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Art [367]
3 years ago
5

Does peer pressure affect your physical activity routine? Describe this peer pressure and tell whether it is positive or negativ

e.
Physics
2 answers:
Alexxx [7]3 years ago
4 0
Yes, peer pressure affects one's physical activity routine. It can do so both negatively and positively. For instance, if one is pressured to do drugs when around their peers, it would most likely lead to an addiction that lasts even when they are not with those people anymore. However, from a positive viewpoint, one's peers could also pressure them to do something productive, such as trying a new beneficial activity that they are afraid of (ex. trying out for a talent show.) This could lead to a disruption in routine as that individual would begin practicing for said talent show. Hence, peer pressure can be both negative and positive, but in both instances, it changes the routine of the individual effected. 
Serga [27]3 years ago
3 0

Yes, peer pressure does affect your physical activity routine. Peer pressure can do both negative and positive. A negative way of peer pressure is having an addiction to drugs after these people have been leaned into drugs they will now do them even when the people that gave it to them are not around. A positive way of peer pressure is someone pressuring you to do something positive like to start working out. This would end up to you to start a workout routine.  

Hope it helps :)

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The following statements address the science behind the pulley system illustrated:
Leni [432]

Answer:

i. Statements A and B

Explanation:

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7 0
3 years ago
Problem: The frequency of an FM radio station is 89.3 MHz. Calculate its period. Part B: From the Library, select the general eq
vekshin1

Answer:

Time period, T=1.11\times 10^{-8}\ s

Explanation:

We have,

The frequency of an FM radio station is 89.3 MHz.

It is required to find the period of the wave.

The reciprocal of frequency is called time period of a wave. It can be given by :

T=\dfrac{1}{f}\\\\T=\dfrac{1}{89.3\times 10^{6}\ Hz}\\\\T=1.11\times 10^{-8}\ s

So, the period of the wave is 1.11\times 10^{-8}\ s.

5 0
3 years ago
if charge is located at center of spherical volume and electric flux through surface of sphere is Φ what would be flux through s
jonny [76]

Answer:

The flux will be nine times as great.

Explanation:

The electric flux due to a charge Q located in the center of a sphere can be obtained using Gauss's law. Considering a Gaussian surface in the form of a sphere of radius r:

\Phi_E=\int\limits {Ecos\theta dS} \,

The electric field (E) is parallel to the surface vector (dS), so \theta=0

\Phi_E=\int\limits{Ecos(0)dS} \,\\\Phi_E=E\int\limits{dS} \,\\\Phi_E=ES\\\Phi_E=E4\pi r^2

Since the electric flux is proportional to the square of the sphere's radius,  if radius of sphere were tripled, the flux will be nine times as great.

7 0
3 years ago
Mary starts from her house, walks 80 meters south, and stops to chat with her aunt on the sidewalk. After chatting for a few min
Vaselesa [24]
The average speed will be:
Total distance travelled divided by time taken
Total distance (in metres)= 80+125+45=250
Total time (minutes) =10
250/10
=25
Thus Mary's average speed is 25 metres per minute.
6 0
4 years ago
Read 2 more answers
A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C
mario62 [17]

Answer:

3.62 m  and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}

\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}

\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}

x = - 1.4 m

6 0
4 years ago
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