3. The soccer kicker kicks a 2 kg football with a force of 68 N. How fast will the ball accelerate down the field?

Physics

1 answer:

Katen [24]3 years ago

6 0

Answer:

$\boxed {\boxed {\sf m= 2 \ kg , \\a= 34 \ m/s^2, \\\F= 68 \ N}}$

Explanation:

The formula for force is:

$F=m*a$

If we rearrange the formula to solve for a (acceleration), the formula becomes

$\frac{F}{m} =a$

The force is 68 Newtons. Let's convert the units to make the problem easier later on. 1 N is equal to 1 kg*m/s², so the force of 68 N is equal to 68 kg*m/s².

The mass is 2 kilograms.

$F=68 \ kg*m/s^2 \\m= 2\ kg$

Substitute the values into the formula.

$\frac{68 \ kg*m/s^2}{2 \ kg} =a$

Divide. Note that the kilograms will cancel each other out (hence why we changed the units).

$\frac{68 \ m/s^2}{2}=a$

$34 \ m/s^2=a$

The acceleration is<u> </u><u>34 meters per second squared.</u>

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You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .