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vodomira [7]
3 years ago
13

5. During an anti-bullying club meeting after school, a student mentions that a new student has joined your class at school. The

individual wears clothes that are not very common at your school and are religious in nature. Some of the other students have been talking negatively about the new student.
a. How can this scenario lead to bullying?


b. How should the club advise students to respond in this situation?


c. How will the advice help prevent potential bullying in this situation
Physics
1 answer:
Oksana_A [137]3 years ago
5 0

Answer:

a. Constant gossiping, and talking negative about a student could cause them to feel very very ill about their beliefs, their familly, school, clothes, etc. this  scenario will make the student feel alienated.

b. The club should advise the students that they should not participate in these talks, allow them into their groups, and should try to talk people out of those conversations.

c. The advice will keep several kids pull away from talking negatively about the kid, help stop the ill intended gossip, and hopefully make them feel better.

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What is the same for all types of electromagnetic radiation?
Mariulka [41]
<span>Radio Waves, Microwaves, infra-Red, Visible spectrum, Ultraviolet radiation, x-rays, Gamma Rays. Then again I could be wrong.</span>
5 0
3 years ago
(help please no links )
trapecia [35]

Answer:

Hello, how's your day going?

if humanity came together and made a base on the moon, it would be revolutionary. The point of a base on the moon would have multiple purposes. for example, some think that the moon contains valuable metals such as iron and titanium. a base would serve as a place for workers harvesting metals to rest. Obviously or not most of the iron harvesting would be done automatically by robots and such.

If such a base were constructed on the moon, it would be the begining of people living on other worlds and would be a great start for a base on Mars.

Hope it helped

Spiky Bob

5 0
3 years ago
A car of mass 650 kg slows from a speed of 20.0 m/s to a speed of 10.0 m/s. How much kinetic energy has it lost. How much force
Natalija [7]
Formulas you need for this problem:
F= mass•acceleration
KE= mass•velocity^/2
Acceleration= final velocity-intial velocity/time
time= distance/speed

t= 97.5\10= 9.75 seconds
Acc= 10-20/9.75= -1.03 m/s/s

These are the answers :)
F=650•-1.03= -669.5N
KE= 650•100/2= 32,500J or 32.5KJ
7 0
4 years ago
A rock with a mass of 2.0 kg falls from a cliff. At a height of 5.0 meters, its velocity is 6.0 m/s. What is the mechanical ener
REY [17]

Answer : The mechanical energy of the rock is, 134 J

Solution :

As we know that the mechanical energy is equal to the sum of the kinetic energy and the potential energy.

The formula used for mechanical energy is,

M.E=K.E+P.E\\\\M.E=(\frac{1}{2}mv^2)+(mgh)

where,

M.E = mechanical energy

K.E = kinetic energy

P.E = potential energy

m = mass of an object = 2 Kg

v = velocity of an object = 6 m/s

h = height = 5 m

g = acceleration due to gravity = 9.8m/s^2

Now put all the given values in the above formula, we get the mechanical energy of the rock.

M.E=[\frac{1}{2}(2Kg)\times (6m/s)^2]+(2Kg\times 9.8m/s^2\times 5m)=134\text{ Kg }m^2s^{-2}=134J

Therefore, the mechanical energy of the rock is, 134 J

6 0
3 years ago
Read 2 more answers
If 5.4 J of work is needed to stretch a spring from 15 cm to 21 cm and another 9 J is needed to stretch it from 21 cm to 27 cm,
qaws [65]

Answer:

the natural length of the spring is 9 cm

Explanation:

let the natural length of the spring = L

For each of the work done, we set up an integral equation;

5.4 = \int\limits^{21-l}_{15-l} {kx} \, dx \\\\5.4 = [\frac{1}{2}kx^2 ]^{21-l}_{15-l}\\\\5.4 = \frac{k}{2} [(21-l)^2 - (15-l)^2]\\\\k = \frac{2(5.4)}{(21-l)^2 - (15-l)^2}  \ \ \ -----(1)

The second equation of work done is set up as follows;

9 = \int\limits^{27-l}_{21-l} {kx} \, dx \\\\9 = [\frac{1}{2}kx^2 ]^{27-l}_{21-l}\\\\9 = \frac{k}{2} [(27-l)^2 - (21-l)^2] \\\\k = \frac{2(9)}{(27-l)^2 - (21-l)^2} \ \ \ -----(2)

solve equation (1) and equation (2) together;

\frac{2(9)}{(27-l)^2 - (21-l)^2} = \frac{2(5.4)}{(21-l)^2 - (15-l)^2}\\\\\frac{2(9)}{2(5.4)} = \frac{(27-l)^2 - (21-l)^2}{(21-l)^2 - (15-l)^2}\\\\\frac{9}{5.4} = \frac{(729 - 54l+ l^2) - (441-42l+ l^2)}{(441-42l+ l^2) - (225 -30l+ l^2)} \\\\\frac{9}{5.4 } = \frac{288-12l}{216-12l} \\\\\frac{9}{5.4 } =\frac{12}{12}  (\frac{24-l}{18 -l})\\\\\frac{9}{5.4 } = \frac{24-l}{18 -l}\\\\9(18-l) = 5.4(24-l)\\\\162-9l = 129.6-5.4l\\\\162-129.6 = 9l - 5.4 l\\\\32.4 = 3.6 l\\\\l = \frac{32.4}{3.6} \\\\

l = 9 \ cm

Therefore, the natural length of the spring is 9 cm

4 0
3 years ago
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