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Pani-rosa [81]
3 years ago
15

Describe why we fear flying in a plane, nuclear power plants, being struck by lightning, or exposure to BPA in plastics MORE tha

n we fear the risks of overeating and lack of exercise.
Physics
1 answer:
Phantasy [73]3 years ago
3 0

Answer:

Flying in a plane gives the notion of one exiting the earths atmosphere and also, there is the risk of exposure to dangerous cosmic radiation.

Nuclear power plants possesses great forms of chemicals and toxic elements that could lead to chemical radiation and further more, cause mutation of human cells which invariably brings cancer. There is the extreme health hazard to be weary of when in a nuclear power plant.

Being struck by lightning can be hazardous to ones health and could bring forth a number of health problems like; hearing and/or vision loss, irregular heartbeat,  headache, chest pain, hypersomnia, confusion,  shortness of breath, weakness, stiffness, and/or temporary paralysis.

Exposure to BPA in plastics has been connected to be cause of a number of fertility problems, male impotence, heart disease, and other conditions.

What do all of these problems have in common? They all are things you never see coming, you can't literally control the particles from nuclear plants, the forces handling the elevation and descending of planes, the chemical and physical nature of compounds involved in lightning production as compared to not exercising and knowing what could come from that. The side effects of not exercising can not be compared to the latter explained and those are life threatening processes.

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Answer:

The answer is 73.8 J

Explanation:

5 0
3 years ago
A body weighs 100newtons when submerged in water. calculate the upthrust action on the body​
Andrews [41]

Answer:

Upthrust = 20 N

Explanation:

The question says that "A body weighs 100N in air and 80N when submerged in water. Calculate the upthrust acting on the body ?"

Upthrust is defined as the force when a body is submerged in liquid, then liquid applies a force on it.

ATQ,

Weight of body in air is 100 N

Weight of body in water is 80 N

Upthrust is equal to the weight of body in air minus weight of body in water.

Upthrust = 100 N - 80 N

Upthrust = 20 N

So, 20 N of upthrust is acting on the body.

7 0
3 years ago
Boxes A and B are being pulled to the right on a frictionless surface. Box A has a larger mass than B. How do the two tension fo
Vlad1618 [11]

Answer:

Tension T1 is less than tension T2.

T1 < T2

Explanation:

According to given data,

mass of box A ( mA) is grater than mass of box B (mB)

we can write,

m(A) > m(B)

Newton's second law states that:

Tension of object is directly proportional to the mass of the system.

T ∝ m

here Boxes A and B are being pulled to the right on a frictionless surface,

so Tension T1 generates due to the mass of box A m(A)

and Tension T2 arises due to mass of the system m(A) + m(B)

Thus tension T1 will be less than tension T2

T1 < T2

learn more about Tension force here:

<u>brainly.com/question/13175014</u>

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8 0
2 years ago
a wall switch is connected to a wall outlet. A lamp is plugged into the same outlet. You turn the switch on, but the lamp stays
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2:  the lamp is turned off.
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5 0
3 years ago
A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma
raketka [301]

  • Let, the maximum height covered by projectile be \sf{H_m}

\purple{ \longrightarrow  \bf{h_m =  \dfrac{ {v}^{2} \: {sin}^{2} \theta  }{2g} }}

  • Projectile is thrown with a velocity = v
  • Angle of projection = θ

  • Velocity of projectile at a height half of the maximum height covered be \sf{v_0}

\qquad______________________________

Then –

\qquad \pink{  \longrightarrow \bf{ \dfrac{h_m}{2}  = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }}

\qquad \longrightarrow \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2g} \times  \dfrac{1}{2}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{4g}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2}  =   {v_0}^{2} \: {sin}^{2} \theta }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2} }{2}  =   {v_0}^{2} }

\qquad\longrightarrow \bf{v_0 =   \sqrt{ \dfrac{ {v}^{2} }{2} } =  \dfrac{v}{ \sqrt{2} }  }

  • Now, the vertical component of velocity of projectile at the height half of \sf{h_m} will be –

\qquad \longrightarrow   \bf{v_{(y)}=v_0 \: sin \theta }

\qquad \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} }  \: sin \theta =  \dfrac{v \: sin \: \theta}{ \sqrt{2} }  }

Therefore, the vertical component of velocity of projectile at this height will be–

☀️\qquad\pink {\bf{ \dfrac{v \: sin \:  \theta}{ \sqrt{2} }} }

6 0
3 years ago
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