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Norma-Jean [14]
3 years ago
8

Which of the following elements is in Group 2?

Physics
1 answer:
Rzqust [24]3 years ago
6 0

Answer:

calcium

Explanation:

if you have a periodic table that would be very helpful to you

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The thermal emission of the human body has maximum intensity at a wavelength of approximately 9.5 μm.What photon energy correspo
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Explanation:

2.1 x 10^2 - 20J

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Which best describes a difference between electric current and static electricity
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Static electricity is caused by the build up of charges on surfaces of objects while electric current is the constant flow of electrons along a conductor (the rate of flow of charge).
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Read each scenario and then answer the questions. Scenario A: A shopping cart is pushed a distance of 3 m by a force of 15 N. Sc
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To solve this we are going to use the formula for work: W=Fd
where
W is the work done
F is the force 
d is the distance 

Scenario A. We know for our problem that F=15N and d=3m. Lets replace those values in our formula to find W:
W=15N*3m
W=45J

Scenario B. We know for our problem that F=12N and d=4. Lest replace those value in our formula:
W=12N*4m
W=48J

Scenario C. We know for our problem that F=10N and d=6m. Lets replace the values in our formula:
F=10N*6m
F=60J

We can conclude that:
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- The scenario that requires the least work is A.
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1. A children’s roller coaster is released from the top of a track. If its maximum speed at ground level is 3 m/s, find the heig
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3 years ago
An electric fan is turned off, and its angular velocity decreases uniformly from 550 rev/min to 180 rev/min in a time interval o
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Answer:

(a) \alpha =-1.43rev/sec^2

(b) 4.17 rev

(c) 6.40 sec

Explanation:

We have given the angular velocity of fan decreases from 550 rev/min to 180 rev/min

So initial angular speed \omega _0=550rev/min=\frac{550}{60}=9.166rev/sec

Final angular speed \omega =180rev/min=\frac{180}{60}=3rev/sec

Time is given as t = 4.3 sec

(a) We know that angular acceleration is given by \alpha =\frac{\omega -\omega _0}{t}=\frac{3-9.166}{4.3}=-1.43rev/sec^2

(b) We know the relation \Theta =\omega _0t+\frac{1}{2}\alpha t^2=9.166\times 4.3-\frac{1}{2}\times 1.43\times 4.3^2=26.1935rad=\frac{26.1935}{2\pi }=4.17rev

(c) Now final angular velocity \omega =0rev/sec

So time t=\frac{0-9.166}{-1.43}=6.40sec

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