If I am throwing a bowling ball off a roof and a watermelon off a roof at 30

Hydrogen is a nonmetal with many nonmetal properties. Why is it on the periodic table with the metals in Group 1?

mrs_skeptik [129]

Answer:

This is because it has ns1 electron configuration like the alkali metals.

Pls Mark Brainiest, am trying to become Virtuoso

6 0

3 years ago

A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute

levacccp [35]

Explanation:

The given data is as follows.

Mass, m = 75 g

Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

where, $m_{1}$ = mass of the projectile

$m_{2}$ = mass of block

v = velocity after the impact

Now, putting the given values into the above formula as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

$75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v$

= $\frac{45}{50.075}$

v = 0.898 m/s

Now, equation for energy is as follows.

E = $\frac{1}{2}mv^{2}$

= $\frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}$

= 13500 J

Now, energy after the impact will be as follows.

E' = $\frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}$

= 20.19 J

Therefore, energy lost will be calculated as follows.

$\Delta E$ = E E'

= (13500 - 20) J

= 13480 J

And, n = $\frac{\Delta E}{E}$

= $\frac{13480}{13500} \times 100$

= 99.85

= 99.9%

Thus, we can conclude that percentage n of the original system energy E is 99.9%.

7 0

3 years ago

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What is the prefix for oxygen in As2O5? tri- di- mono- penta-

ikadub [295]

The prefix for oxygen in As2O5 is PENTA.
Penta is used to show that the number of oxygen atoms present in the compound is five. Penta means five, while mono mean one, di means two, tri means three and tetra mans four and so on. The chemical name for the given compound is Arsenic pentoxide.

3 0

3 years ago

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Calculate the distance traveled by a projectile as a function of launch angle. Compare the distances for two projectiles launche

DaniilM [7]

Answer:

$R = x_{max} = \frac{v^2\sin(2\theta)}{g}\\\frac{R_1}{R_2} = \frac{\sin(2\theta_1}{\sin(2\theta_2}$

Explanation:

Using kinematics equations:

$\Delta x = v_{0x}t\\\Delta y = -\frac{1}{2}gt^2+v_{0y}t$

Use $\Delta y = 0$ due to condition of distance traveled.

Solving second equation for time, there are two solutions. t=0 and

$t=\frac{2v_{0y}}{g}$

Use the expression in the first equation to have

$R = \frac{2v^2 \cos\theta\sin\theta}{g}$

Using trigonometric identities, you have the answer of the distance.

By doing the ratio for two different angles, you have the second answer. Due to sine function properties, the distances can be the same to complementary angles. Example, for 20° and 70°, the distance is the same.

5 0

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?

timurjin [86]

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).

5 0

3 years ago