2225.6 grams =_______

The trajectory of a rock ejected from the Kilauea volcano, with a velocity of magnitude 6.4 m/s and at angle 2.2 degrees above t

Vika [28.1K]

Answer:

v = 8.03 m / s

Explanation:

This is a missile throwing exercise.

y = y₀ + $v_{oy}$ t - ½ g t²

indicates that y = -1.2 m and the initial velocity is

v_{oy} = v₀ sin θ

v_{oy} = 6.4 sin 2.2

v_{oy} = 0.2457 m / s

we substitute in the equation

-1.2 = 0.2457 t - ½ 9.8 t²

4.9 t² - 0.2457 t - 1.2 = 0

t² - 0.05014 t -0.2449 = 0

we solve the quadratic equation

t = [0.05014 ±√ (0.05014² + 4 0.2449)] / 2

t = [0.05014 ± 0.9910] / 2

t₁ = 0.5206 s

t₂ = -0.47 s

time must be a positive magnitude therefore the correct answer is

t = 0.5206 s

with this time we can calculate the vertical speed when the rock hits the ground

v_{y} = v_{oy} - gt

v_{y} = 0.2457 - 9.8 0.5206

v_{y} = -4.856 m / s

the negative sign indicates that the speed is down

horizontal velocity is constant, due to no acceleration

vₓ = v₀ₓ = v₀ cos 2,2

v₀ₓ = 6.4 cos 2.2

v₀ₓ = 6.395 m / s

therefore let's use Pythagoras' theorem to find the velocity

v = √ (vₓ² + $v_{y}^{2}$ )

v = √ (6,395² + 4,856²)

v = 8.03 m / s

the direction can be found with trigonometry

tan θ = v_{y} / vₓ

θ = tan⁻¹ (-4,856 / 6,395)

θ = - 37

the negative sign indicates that it is half clockwise from the x axis

4 0

4 years ago

Define speed and record the formula for calculating speed. Be sure to label variables.

hoa [83]

Answer:speed is how fast you or something goes

Explanation:

3 0

3 years ago

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What are the advantages of using a telescope compared to making observations with the naked eye?

lara31 [8.8K]

The naked eye is very limited. Unlike the telescope which can see far away.

4 0

4 years ago

Water is flowing into a factory in a horizontal pipe with a radius of 0.0183 m at ground level. This pipe is then connected to a

timama [110]

Answer:

$0.0168 m^3/s$

Explanation:

We are given that

$r_1=0.0183 m$

$h_1=0$

$r_2=0.0420 m$

$h_2=12.6 m$

Let $P_1=P_2=P$

By using Bernoulli theorem

$P+\frac{1}{2}\rho v^2_1+\rho gh_1=P+\frac{1}{2}\rho v^2_2+\rho gh_2$

$\frac{1}{2}\rho v^2_1+\rho gh_1=\frac{1}{2}\rho v^2_2+\rho gh_2$

$v^2_1+2gh_1=v^2_2+2gh_2$

$A_1v_1=A_2v_2$

$v_1=\frac{A_2v_2}{A_1}$

$(\frac{A_2}{A_1})^2v^2_2+2g\times 0=v^2_2+2\times 9.8\times 12.6$

$(\frac{\pi r^2_2}{\pi r^2_1})^2v^2_2-v^2_2=246.96$

$v^2_2((\frac{r^2_2}{r^2_1})^2-1)=246.96$

$v^2_2=246.96\frac{r^4_1}{r^2_4-r^4_1}$

$v_2=\sqrt{246.96\frac{r^4_1}{r^4_2-r^4_1}}$

$v_2=\sqrt{246.96\times \frac{(0.0183)^4}{(0.042)^4-(0.0183)^4}}$

$v_2=3.038 m/s$

Volume flow rate = $A_2v_2$

Volume flow rate = $\pi r^2_2v_2=\pi (0.042)^2\times 3.038=0.0168 m^3/s$

3 0

3 years ago

100 POINTS. Help Please.

soldi70 [24.7K]

Answer:

The electrical force between two balloons is 67.5N.

Explanation:

There are two charged balloons, let's say a and b.

The charge on the balloon a = $Q_a = 3.0 * 10^{-6}$ C

The charge on the balloon b = $Q_b = 2.5 * 10^{-7}$ C

Both balloons are 1 cm apart; it means that the distance<em> r</em> between the balloon a and the balloon b is 0.01 m (since 1 cm = 0.01 m).

We need to find the electrical force between them. By using the Coulomb's law, the magnitude of the electrical force between both the balloon is given as follows:

$F = \frac{kQ_aQ_b}{r^2}$ --- (A)

Where,

k = Coulomb's constant = $\frac{1}{4\pi\epsilon_o}$ = $9 * 10^9 \frac{Nm^2}{C^2}$

Plug all the values in the equation (A):

$F = \frac{(9*10^9)(3.0*10^{-6})(2.5*10^{-7})}{(0.01)^2} \\F = 67.5 N$

Hence, the electrical force between two balloons is 67.5N (three significant figures).

4 0

3 years ago

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2225.6 grams =________ ft