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3 years ago

Help?

Physics

1 answer:

ankoles [38]3 years ago

8 0

center, power forward, small forward, point guard and shooting guard

1896

the University of Chicago and the University of Iowa

James Naismith was a physical education teacher

Sprinfield College

Upon the request of his boss, Naismith was tasked to create an indoor sports game to help athletes keep in shape as they endured the cold New England winters. Naismith's boss also stipulated that this new game should be "fair for all players and not too rough."

I just looked it up hope this helps tho.

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In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin

alexandr402 [8]

Answer:

a) $v=0.999124c$

b) $E=7.566*10^{22}$

c) $E_a=760 times\ larger$

Explanation:

From the question we are told that

Distance to Betelgeuse $d_b=430ly$

Mass of Rocket $M_r=20000$

Total Time in years traveled $T_d=36years$

Total energy used by the United States in the year 2000 $E_{2000}=1.0*10^20$

Generally the equation of speed of rocket v mathematically given by

$v=\frac{2d}{\triangle t}$

$v=860ly/ \triangle t$

where

$\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\sqrt{(860)^2+(36)^2}$

$\triangle t=860.7532$

Therefore

$v=\frac{860ly}{ 860.7532}$

$v=0.999124c$

Generally the equation of the energy E required to attain prior speed mathematically given by

$E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2$

$E=7.566*10^{22}$

c)Generally the equation of the energy $E_a$ required to accelerate the rocket mathematically given by

$E_a=\frac{E}{E_{2000}}$

$E_a=\frac{7.566*10^{22}}{1.0*10^{20}}$

$E_a=760 times\ larger$

8 0

3 years ago

Helps please people

serg [7]

1. The iris regulates the amount of light entering the eye
2. The retina receives and organises visual information
3. The lens refracts light rays in a camera

7 0

2 years ago

An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t

elixir [45]

To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

$\mu = \frac{f_0}{f_e}$

Here,

$\mu$ = Magnification

$f_e$ = Focal length eyepieces

$f_0$ = Focal length of the Objective

Rearranging to find the focal length of the objective

$f_0 = \mu f_e$

Replacing with our values

$f_0 = 7.5* 3.7cm$

$f_0 = 27.75cm$

Therefore the focal length of th eobjective lenses is 27.75cm

5 0

3 years ago

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00

vivado [14]

Answer:

7.5 m

Explanation:

$v$ = initial speed of the ball = 8 m/s

$\theta$ = angle of launch = 40° deg

Consider the motion along the vertical direction :

$v_{oy}$ = initial velocity along vertical direction = $v Sin\theta$ = 8 Sin40 = 5.14 m/s

$a_{y}$ = acceleration along vertical direction = - 9.8 m/s²

$t$ = time of travel

$y$ = vertical displacement = - 1 m

Using the kinematics equation

$y = v_{oy}t + (0.5)a_{y}t^{2}$

$- 1 = (5.14)t + (0.5)(- 9.8)t^{2$

$t$ = 1.22 sec

Consider the motion along the horizontal direction :

$v_{ox}$ = initial velocity along horizontal direction = $v Sin\theta$ = 8 Cos40 = 6.13 m/s

$a_{x}$ = acceleration along vertical direction = 0 m/s²

$t$ = time of travel = 1.22 sec

$x$ = horizontal displacement = ?

Using the kinematics equation

$x = v_{ox}t + (0.5)a_{x}t^{2}$

$x = (6.13)(1.22) + (0.5)(0)(1.22)^{2$

$x$ = 7.5 m

4 0

3 years ago

How are weather satellites and weather stations similar?

alexandr402 [8]

<span>Weather satellites and weather stations are similar, because they both have the same purpose. They are used to help predict future weather as well as as current conditions. The satellites are viewing the weather from a distance at a large scope, but stations are using data they collect on earth to help with the same task.</span>

6 0

3 years ago