.

8 0
2 years ago
Help asap please!! An aluminum block of mass 12.00 kg is heated from 20 C to 118 C. If the specific heat of aluminum is 913 J-1
professor190 [17]
Q = mCΔT, where Q = Amount of energy required, m = mass of the blcok, C = specific heat, ΔT = change in temperature.

Using the given values;

Q = 12*913*(118-20) = 1073688 J = 1073.688 kJ.

The correct answer in B.
7 0
2 years ago
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