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3 years ago

While running at a constant velocity, how should you throw a ball with respect to you so that you can catch it yourself?

1 answer:

5 0

You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x).

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Which of the following statement is not true about machines?

viva [34]

The answer to this question is D or the last one

4 0

3 years ago

Read 2 more answers

A basketball rolls without slipping (starting from rest) down a ramp. If the ramp is sloped by an angle of 4 degrees above the h

slavikrds [6]

Answer:

11.7 m/s

Explanation:

To find its speed, we first find the acceleration of the center of mass of a rolling object is given by

a = gsinθ/(1 + I/MR²) where θ = angle of slope = 4, I = moment of inertia of basketball = 2/3MR²

a = 9.8 m/s²sin4(1 + 2/3MR²/MR²)

= 9.8 m/s²sin4(1 + 2/3)

= 9.8 m/s²sin4 × (5/3)

= 1.14 m/s²

To find its speed v after rolling for 60 m, we use

v² = u² + 2as where u = initial speed = 0 (since it starts from rest), s = 60 m

v = √(u² + 2as) = √(0² + 2 × 1.14 m/s × 60 m) = √136.8 = 11.7 m/s

4 0

3 years ago

An observer stands 24.7 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of

GaryK [48]

Explanation:

The given data is as follows.

Velocity of bullet, $c_{p}$ = 814.8 m/s

Observer distance from marksman, d = 24.7 m

Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.

t = $\frac{24.7}{343}$ (velocity in air = 343 m/s)

= 0.072 sec

Now, before the observer hears the report the distance traveled by the bullet is as follows.

$d_{b} = c_{b} \times t$

= $814.8 \times 0.072$

= 58.66

= 59 (approx)

Thus, we can conclude that each bullet will travel a distance of 59 m.

8 0

3 years ago

A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th

Stells [14]

Answer:

The relationship between the initial stored energy $PE_{i}$ and the stored energy after the dielectric is inserted $PE_{f}$ is:

c) $PE_{f} =0.5\ PE_{i}$

Explanation:

A parallel plate capacitor with $C_{o}$ that is connected to a voltage source $V_{o}$ holds a charge of $Q_{o} =C_{o} V_{o}$ . Then we disconnect the voltage source and keep the charge $Q_{o}$ constant . If we insert a dielectric of $\kappa=2$ between the plates while we keep the charge constant, we found that the potential decreases as: