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creativ13 [48]
2 years ago
7

y’all I would rlly appreciate if y’all gave me the answers to all my questions. My cousin has over 200 assignments she hasn’t do

ne so I’m trying to help her and tbh I don’t know this stuff bc I’ve never done it in school.

Physics
1 answer:
Delicious77 [7]2 years ago
6 0

Answer:

BB×bb cross is the answer

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Imagine that Earth stops orbiting the Sun but continues to rotate in place about its own axis at its current rate. In this case,
otez555 [7]

Answer:

The length of the solar day will get shorter.

Explanation:

  • The blue planet Earth not only rotates around it's own axis but also rotates around the Sun and everyday it moves a little bit around the axis.
  • Since the speed of the Earth's rotation on it's own axis and around the Sun is constant we don't feel the effects of the rotation.We can only feel the motion if the earth changes it's rotation speed.
  • If by any means or chance the Earth stopped spinning (stopped rotation) then the atmosphere surrounding the Earth would be in motion and all the Earth's land would be scoured clean.

7 0
3 years ago
A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?
rosijanka [135]

Answer:

a

     l_o  =52 \  m

b

      l = 37.13 \ LY

Explanation:

From the question we are told that

    The  speed of the spaceship is  v  =  0.800c

    Here  c is the speed of light with value  c =  3.0*10^{8} \ m/s

    The  length is  l = 31.2 \  m

     The  distance of the star for earth is d = 145 \  light \  years

     The  speed is v_s = 2.90 *10^{8}

     

Generally the from the length contraction equation we have that

       l  =  l_o  \sqrt{1 -[\frac{v}{c } ]}

Now the when at rest the length is  l_o

So  

      l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }

      l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }

      l_o=52 \  m

Considering b  

  Applying above equation

            l  =l_o \sqrt{1 -  [\frac{v}{c } ]}

Here l_o  =145 \  LY(light \ years )

So

           l=145 *  \sqrt{1 -  \frac{v_s^2}{c^2 } }

            l =145 *  \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }

            l = 37.13 \ LY

4 0
3 years ago
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
A chinook salmon needs to jump a waterfall that is 1.5 m high. If the fish starts from a distance of 1.00 m from the base of the
____ [38]
 <span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal 

it can cross and reach top of trajectory if its top height h = 1.5m 
and horizontal distance d = (1/2) Range 
--------------------------------------... 
let t be top height time 
at top height, vertical component of its velocity =0 
vy = 0 = u sin p - gt 
t = u sin p/g 
h = [u sin p]*t - 0.5 g[t[^2 
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g 
u^2 sin^2 p/2g = 1.5 
u^2 sin^2 p = 1.5*2*9.8 = 29.4 
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component 
===================== 
t = HALF the time of flight 
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g] 
1 = u^2 sin p cos p/g 
u sin p * u cos p = 9.8 
5.42 * u cos p = 9.8 
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component 
check>> 
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s 
u < less than fish's potential jump speed 6.26 m/s 

so it will able to cross</span>
8 0
2 years ago
The message refers to which of the following?
Oliga [24]

The message is the information being communicated from one place to another.

It used to be called the "intelligence".  But as time went on, it became
harder to ignore the obvious fact that that was going too far, and the
label was changed to the more IQ-neutral "message".    

7 0
2 years ago
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