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Strike441 [17]
3 years ago
7

Can someone help me with this I'm-- never mind I would just like help..

Physics
2 answers:
Alik [6]3 years ago
6 0
Silicon: Si
Iron: Fe
K: Potassium
Hg: Mercury
I: Iodine
Copper: Cu

A. Bromine
B.Beryllium
C. Gold
D. Tin

A. Alkaline Metals
B. Transition Metals
C. Reactive Nonmetals
D. Alkaline Earth Metals
serg [7]3 years ago
6 0
Dang it I love elements :(
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Suppose a current flows through a copper wire. Which two things occur?
Bingel [31]

Answer:

a

Explanation:

as the copper wire is very dangerous so now if these two thing happens then it would easily help the current flows through it so it might be a little bit easy for the current to flow through it

4 0
2 years ago
Plzzzzz help me plzzz
yarga [219]
8 miles per hour

(extra space)
6 0
3 years ago
Read 2 more answers
A feris wheel is turning at a constant speed of 5m/s is it not accelerating ,true or false
erma4kov [3.2K]

Answer:no it is staying the same speed

Explanation:

6 0
3 years ago
An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e
VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

r = \frac{8.26}{2} = 4.13m

And the distance 'd' is

d = lsin\theta

d = 1.14 sin 16.2\°

d = 0.318m

Through the free-body diagram the tension components are given by

Tcos\theta = mg

Tsin\theta = \frac{mv^2}{R}

Here we can watch that,

R = r+d

Dividing both expression we have that,

tan\theta = \frac{v^2}{Rg}

Replacing the values,

tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}

v = 4.83371m/s

PART B) Using the vertical component we can find the tension,

Tcos\theta = mg

T = \frac{mg}{cos\theta}

T = \frac{(10+26.2)(9.8)}{cos(16.2)}

T = 369.42N

6 0
3 years ago
Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller
jekas [21]

Answer:

F_2 = 29.54 N

Explanation:

As we know that the combination is maintained at rest position

So we will take net torque on the system to be ZERO

so we know that

\tau = \vec r \times \vec F

here we will have

\vec r_1 \times F_1 = \vec r_2 \times F_2

so we have

13 \times 50 = 22 \times F_2

so we have

F_2 = \frac{13 \times 50}{22}

F_2 = 29.54 N

8 0
3 years ago
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