If you double the mass of an object what happens to the acceleration

What is playing catch? ( in terms of the force of the air interacting with ball )

Ugo [173]

Playing catch involves the ball moving in one direction in the air when you exert force. When the ball is in air, there is a gravitational force and a little it of backward air resistance until the other person catches it.

6 0

2 years ago

The ideal mechanical advantage of a pulley system is equal to the?

marishachu [46]

Answer: The correct answer is "Number of rope segments supporting the load".

Explanation:

Mechanical advantage: It is defined as the ratio of the force produced by a machine to the force applied on the machine. The ideal mechanical advantage of a machines is mechanical advantage in the absence of friction.

The ideal mechanical advantage of a pulley system is equal to the number of rope segments which is supporting the load. More the rope segments, It is more helpful to do the lifting the work.

It means that less force is needed for this task to complete.

Therefore, the correct option is (C).

7 0

2 years ago

Read 2 more answers

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:

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