52.887 in significant figures

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

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

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

=> $R = 590.2 \ m$

8 0

3 years ago

PLEASE HELP!

WINSTONCH [101]

So you would first multiply 400 by 2 which equals 800, then add 30 which is 830.
Then you would subtract 1000-830=170.
The total force of the 6 other players would be 170N.
Hoped this helped ☺️

7 0

3 years ago

Which forces are shown on a free body diagram?

notsponge [240]

Answer:

What is a Free Body Diagram?

The free body diagram helps you understand and solve static and dynamic problem involving forces. It is a diagram including all forces acting on a given object without the other object in the system. You need to first understand all the forces acting on the object and then represent these force by arrows in the direction of the force to be drawn.

Explanation:

5 0

3 years ago

You would like to store 8.1 J of energy in the magnetic field of a solenoid. The solenoid has 620 circular turns of diameter 6.6

marta [7]

Answer:

(a) The current needed is 56.92 A

(b) The magnitude of the magnetic field inside the solenoid is 0.134 T

(c) The energy density inside the solenoid is 7.144 kJ/m³

Explanation:

Given;

energy stored in the magnetic field of solenoid, E = 8.1 J

number of turns of the solenoid, N = 620 turns

diameter of the solenoid, D = 6.6 cm = 0.066 m

radius of the solenoid, r = D/2 = 0.033 m

length of the solenoid, L = 33 cm = 0.33 m

Inductance of the solenoid is given as;

$L= \frac{\mu_o N^2 A}{l}$

where;

A is the area of the solenoid = πr² = π (0.033)² = 0.00342 m²

μ₀ is permeability of free space = 4π x 10⁻⁷ H/m

$L= \frac{4\pi*10^{-7} *620^2 *0.00342}{0.33} \\\\L = 0.005 \ H$

(A). How much current needed

Energy stored in magnetic field of solenoid is given as;

$E = \frac{1}{2} LI^2\\\\$

Where;

I is the current in the solenoid

$E = \frac{1}{2} LI^2\\\\I^2 = \frac{2E}{L}\\\\I = \sqrt{\frac{2*8.1}{0.005}}\\\\ I = 56.92 \ A$

(B) The magnitude of the magnetic field inside the solenoid

B = μ₀nI

where;

n is number of turns per unit length

B = μ₀(N/L)I

B = (4π x 10⁻⁷)(620/0.33)(56.92)

B = 0.134 T

(C) The energy density (energy/volume) inside the solenoid

$U_B = \frac{B^2}{2\mu_0} \\\\U_B = \frac{(0.134)^2}{2*4\pi*10^{-7}} \\\\U_B = 7143.54 \ J/m^3\\\\U_B = 7.144 \ kJ/m^3$

3 0

3 years ago

Its economics. Its not b. What is it?

Yuliya22 [10]

Answer:

A

I say this because exponential growth takes place when a population per Capita growth stays the same

5 0

3 years ago