3. One cubic metre is equal to

Height of cannon 5 m, initial speed of projectile 15m/s, angle of launch 0 degrees. What is the range and time in the air? Pleas

Westkost [7]

Answer:

The range is 15.15 m and the time in the air is 1.01 s

Explanation:

Horizontal Motion

When an object is thrown horizontally (with angle 0°) with a speed v from a height h, it follows a curved path ruled exclusively by gravity until it eventually hits the ground.

The range or maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$

To calculate the time the object takes to hit the ground, we use the equation below:

$\displaystyle t=\sqrt{\frac{2h}{g}}$

The cannon is shot from a height of h=5 m with an initial speed of v=15 m/s. The range is calculated below:

$\displaystyle d=15\cdot\sqrt{\frac {2*5}{9.8}}=15*1.01$

d = 15.15 m

The time in the air is:

$\displaystyle t=\sqrt{\frac{2*5}{9.8}}$

t = 1.01 s

The range is 15.15 m and the time in the air is 1.01 s

8 0

3 years ago

How many neutrons does the most common isotope of hydrogen have?

kipiarov [429]

The most common hydrogen isotope, protium, has no neutrons in the nucleus.

Isotopes of

8 0

3 years ago

two charges + 2.6 uc and -5.4uc experience an attractive force of 6.5mN. What is the separation between the charges?

djyliett [7]

Answer:

...

Explanation:

....

7 0

4 years ago

What is the magnitude of the force acting on a spring with a spring constant of 275 N/m that is stretched 14.3 cm?

Travka [436]

15 i thank
if u need more help just ask me

8 0

3 years ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the x-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its y-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the x- and y-components of the copter's position vector after t = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

3 years ago

3. One cubic metre is equal to​

3. One cubic metre is equal to