The purple arrow shows speed and direction of the ball. This arrow is called what?

If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun

zvonat [6]

Answer:

H = 1/2 g t^2 where t is time to fall a height H

H = 1/8 g T^2 where T is total time in air (2 t = T)

R = V T cos θ horizontal range

3/4 g T^2 = V T cos θ 6 H = R given in problem

cos θ = 3 g T / (4 V) (I)

Now t = V sin θ / g time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V) from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97 6 within rounding

3 0

2 years ago

Around which months does el nino end and la nina start?

ad-work [718]

Answer:

They both tend to develop during the spring (March-June), reach peak intensity during the late autumn or winter (November-February), and then weaken during the spring or early summer (March-June)

3 0

2 years ago

Read 2 more answers

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

$70.67\frac{m^2}{s^2}=v^2$

Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

Therefore, the velocity is 8.4 m/s

7 0

10 months ago

Examine the scenario.

zhannawk [14.2K]

The electrons in the positive object are attracted to the negatively charged object. Some of the electrons move from the positive object to the negative object.

an example of this is lightning because the positive electrons on/in the earth are attracted to the negative electrons in the clouds and sky so the positive move to the negative charge.

6 0

3 years ago

Light passes through a greenhouse's glass walls due to the process of _____.

likoan [24]

Answer:

B..

Explanation:

4 0

3 years ago