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3 years ago

2. A projectile is launched at a speed of 35 m/s and an angle of 40 above the horizontal.

Physics

2 answers:

IrinaVladis [17]3 years ago

6 0

Answer:

hello hi how are you glad you're good goodbye

PolarNik [594]3 years ago

5 0

Answer:

hshsbwvgwgwhwhwhwhwhw

Explanation:

nsvavsvahshwhwbwgeg

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The genes for alpha-goblin and beta-globin are found:\

Luba_88 [7]

Actually the third one is introns, trust me I did the test

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4 years ago

Read 2 more answers

The hydraulic oil in a car lift has a density of 8.30 102 kg/m³. The weight of the input piston is negligible. The radii of the

denpristay [2]

Answer:

(a) $F_i=68.58\ N$

(b) $F_i=69.903\ N$

Explanation:

Given:

density of hydraulic oil, $\rho=830\ kg.m^{-3}$

radius of input piston, $r_i=6.3\times 10^{-3}\ m$

radius of output plunger, $r_o=0.125\ m$

force to be supported, $F_o=27000\ N$

(a)

<u><em>Condition:</em></u><em>The bottom surfaces of piston and plunger at the same level.</em>

According to Pascal's law the pressure of a fluid is exerted equally in all directions against the walls of its container.

Mathematically:

$\frac{F_i}{A_i} =\frac{F_o}{A_o}$

putting respective values

$\frac{F_i}{\pi\times r_i^2} =\frac{27000}{\pi\times r_o^2}$

$\frac{F_i}{\pi\times (6.3\times 10^{-3})^2} =\frac{27000}{\pi\times 0.125^2}$

$F_i=68.58\ N$

(b)

<u><em>Condition:</em></u><em>The bottom surface of the output plunger is 1.30 m above that of the input piston.</em>

Given:

$h=1.3\ m$

Now,

$P_i=P_o+\rho.g.h$

$\frac{F_i}{\pi\times (6.3\times 10^{-3})^2} =\frac{27000}{\pi\times 0.125^2} +830\times 9.8\times 1.3$

$F_i=69.903\ N$

7 0

3 years ago

two astronauts are taking a spacewalk outside the International Space Station the first astronaut has a mass of 64 kg the second

Fittoniya [83]

Answer:

Approximately $0.88\; {\rm m \cdot s^{-1}}$ to the right (assuming that both astronauts were originally stationary.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Since momentum of this system (of the astronauts) conserved:

$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\end{aligned}$ .

Assuming that both astronauts were originally stationary. The total initial momentum of the two astronauts would be $0$ since the velocity of both astronauts was $0\!$ .

Therefore:

$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

The final momentum of the first astronaut ( $m = 64\; {\rm kg}$ , $v = 0.8\; {\rm m\cdot s^{-1}}$ to the left) would be $p_{1} = m\, v = 64\; {\rm kg} \times 0.8\; {\rm m\cdot s^{-1}} = 51.2\; {\rm kg \cdot m \cdot s^{-1}}$ to the left.

Let $p_{2}$ denote the momentum of the astronaut in question. The total final momentum of the two astronauts, combined, would be $(p_{1} + p_{2})$ .

$\begin{aligned} & p_{1} + p_{2} \\ &= (\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

Hence, $p_{2} = (-p_{1})$ . In other words, the final momentum of the astronaut in question is the opposite of that of the first astronaut. Since momentum is a vector quantity, the momentum of the two astronauts magnitude ( $51.2\; {\rm kg \cdot m \cdot s^{-1}}$ ) but opposite in direction (to the right versus to the left.)

Rearrange the equation $p = m\, v$ to obtain an expression for velocity in terms of momentum and mass: $v = (p / m)$ .

$\begin{aligned}v &= \frac{p}{m} \\ &= \frac{51.2\; {\rm kg \cdot m \cdot s^{-1}}}{64\; {\rm kg}} && \genfrac{}{}{0}{}{(\text{to the right})}{} \\ &\approx 0.88\; {\rm m\cdot s^{-1}} && (\text{to the right})\end{aligned}$ .

Hence, the velocity of the astronaut in question ( $m = 58.2\; {\rm kg}$ ) would be $0.88\; {\rm m \cdot s^{-1}}$ to the right.

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2 years ago

To construct an oscillating LC system, you can choose from a 11 mH inductor, a 6.0 μF capacitor, and a 4.2 μF capacitor. What ar

Free_Kalibri [48]

Answer:

a. 475.14 Hz

b. 1959 Hz

c. 2341.53 Hz , 3053.34 Hz

Explanation:

$f = \frac{1}{2\pi*\sqrt{C*L}}$

a. smallest use the capacitive 4.2 uF + 6.0 uF = 10.2uF replacing:

$f = \frac{1}{2\pi*\sqrt{C*L}}f=\frac{1}{2\pi*\sqrt{10.2uF*11mH}}$

$f = 475.14 Hz$

b. second smallest use the capacitive 6 uF so:

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{6uF*11mH}}$

$f = 1959Hz$

c. second largest and largest oscillation first combination so:

Use 4.2 uF

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{4.2uF*11mH}}$

$f = 2341.53 Hz$

And finally largest oscillation cap in serie so:

$C=\frac{c_1*c_2}{c_1+c_2}=\frac{4.2uF*6.0uf}{4.2uf+6.0uF}$

$C=2.47 uF$

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{2.47uF*11mH}}$

$f = 3053.34 Hz$

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3 years ago

what helps differentiate between the sound of a fire truck, an ambulance and an 18 wheeler? imbre form dynamics sound

DENIUS [597]

Dynamics sound helps differentiate between the sound of a fire truck, an ambulance and an 18-wheeler.

<h3 /><h3>What is dynamics sound?</h3>

Elements allude to the din or delicateness of music. Elements offer a method for showing articulation in printed music. They help to drive the profound substance of music through volume and force. Elements can likewise be shown at the large-scale level for a piece of music in general. This may be just a single time toward the beginning, or a few times all through in the event that the din changes during various segments. Static elements are melodic directions that advise us to play the music at a specific volume that doesn't change. As such, don't get stronger or calmer, play each note at a similar volume as the final remaining one.

Learn more about dynamics sound, refer:

brainly.com/question/760557

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1 year ago