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3 years ago

Please sb help me fr

Physics

2 answers:

marissa [1.9K]3 years ago

7 0

me lo traduses plis no se ingles solo escribe lo que dice en la guia

DIA [1.3K]3 years ago

5 0

Answer:

I think it might be box 1. Hope this is right have a good day

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Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of th

Illusion [34]

Complete Question

Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of the moon’s surface, making what some consider the longest golf drive in history. Assume one of the golf balls was struck with initial velocity v0 = 32.75 m/s at an angle θ = 32° above the horizontal. The gravitational acceleration on the moon’s surface is approximately 1/6 that on the earth’s surface. Use a Cartesian coordinate system with the origin at the ball's initial position.

Randomized Variables

vo 32.75 m/s

theta 32 degrees

What horizontal distance, R in meters, did this golf ball travel before returning to the lunar surface?

Answer:

The horizontal distance is $R = 590.2 \ m$

Explanation:

From the question we are told that

The initial velocity is $v_o = 32.75 \ m/s$

The angle is $\theta = 26^o$

The gravitational acceleration of the moon is $g_m = \frac{1}{6} * 9.8 = 1.633 m/s^2$

Generally the distance traveled is mathematically represented as

$R = \frac{v_o^2 sin 2(\theta)}{g_m}$

=> $R = \frac{32.75^2 sin 2(32)}{1.633}$

=> $R = 590.2 \ m$

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3 years ago

What are the three categories of adult development?

andreev551 [17]

Physical development , Cognitive Development , psychosocial development I think!

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4 years ago

What can you conclude about the velocity of the object on the graph below

Mars2501 [29]

Answer:

The more time goes on the higher in meters. So most likely more time = closer to terminal velocity. -hope this helps.

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3 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?

SpyIntel [72]

Answer:

U= $0.112 x10^{-6}$ J

u= 0.708 $\frac{J}{m^{3} }$

Explanation:

diameter= 2 cm

$r= 1 cm * \frac{1m}{100cm} = 0.01 m$

distance= 0.5 mm

$d= 0.5 mm * \frac{1m}{1000m}= 0.5 x10^{-3}$

Area A= $\pi *r^{2}= 0.314x10^{-3} m^{2}$

Volume v= $A*d = 0.314 x10^{-3}m^{2}*0.5x10^{-3} m =0.157x10^{-6} m^{3}$

$v= 0.157 x10^{-6} m^{3}$

Constant vacuum permittivity

$E_{o}= 8.85x10^{-12}$

a).

$C= \frac{A*E_{o} }{d} = \frac{0.314x10^{-3} *8.85x10^{-12} }{0.5x10^{-3} } \\C= 5.56 x10^{-12}$ F

$U= \frac{1}{2} *C *V^{2}\\ U=\frac{1}{2} * 5.56x10^{-12}*(200)^{2} \\U=0.112 x^{-6} J\\$

b).

$u=\frac{U}{v}$

$u=\frac{0.112 x10^{-6}J}{0.157x10^{-6}m^{3} } \\u=0.708 \frac{J}{m^{3} }$

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4 years ago

The forces F 1, F 2, ..., F n acting at the same point P are said to be in equilibrium if the resultant force is zero, that is,

allochka39001 [22]

For the forces to be in equilibrium:
F 1 + F 2 + F 3 + F 4 = 0
i + j + i - j + 9 i + 2 j + F 4 = 0
11 i + 2 j + F 4 = 0
F 4 = - 11 i + 2 j
F 4 = ( -11, - 2 )

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3 years ago

Read 2 more answers