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Vlada [557]
2 years ago
10

A place kicker applies an average force of 2400 N to a football of .040 kg. The force is applied at an angle of 20.0 degrees fro

m the horizontal. Contact time is .010 sec. a. Find the velocity of the ball upon leaving the foot. b. Assuming no air resistance find the time to reach the goal posts 40.0 m away. c. The posts are 4.00 m high. Is the kick good? By how much?
Physics
1 answer:
Wewaii [24]2 years ago
4 0

Answer:

a)  The velocity of the ball upon leaving the foot = 600 m/s

b)  Time to reach the goal posts 40.0 m away = 0.07 seconds

c)  The kick won't e going inside goal post, it is higher by 10.34m.

Explanation:

a) Rate of change of momentum = Force

   \frac{\texttt{Final momentum - Initial momentum}}{\texttt{Time}}=\texttt{Force}\\\\\frac{0.040v-0.040\times 0}{0.010}=2400\\\\v=600m/s

  The velocity of the ball upon leaving the foot = 600 m/s

b) Horizontal velocity = 600 cos20 = 563.82 m/s

   Horizontal displacement = 40 m

   Time

            t=\frac{40}{563.82}=0.07s

   Time to reach the goal posts 40.0 m away = 0.07 seconds

c) Vertical velocity = 600 sin20 = 205.21 m/s

    Time to reach the goal posts 40.0 m away = 0.07 seconds

    Acceleration = -9.81m/s²

    Substituting in s = ut + 0.5at²

             s = 205.21 x 0.07 - 0.5 x 9.81 x 0.07²= 14.34 m

    Height of ball = 14.34 m

    Height of post = 4 m

    Difference in height = 14.34 - 4 = 10.34 m

    The kick won't e going inside goal post, it is higher by 10.34m.

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allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

7 0
2 years ago
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 4.15 m/s . Her husband Bruce suddenly reali
aleksandr82 [10.1K]

Answer:

a 15.22 m/s

b 45.65 m

Explanation:

Using the same formula,

x = vt, where

x is now 45.65, and

t is 3 s, then

45.65 = 3v

v = 45.65/3

v = 15.22 m/s

See the attachment for the part b. We used the distance gotten in part B, to find question A

5 0
2 years ago
A Wooden block has a mass of 0.200kg, a specific heat of 710 J (kg times degrees Celsius and is at a temperature of 20.0 degrees
olchik [2.2K]

Answer:

35°C

Explanation:

q = mCΔT

2130 J = (0.200 kg) (710 J/kg/°C) (T − 20.0°C)

T = 35°C

8 0
2 years ago
A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is
Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

F = AY\alpha \Delta T

Where

A = Cross-sectional Area

Y = Young's modulus

\alpha= Coefficient of linear expansion for steel

\Delta T= Temperature Raise

Our values are given as,

A = 4.45cm^2

T = 37K

\alpha = 1.17*10^{-5}K^{-1}

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Replacing we have,

F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)

F = 38526.1N

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

7 0
2 years ago
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block
melomori [17]

This question is incomplete, the complete question is;

A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.

How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°

Answer:

the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J

Explanation:

Given that;

m = 9.9 kg

h = 4.9 m

d = 5 m

μ = 0.3

θ = 36.87°

Now from conservation of energy, the energy is;

Et = mgh

we substitute

Et = 9.9 × 9.8 × 4.9

= 475.398 J

Also the loss of energy i

E_loss = (umg cosθ) d

we substitute

E_loss  = 0.3 × 9.9 × 9.8 × cos36.87°  × 5

= 116.423 J

so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be

E = Et - E_loss

E = 475.398 J - 116.423 J

E = 358.975 J

5 0
2 years ago
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