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rjkz [21]
3 years ago
8

Need help on these questions

Physics
1 answer:
Eddi Din [679]3 years ago
6 0
1. V= 110
I= 11
R=10

2. V= 110
I= 5.5
R= 20

3. V= 100
I= 5
R= 20

4. V= 500
I= 5
R= 100

To show work plug in those numbers to V=IR
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All of the following are important reasons to take notes except.
jeka57 [31]

Its a waste of time, you have to not only write it down, but study it after too . other than that notes are great.

4 0
3 years ago
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Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orb
Maslowich

Answer: 58,045,522,878.8 meters

Explanation:

Ok, the data we have is

Period = T = 88 days

Radial acceleration = ar = 3.96x10^-2 m/s^2

And we know that the equation for the radial acceleration is:

ar = v^2/r = r*w^2

Where v is the velocity. r is the radius and w is the angular velocity.

And we know that:

w = 2*pi*f

where f is the frequency, and:

T = 1/f.

Then we can write:

w = 2*pi/T

and our equation becomes:

ar = r*(2*pi/T)^2

Now we solve this for r.

First we need to use the same units in both equations, so we want to write T in seconds.

T = 88 days,

A day has 24 hours, and one hour has 3600 seconds:

T = 88*24*3600 s =7,603,200s

Then:

3.96x10^-2 m/s^2 = r*(2*3.14/7,603,200s)^2

r = (3.96x10^-2 m/s^2) /(2*3.14/7,603,200s)^2 = 58,045,522,878.8 meters

5 0
3 years ago
A rock is rolling down a hill. At position 1, its velocity is 2.0 m/s. Twelve seconds later, as it passes position 2, its veloci
alukav5142 [94]
As this happens over twelve seconds, you would take the total difference in velocities and divide it by twelve to find the change per second

44.0 m/s - 2.0 m/s = 42.0 m/s 

42.0 m/s / 12 s = 3.5 m/s2

the acceleration of the rock would be 3.5 m/s2
4 0
3 years ago
Read 2 more answers
If an object is placed between the focal point and twice the focal length of a convex lens, which type of image will be produced
valentinak56 [21]
<span>The image produced is real and enlarged.

Check for various positions of objects and Images for convex lens.

Note at position of 2F, the image is same as the object, and once it is between 2F and F, the image becomes bigger than the object. </span>
3 0
2 years ago
Read 2 more answers
A fan blade, initially at rest, rotates with a constant acceleration of 0.029 rad/s2. What is the time interval required for it
LenKa [72]

Answer:

The time interval is  t = 21.30 \ s

Explanation:

From the question we are told that

    The constant acceleration is \alpha  = 0.029 \ rad / s^2

    The displacement is  \theta  =  6.58 \ rad

     

According to the second equation of motion we have that

    \theta  =  w_i* t  +  \frac{1}{2} *  \alpha  t^2

given that the blade started from rest

     w_i which is the initial angular velocity is 0

 So  

       \theta  = 0 +  \frac{1}{2} *  \alpha  t^2

 =>  t = \sqrt{ \frac{2 * \theta }{\alpha } }

substituting values  

=>    t = \sqrt{ \frac{2 * 6.58 }{0.029 } }

=>    t = 21.30 \ s

6 0
3 years ago
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