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Klio2033 [76]
2 years ago
8

1. A rocket is launched from a 300 cm rail. The upper Launch Lug is placed 1 point 150 cm from the bottom of the rocket. What is

the effective distance of the Launch Rail? 300 cm 150 cm 3 m 300 m O 1m O 1.5 m​
Physics
1 answer:
patriot [66]2 years ago
4 0

Answer:

i think its going to be 150 because its half of 300

Explanation:

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Answer please need help
Basile [38]

the answer is C. It allows citizens to submit anonymous tips to the police.

8 0
3 years ago
Which of the following is characteristic of proficient catching?
uysha [10]

Answer:

The correct answer is option D i.e. A and C

Explanation:

The correct answer is option D i.e. A and C

for proficient catching player must

- learn to absorbed the ball force

- moves the hang according to ball direction to hold the ball

- to catch ball at high height move the finger at higher position

- to catch ball at low height move the finger at lower position

5 0
3 years ago
A certain ideal gas has molar heat capacity at constant volume CV. A sample of this gas initially occupies a volume V0 at pressu
ANTONII [103]

Answer:

Explanation:

The processes are described on the image attached below. The isobaric process consists of an horizontal line, the adiabatic expansion is described by a polytropic curve:

P_{2} \cdot V_{2}^{\gamma} = P_{3} \cdot V_{3}^{\gamma}

Where:

\gamma = \frac{c_{p}}{c_{v}}

\gamma = 1 + \frac{R}{c_{v}}

Final pressure is:

P_{3} = P_{2}\cdot \left(\frac{V_{2}}{V_{3}}  \right)^{\gamma}

P_{3} = P_{o}\cdot \left(\frac{1}{2}\right)^{\gamma}

P_{3} = \frac{P_{o}}{2^{\gamma}}

8 0
3 years ago
Which sphere is NOT a part of the cycling of oxygen through Earths systems ?
stepan [7]
The answer would be Exosphere because, there are 3 main regions that circulate oxygen through the Earths system, which are the Biosphere, Atmosphere, and the Lithosphere. 
5 0
3 years ago
A projectile falls beneath the straight-line path it would follow if there were no gravity. How many meters does it fall below t
maria [59]

Answer:

Explanation:

Suppose v is the initial velocity and \theta is the angle of inclination

distance traveled in vertical direction in t=1 s

When gravity is present

y=vt+\frac{1}{2}at^2

where y=vertical\ distance

a=acceleration

t=time

v=initial\ velocity

here initial velocity is v\sin \theta [/tex] so

y=v\sin \theta \times 1-\frac{1}{2}gt^2

y=v\sin \theta -0.5g

In absence of gravity

y_2=v\sin \theta \times t

y_2=v\sin \theta \times 1

\Delta y=y_2-y=v\sin \theta -v\sin \theta +0.5 g=4.9\ m

         

4 0
3 years ago
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